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## Equipartition Theorem
Using the equipartition theorem determine the specific heat capacity at constant volume for argon and nitrogen. Compare the resulting values with table values.
• #### Hint – Equipartition Theorem
The equipartition theorem is a statement from statistical physics and states that in equilibrium for every generalized momentum or generalized coordinate acting in the formula for the energy of the system quadraticaly there is a mean energy of
$\frac{1}{2}kT,$
where k is the Boltzmann constant and T thermodynamic temperature of the system.
How is this fact related to a specific heat capacity (or molar heat capacity) and degrees of freedom of gas?
• #### Hint 2
Argon is a one-atom gas and nitrogen is a two-atom gas. What is their number of degrees of freedom?
• #### Given Values
cVi = ? specific heat capacity of gases at constant volume
Table values:
MmN = 28 g mol−1 = 0.028 kg mol−1 molar mass of nitrogen N2 (two-atom molecule) MmAr = 36 g mol−1 = 0.039 kg mol−1 molar mass of argon Ar (one-atom molecule) R = 8.31 Jmol−1K−1 molar gas constant
• #### Analysis
The equipartition theorem is a statement from statistical physics and states that in equilibrium, every generalized momentum or generalized coordinate acting in the formula for the energy of the system quadratically corresponds to mean energy of 1/2kT, where k is the Boltzmann constant and T is thermodynamic temperature of the system.
This corresponds to the contribution of 1/2R to molar heat capacity CV at constant pressure.
This contribution belongs to each translational and rotational degree of freedom. It is different with vibrational degrees of freedom. The contribution is twice as big, i.e. R. This is due to the fact that the energy of vibrational motion is composed of kinetic and potential energy, and not only kinetic energy as it is with translational and rotational motion.
The task is to determine the degrees of freedom for one-atom molecules of argon and two-atom molecules of nitrogen.
• #### Solution
The equipartition theorem is a statement from statistical physics according to which to each generalized momentum and each generalized coordinate acting in the formula for energy of the system quadratically corresponds to mean energy of $$\frac{1}{2}kT,$$ where k is the Boltzmann constant and T is thermodynamic temperature.
In other words each generalized momentum and generalized coordination acting in the formula for the energy of the system quadratically corresponds to a contribution of $$\frac{1}{2}R$$ to molar heat capacity CV.
This means that the state contribution corresponds to each translational and rotational degree of freedom. It is different with vibrational degrees of freedom. The contribution is twice as big, i.e. R. This is due to the fact that the energy of vibrational motion constitutes of kinetic and potential energy, and not only kinetic energy as it is with translational and rotational motion.
Now we can easily determine heat capacities of given gases.
Argon is a one-atom gas which means that its molecules have only three translational degrees of freedom. Molar heat capacity CV at constant volume is therefore
$C_{VAr}=3\cdot\frac{1}{2}R=\frac{3}{2}R.$
Related specific heat capacity cV is determined from molar mass MmAr as follows:
$c_{VAr}=\frac{C_{VAr}}{M_{mAr}}=\frac{3R}{2M_{mAr}}.$
Nitrogen is a two-atom gas. It has six degrees of freedom in total. Three of them are translational, two are rotational and one is vibrational. According to the equipartition theorem the molar heat capacity CVN at constant volume is given by
$C_{VN}=3\cdot\frac{1}{2}R+2\cdot\frac{1}{2}R+R=\frac{7}{2}R.$
For a corresponding specific heat capacity cVN we then obtain:
$c_{VN}=\frac{C_{VN}}{M_{mN}}=\frac{7R}{2M_{mN}},$
where MmN is molar mass of nitrogen.
• #### Numerical Solution
$c_{VAr}=\frac{3R}{2M_{mAr}}=\frac{3\cdot{8.31}}{2\cdot{0.039}}\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}$ $c_{VAr}=319.6\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}=0.32\,\mathrm{kJkg}^{-1}\mathrm{K}^{-1}$
$c_{VN}=\frac{7R}{2M_{mN}}=\frac{7\cdot{8.31}}{2\cdot{0.028}}\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}$ $c_{VN}=1038.8\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}\dot{=}1.04\,\mathrm{kJkg}^{-1}\mathrm{K}^{-1}$
• #### Answer and Result Discussion
Calculated value Table value for t = 20 °C cVAr 0.32 kJ kg−1 K−1 0.32 kJ kg−1 K−1 cVN 1.04 kJ kg−1 K−1 0.74 kJ kg−1 K−1
By comparing calculated results with table values we can see that while the values for argon are the same, there is a big difference in the values for nitrogen. This is due to the fact that the equipartition theorem is derived within classical physics and thus does not respect the specifics of quantum mechanics that are manifested quite profoundly when dedscribing the vibrational motion at normal temperatures. In reality, the vibrations practically do not contribute to heat capacity under normal conditions. Therefore, to determine molar heat capacity CVN of nitrogen, we should use a more precise formula
$C_{VN}=3\cdot\frac{1}{2}R+2\cdot\frac{1}{2}R=\frac{5}{2}R$
that involves only the translation and rotation.
If we calculate corresponding specific heat capacity cVN
$c_{VN}=\frac{C_{VN}}{M_{mN}}=\frac{5R}{2M_{mN}},$
we obtain a numerical value of
$c_{VN}\dot{=}742\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}\dot{=}0.74\,\mathrm{kJkg}^{-1}\mathrm{K}^{-1}.$
This result is in a good agreement with the table value.
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## Equipartition Theorem. Using the equipartition theorem determine the specific heat capacity at constant volume for argon and nitrogen. Compare the resulting values with table values.. • #### Hint – Equipartition Theorem. The equipartition theorem is a statement from statistical physics and states that in equilibrium for every generalized momentum or generalized coordinate acting in the formula for the energy of the system quadraticaly there is a mean energy of. $\frac{1}{2}kT,$. where k is the Boltzmann constant and T thermodynamic temperature of the system.. How is this fact related to a specific heat capacity (or molar heat capacity) and degrees of freedom of gas?. • #### Hint 2. Argon is a one-atom gas and nitrogen is a two-atom gas. What is their number of degrees of freedom?. • #### Given Values. cVi = ? specific heat capacity of gases at constant volume. Table values:. MmN = 28 g mol−1 = 0.028 kg mol−1 molar mass of nitrogen N2 (two-atom molecule) MmAr = 36 g mol−1 = 0.039 kg mol−1 molar mass of argon Ar (one-atom molecule) R = 8.31 Jmol−1K−1 molar gas constant. • #### Analysis. The equipartition theorem is a statement from statistical physics and states that in equilibrium, every generalized momentum or generalized coordinate acting in the formula for the energy of the system quadratically corresponds to mean energy of 1/2kT, where k is the Boltzmann constant and T is thermodynamic temperature of the system.. This corresponds to the contribution of 1/2R to molar heat capacity CV at constant pressure.. This contribution belongs to each translational and rotational degree of freedom. It is different with vibrational degrees of freedom. The contribution is twice as big, i.e. R. This is due to the fact that the energy of vibrational motion is composed of kinetic and potential energy, and not only kinetic energy as it is with translational and rotational motion.. The task is to determine the degrees of freedom for one-atom molecules of argon and two-atom molecules of nitrogen.. • #### Solution. The equipartition theorem is a statement from statistical physics according to which to each generalized momentum and each generalized coordinate acting in the formula for energy of the system quadratically corresponds to mean energy of $$\frac{1}{2}kT,$$ where k is the Boltzmann constant and T is thermodynamic temperature.. In other words each generalized momentum and generalized coordination acting in the formula for the energy of the system quadratically corresponds to a contribution of $$\frac{1}{2}R$$ to molar heat capacity CV.. This means that the state contribution corresponds to each translational and rotational degree of freedom. It is different with vibrational degrees of freedom. The contribution is twice as big, i.e. R. This is due to the fact that the energy of vibrational motion constitutes of kinetic and potential energy, and not only kinetic energy as it is with translational and rotational motion.
|
Now we can easily determine heat capacities of given gases.. Argon is a one-atom gas which means that its molecules have only three translational degrees of freedom. Molar heat capacity CV at constant volume is therefore. $C_{VAr}=3\cdot\frac{1}{2}R=\frac{3}{2}R.$. Related specific heat capacity cV is determined from molar mass MmAr as follows:. $c_{VAr}=\frac{C_{VAr}}{M_{mAr}}=\frac{3R}{2M_{mAr}}.$. Nitrogen is a two-atom gas. It has six degrees of freedom in total. Three of them are translational, two are rotational and one is vibrational. According to the equipartition theorem the molar heat capacity CVN at constant volume is given by. $C_{VN}=3\cdot\frac{1}{2}R+2\cdot\frac{1}{2}R+R=\frac{7}{2}R.$. For a corresponding specific heat capacity cVN we then obtain:. $c_{VN}=\frac{C_{VN}}{M_{mN}}=\frac{7R}{2M_{mN}},$. where MmN is molar mass of nitrogen.. • #### Numerical Solution. $c_{VAr}=\frac{3R}{2M_{mAr}}=\frac{3\cdot{8.31}}{2\cdot{0.039}}\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}$ $c_{VAr}=319.6\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}=0.32\,\mathrm{kJkg}^{-1}\mathrm{K}^{-1}$. $c_{VN}=\frac{7R}{2M_{mN}}=\frac{7\cdot{8.31}}{2\cdot{0.028}}\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}$ $c_{VN}=1038.8\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}\dot{=}1.04\,\mathrm{kJkg}^{-1}\mathrm{K}^{-1}$. • #### Answer and Result Discussion. Calculated value Table value for t = 20 °C cVAr 0.32 kJ kg−1 K−1 0.32 kJ kg−1 K−1 cVN 1.04 kJ kg−1 K−1 0.74 kJ kg−1 K−1. By comparing calculated results with table values we can see that while the values for argon are the same, there is a big difference in the values for nitrogen. This is due to the fact that the equipartition theorem is derived within classical physics and thus does not respect the specifics of quantum mechanics that are manifested quite profoundly when dedscribing the vibrational motion at normal temperatures. In reality, the vibrations practically do not contribute to heat capacity under normal conditions. Therefore, to determine molar heat capacity CVN of nitrogen, we should use a more precise formula. $C_{VN}=3\cdot\frac{1}{2}R+2\cdot\frac{1}{2}R=\frac{5}{2}R$. that involves only the translation and rotation.. If we calculate corresponding specific heat capacity cVN. $c_{VN}=\frac{C_{VN}}{M_{mN}}=\frac{5R}{2M_{mN}},$. we obtain a numerical value of. $c_{VN}\dot{=}742\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}\dot{=}0.74\,\mathrm{kJkg}^{-1}\mathrm{K}^{-1}.$. This result is in a good agreement with the table value.
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# Permutations and Combinations | Counting | Infinity Learn | Summary and Q&A
3.5M views
July 13, 2015
by
Infinity Learn NEET
Permutations and Combinations | Counting | Infinity Learn
## TL;DR
Learn the key rules of counting to understand permutations and combinations, making problem-solving easier.
## Key Insights
• 🔄 The understanding of counting principles is crucial for mastering permutations and combinations.
• 📏 The 'OR' rule represents addition, while the 'AND' rule represents multiplication in counting techniques.
• 👻 Counting techniques can be applied to various scenarios, allowing for the calculation of different combinations and permutations.
• 🔄 Visual representations, such as the example of counting circles in a grid, can simplify counting and understanding permutations and combinations.
• 🧑🎓 Many students find permutations and combinations challenging, but with a solid understanding of counting principles, it becomes a manageable topic.
• 🔄 Formulae for permutations and combinations are useful but not as important as understanding counting principles.
• 👻 Developing counting skills allows for efficient problem-solving in permutations and combinations.
## Transcript
The formula for N C R is 'N factorial, divided by R' factorial times 'N minus R' factorial. What is this? This is the formula for combinations, but it is not the best way to understand it ! And what is NPR. It's N factorial over 'N minus R' factorial. And this again, is not the best way to understand permutations! In many books, and videos you woul... Read More
### Q: What is the formula for calculating permutations and combinations?
The formula for permutations is N! / (N-R)! while combinations use N! / (R!(N-R)!). However, understanding counting principles is more important than relying solely on these formulas.
### Q: How do counting techniques relate to permutations and combinations?
Counting techniques, such as understanding 'OR' and 'AND' principles, are essential to solving permutations and combinations problems. 'OR' represents addition while 'AND' represents multiplication.
### Q: How can counting principles be applied to simple scenarios?
Counting principles can be applied to various scenarios. For example, if you have 3 pens and 2 markers, you can count the total number of ways to select any one item as 3 (pens) + 2 (markers) = 5.
### Q: How can counting principles be used to determine the ways to select one pen and one marker?
To calculate the ways to select one pen and one marker, you multiply the number of ways to select a pen (3) with the number of ways to select a marker (2), resulting in 3 (pens) x 2 (markers) = 6.
## Summary & Key Takeaways
• Understanding permutations and combinations requires a solid grasp of counting techniques.
• Permutations can be calculated using the formula N! / (N-R)! while combinations use N! / (R!(N-R)!).
• The key to mastering permutations and combinations is to understand the rules of counting, which involve addition (OR) and multiplication (AND) principles.
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# Permutations and Combinations | Counting | Infinity Learn | Summary and Q&A. 3.5M views. July 13, 2015. by. Infinity Learn NEET. Permutations and Combinations | Counting | Infinity Learn. ## TL;DR. Learn the key rules of counting to understand permutations and combinations, making problem-solving easier.. ## Key Insights. • 🔄 The understanding of counting principles is crucial for mastering permutations and combinations.. • 📏 The 'OR' rule represents addition, while the 'AND' rule represents multiplication in counting techniques.. • 👻 Counting techniques can be applied to various scenarios, allowing for the calculation of different combinations and permutations.. • 🔄 Visual representations, such as the example of counting circles in a grid, can simplify counting and understanding permutations and combinations.. • 🧑🎓 Many students find permutations and combinations challenging, but with a solid understanding of counting principles, it becomes a manageable topic.. • 🔄 Formulae for permutations and combinations are useful but not as important as understanding counting principles.. • 👻 Developing counting skills allows for efficient problem-solving in permutations and combinations.. ## Transcript. The formula for N C R is 'N factorial, divided by R' factorial times 'N minus R' factorial.
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What is this? This is the formula for combinations, but it is not the best way to understand it ! And what is NPR. It's N factorial over 'N minus R' factorial. And this again, is not the best way to understand permutations! In many books, and videos you woul... Read More. ### Q: What is the formula for calculating permutations and combinations?. The formula for permutations is N! / (N-R)! while combinations use N! / (R!(N-R)!). However, understanding counting principles is more important than relying solely on these formulas.. ### Q: How do counting techniques relate to permutations and combinations?. Counting techniques, such as understanding 'OR' and 'AND' principles, are essential to solving permutations and combinations problems. 'OR' represents addition while 'AND' represents multiplication.. ### Q: How can counting principles be applied to simple scenarios?. Counting principles can be applied to various scenarios. For example, if you have 3 pens and 2 markers, you can count the total number of ways to select any one item as 3 (pens) + 2 (markers) = 5.. ### Q: How can counting principles be used to determine the ways to select one pen and one marker?. To calculate the ways to select one pen and one marker, you multiply the number of ways to select a pen (3) with the number of ways to select a marker (2), resulting in 3 (pens) x 2 (markers) = 6.. ## Summary & Key Takeaways. • Understanding permutations and combinations requires a solid grasp of counting techniques.. • Permutations can be calculated using the formula N! / (N-R)! while combinations use N! / (R!(N-R)!).. • The key to mastering permutations and combinations is to understand the rules of counting, which involve addition (OR) and multiplication (AND) principles.
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https://www.vedantu.com/question-answer/write-one-million-in-scientific-notatio-class-5-maths-cbse-5ff721257c2bba6bf9e5c9ef
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# How do you write one million in scientific notation?
Last updated date: 12th Aug 2024
Total views: 395.1k
Views today: 10.95k
Verified
395.1k+ views
Hint:In the given question, we have been given the scientific notation of a number written in a phrase. To solve this, we must know what a scientific notation is – it is a way of writing extremely large numbers including lots of zeroes, in the form of a single digit followed by a decimal point, followed by the rest of non-zero numbers, followed by multiplication with ten raised to the power of condensed numbers, including the zeroes.
First, one million is one followed by six zeroes, or, $1,000,000$.
$1 \times {10^6}$
Hence, one million can be written as $1 \times {10^6}$.
The way of writing in the scientific notation is especially helpful to physicists, scientists, researchers who are working with very large numbers. For example, writing $1$ followed by twenty-nine zeros is very space and time consuming, so it is better to write it as $1 \times {10^{29}}$. This small notation saved so many zeroes’ space and time.
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Courses. Courses for Kids. Free study material. Offline Centres. More. Store. # How do you write one million in scientific notation?. Last updated date: 12th Aug 2024. Total views: 395.1k. Views today: 10.95k.
|
Verified. 395.1k+ views. Hint:In the given question, we have been given the scientific notation of a number written in a phrase. To solve this, we must know what a scientific notation is – it is a way of writing extremely large numbers including lots of zeroes, in the form of a single digit followed by a decimal point, followed by the rest of non-zero numbers, followed by multiplication with ten raised to the power of condensed numbers, including the zeroes.. First, one million is one followed by six zeroes, or, $1,000,000$.. $1 \times {10^6}$. Hence, one million can be written as $1 \times {10^6}$.. The way of writing in the scientific notation is especially helpful to physicists, scientists, researchers who are working with very large numbers. For example, writing $1$ followed by twenty-nine zeros is very space and time consuming, so it is better to write it as $1 \times {10^{29}}$. This small notation saved so many zeroes’ space and time.
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https://www.thestudentroom.co.uk/showthread.php?t=1622984
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# Core 2 help - AQAWatch
#1
Hey could anybody help me differentiate this:
y = x + 3 + 8/x^4 (x^4 meaning x to the power 4)
I got an answer, which is -24x^-5 +1 however I am not confident with it and I am fairly certain it's wrong
Thanks for any help!
0
7 years ago
#2
Yes, you are slightly wrong, its 1 -32x^-5
because -4 times 8 is -32 not -24 :P.
Multiply by the power, and then reduce the power by 1.
1
7 years ago
#3
(Original post by raveen789)
Hey could anybody help me differentiate this:
y = x + 3 + 8/x^4 (x^4 meaning x to the power 4)
I got an answer, which is -24x^-5 +1 however I am not confident with it and I am fairly certain it's wrong
Thanks for any help!
put more brackets in or write it in latex (google it) so i know what ur asking
0
7 years ago
#4
(Original post by In One Ear)
Yes, you are slightly wrong, its 1 -32x^-5
because -4 times 8 is -32 not -24 :P.
Multiply by the power, and then reduce the power by 1.
ThIs.
0
#5
(Original post by In One Ear)
Yes, you are slightly wrong, its 1 -32x^-5
because -4 times 8 is -32 not -24 :P.
Multiply by the power, and then reduce the power by 1.
Ahh yeah thanks lol I always make stupid mistakes, thanks for the help!
0
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# Core 2 help - AQAWatch. #1. Hey could anybody help me differentiate this:. y = x + 3 + 8/x^4 (x^4 meaning x to the power 4). I got an answer, which is -24x^-5 +1 however I am not confident with it and I am fairly certain it's wrong. Thanks for any help!. 0. 7 years ago. #2. Yes, you are slightly wrong, its 1 -32x^-5. because -4 times 8 is -32 not -24 :P.. Multiply by the power, and then reduce the power by 1.. 1. 7 years ago. #3. (Original post by raveen789). Hey could anybody help me differentiate this:. y = x + 3 + 8/x^4 (x^4 meaning x to the power 4). I got an answer, which is -24x^-5 +1 however I am not confident with it and I am fairly certain it's wrong. Thanks for any help!. put more brackets in or write it in latex (google it) so i know what ur asking. 0. 7 years ago. #4. (Original post by In One Ear). Yes, you are slightly wrong, its 1 -32x^-5. because -4 times 8 is -32 not -24 :P.. Multiply by the power, and then reduce the power by 1.. ThIs.. 0. #5. (Original post by In One Ear). Yes, you are slightly wrong, its 1 -32x^-5. because -4 times 8 is -32 not -24 :P.
|
Multiply by the power, and then reduce the power by 1.. Ahh yeah thanks lol I always make stupid mistakes, thanks for the help!. 0. X. new posts. Latest. My Feed. ### Oops, nobody has postedin the last few hours.. Why not re-start the conversation?. see more. ### See more of what you like onThe Student Room. You can personalise what you see on TSR. Tell us a little about yourself to get started.. ### University open days. • Cranfield University. Cranfield Forensic MSc Programme Open Day Postgraduate. Thu, 25 Apr '19. • University of the Arts London. Open day: MA Footwear and MA Fashion Artefact Postgraduate. Thu, 25 Apr '19. • Cardiff Metropolitan University. Sat, 27 Apr '19. ### Poll. Join the discussion. #### Have you registered to vote?. Yes! (100). 40.16%. No - but I will (12). 4.82%. No - I don't want to (16). 6.43%. No - I can't vote (<18, not in UK, etc) (121). 48.59%.
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# Graphing Linear Functions in Slope-Intercept Form
There are different ways to graph a linear function. Sometimes, the way the rule of the function is written, can dictate the simplest way to graph it. Below, the graphs of linear functions given in slope-intercept form will be explored.
## Slope-Intercept Form
One way to express linear function rules is called slope-intercept form.
y=mx+b
Sometimes, f(x)=mx+b is used. In either case, m and b describe the general characteristics of the line. m indicates the slope, and b indicates the y-intercept. The linear function graphed below can be expressed as
f(x)=2x+1,
because it has a slope of 2 and a y-intercept at (0,1).
## Graphing a Linear Function in Slope-Intercept Form
To graph a line in slope-intercept form, the slope, m, and the y-intercept, b, are both needed. Consider the linear function y=2x3. Since the rule is written as y=mx+b, it can be seen that
To graph the line, plot the y-intercept, then use the slope to find another point on the line. Specifically, from (0,-3), move up 2 units and right 1 unit.
Next, draw a line through both points to create the graph of the linear function.
fullscreen
Exercise
In Clear Lake, Iowa, during a particular evening, there is a 3-inch layer of snow on the ground. At midnight, it begins to snow. Each hour, one inch of snow falls. Graph a function that shows the amount of snow on the ground from midnight to 6 AM.
Show Solution
Solution
To begin, we can define the quantities that x and y represent.
• Let x represent the number of hours it's been snowing.
• Let y represent the number of inches of snow on the ground.
It's been given that there is a 3-inch layer of snow on the ground before it begins to snow. Thus, when x=0,y=3. In other words, the y-intercept is b=3. It is also given that the amount of snow on the ground increases by 1 inch every 1 hour. Thus, the slope of the line is We can write the rule for this function as
To graph the function, we can plot a point at (0,3), then move up 1 unit and right 1 unit to find another point. The line that connects these points is the graph of the function.
The graph above shows the function f(x)=x+3. It can be seen that, at 6AM, there is a total of 9 inches of snow on the ground.
## Describing Key Features of Linear Functions
Key features are certain characteristics of the graphs of functions that are noteworthy. For linear functions, a graph's intercepts, increasing/decreasing intervals, and end behavior are important.
fullscreen
Exercise
The graph shows the function Describe the intercepts, increasing/decreasing intervals, and end behavior. Then, show the features on the graph.
Show Solution
Solution
To begin, we'll describe each of the features. A graph's x- and y-intercepts are the points where the graph intersects the x- and y-axes, respectively. It can be seen that f intersects the x-axis at (6,0) and the y-axis at (0,4). Thus,
Since f is a linear function, it has a constant rate of change. This means the line will always be increasing or always decreasing. Looking from left to right on the graph, it can be seen that, as x increases, f(x) decreases infinitely. Thus,
Since f always decreases, we can say f is negative. Notice that this corresponds with f's negative slope.
Looking at the graph, we can see that the left end extends upward and the right end extends downward. Thus, the end behavior of f can be written as follows.
Lastly, we'll label the features on the graph.
## Solving Linear Equations Graphically
The function y=f(x) is comprised of all the (x,y) points that satisfy its function rule. For example, some of the points that lie on the line y=x+3 are
All of the x-y pairs that satisfy an equation are the points on the corresponding graph. Thus, the solution(s) to an equation can be found using a graph.
fullscreen
Exercise
Use the graph of y=2.5x1.5 to solve the equation 6=2.5x1.5.
Show Solution
Solution
The given graph shows all of the x-y pairs that satisfy y=2.5x1.5. Let's begin by comparing the function rule with the equation.
Since the right-hand side of both the function rule and the equation are the same, it follows then that
y=6.
Thus, solving the equation 6=2.5x1.5 means finding the x-coordinate of the point on the graph of y=2.5x1.5 whose y-coordinate is 6. We'll find this point by first locating the point on the graph where y=6.
We can move down from this point on the graph to find the x-coordinate.
Since the desired point on the graph is (3,6), the solution to 6=2.5x1.5 is x=3. We can verify this value by substituting it into the equation to see if a true statement is made.
6=2.5x1.5
6=6
Since a true statement is made, we can confirm that the solution to the equation is x=3.
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mathleaks.com mathleaks.com Start chapters home Start History history History expand_more Community. Community expand_more. {{ filterOption.label }}. {{ item.displayTitle }}. {{ item.subject.displayTitle }}. arrow_forward. {{ searchError }}. search. {{ courseTrack.displayTitle }}. {{ printedBook.courseTrack.name }} {{ printedBook.name }}. # Graphing Linear Functions in Slope-Intercept Form. There are different ways to graph a linear function. Sometimes, the way the rule of the function is written, can dictate the simplest way to graph it. Below, the graphs of linear functions given in slope-intercept form will be explored.. ## Slope-Intercept Form. One way to express linear function rules is called slope-intercept form.. y=mx+b. Sometimes, f(x)=mx+b is used. In either case, m and b describe the general characteristics of the line. m indicates the slope, and b indicates the y-intercept. The linear function graphed below can be expressed as. f(x)=2x+1,. because it has a slope of 2 and a y-intercept at (0,1).. ## Graphing a Linear Function in Slope-Intercept Form. To graph a line in slope-intercept form, the slope, m, and the y-intercept, b, are both needed. Consider the linear function y=2x3. Since the rule is written as y=mx+b, it can be seen that. To graph the line, plot the y-intercept, then use the slope to find another point on the line. Specifically, from (0,-3), move up 2 units and right 1 unit.. Next, draw a line through both points to create the graph of the linear function.. fullscreen. Exercise. In Clear Lake, Iowa, during a particular evening, there is a 3-inch layer of snow on the ground. At midnight, it begins to snow. Each hour, one inch of snow falls. Graph a function that shows the amount of snow on the ground from midnight to 6 AM.. Show Solution. Solution. To begin, we can define the quantities that x and y represent.. • Let x represent the number of hours it's been snowing.. • Let y represent the number of inches of snow on the ground.. It's been given that there is a 3-inch layer of snow on the ground before it begins to snow. Thus, when x=0,y=3. In other words, the y-intercept is b=3. It is also given that the amount of snow on the ground increases by 1 inch every 1 hour. Thus, the slope of the line is We can write the rule for this function as. To graph the function, we can plot a point at (0,3), then move up 1 unit and right 1 unit to find another point.
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The line that connects these points is the graph of the function.. The graph above shows the function f(x)=x+3. It can be seen that, at 6AM, there is a total of 9 inches of snow on the ground.. ## Describing Key Features of Linear Functions. Key features are certain characteristics of the graphs of functions that are noteworthy. For linear functions, a graph's intercepts, increasing/decreasing intervals, and end behavior are important.. fullscreen. Exercise. The graph shows the function Describe the intercepts, increasing/decreasing intervals, and end behavior. Then, show the features on the graph.. Show Solution. Solution. To begin, we'll describe each of the features. A graph's x- and y-intercepts are the points where the graph intersects the x- and y-axes, respectively. It can be seen that f intersects the x-axis at (6,0) and the y-axis at (0,4). Thus,. Since f is a linear function, it has a constant rate of change. This means the line will always be increasing or always decreasing. Looking from left to right on the graph, it can be seen that, as x increases, f(x) decreases infinitely. Thus,. Since f always decreases, we can say f is negative. Notice that this corresponds with f's negative slope.. Looking at the graph, we can see that the left end extends upward and the right end extends downward. Thus, the end behavior of f can be written as follows.. Lastly, we'll label the features on the graph.. ## Solving Linear Equations Graphically. The function y=f(x) is comprised of all the (x,y) points that satisfy its function rule. For example, some of the points that lie on the line y=x+3 are. All of the x-y pairs that satisfy an equation are the points on the corresponding graph. Thus, the solution(s) to an equation can be found using a graph.. fullscreen. Exercise. Use the graph of y=2.5x1.5 to solve the equation 6=2.5x1.5.. Show Solution. Solution. The given graph shows all of the x-y pairs that satisfy y=2.5x1.5. Let's begin by comparing the function rule with the equation.. Since the right-hand side of both the function rule and the equation are the same, it follows then that. y=6.. Thus, solving the equation 6=2.5x1.5 means finding the x-coordinate of the point on the graph of y=2.5x1.5 whose y-coordinate is 6. We'll find this point by first locating the point on the graph where y=6.. We can move down from this point on the graph to find the x-coordinate.. Since the desired point on the graph is (3,6), the solution to 6=2.5x1.5 is x=3. We can verify this value by substituting it into the equation to see if a true statement is made.. 6=2.5x1.5. 6=6. Since a true statement is made, we can confirm that the solution to the equation is x=3.
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https://music.stackexchange.com/questions/58531/can-we-use-any-note-without-caring-much-on-rule-of-particular-meter-as-long-as-n/58533
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Can we use any note without caring much on rule of particular meter as long as note value fits in the meter?
As we know there are two main common meter :- simple and compound. We all know that simple meter divides a beat into two and compound meter into three. We also know that a measure holding two beat is duple, three beat triple and four beat quadruple so on. Now my question rise here. I saw some where in internet a music sheet using 2/4 meter which seems "simple duple". 2 beat in each measure "duple" and it would have been normal if a quarter note or eighth note use those beat space because they can be divided into two which make sense to be a simple. But a triplet? I found they using two triplet of eight note which we know 1-triplet = 1 beat so, 2-triplet = 2 beat. It fitted fine with the 2/4 time signature in the view of "duple" I think we can say two dotted quarter note made a triplet (if i'm wrong at this point tell me guys). But in view of "simple" a triplet seems to be divided in three part of a note then how is it fitted on "simple" ? If triplet is a note divided in three then isn't it suitable to be in compound? Is 2/4 meter really a compound duple or simple duple? Please help me to clear my misunderstanding.
• Could I ask you to work on the formatting of this question. It is very hard to read as is and that may very easily affect the quality of the answers. – Neil Meyer Jun 19 '17 at 9:58
• I'm not that great on english never mind if you can edit my question keeping in care what I'm trying to say that will be great @NeilMeyer – SOuřaan Gřg Jun 19 '17 at 10:45
• @SOuřaanGřg I think your English is off to a good start; I understood your question no problem! – Richard Jun 19 '17 at 13:20
If ALL the notes relate to triplets, then you will have six quaver triplets per bar, and it would be better written in 6/8 (compound duple). If only some bars have triplets, and others contain the normal 4 quavers, 2 crotchets, or a combination, the time sig will probably stay as simple duple, 2/4. So, it depends on the writer and the tune itself. One will always be a better fit than the other.
• I have less reputation for now but submitted a up vote for your kind help! When we have a dot with any note value is the dot left uncount ? I mean doesn't the dot take effect on the number of beat per measure of a meter? A two "quarter dotted" note is on 2/4 meter which seem only "quarter" note been counted for beat per measure but "dot" doesn't seem to be counted? – SOuřaan Gřg Jun 19 '17 at 2:17
• A dot after a note increases its value or length by an extra half of its original. BUT - triplets are actually three identical notes that are played in the time of two of those same notes. A kind of opposite to what the dot does! – Tim Jun 19 '17 at 6:16
2/4 is a simple duple meter. It doesn't matter how many triplets, quintuplets, septuplets, or any other tuplets are in the music--it's still a simple duple meter.
Now, if you notate all eighth-note triplets and no other tuplets (including bog-standard duplets) in the entire 2/4 piece, people may argue that your music should be in 6/8 (a compound duple meter) instead.
(I'm answering based on my best understanding of your question. I hope I'm interpreting your words right...)
• I have less reputation for now but submitted a up vote for your kind help! It mean as long as a note value fit in the meter we can add it without worrying about "simple or compound" ? Just keeping in mind with duple,triple,quadruple etc.. ? – SOuřaan Gřg Jun 19 '17 at 2:00
• You can put in any notes you want as long as they fit in the meter (such as a 16th-note quintuplet and an 8th-note triplet in the same 2/4 bar). Consistently put triplets instead of any other type of tuplet in a piece, though, and people will wonder why the piece is written in a simple meter instead of a compound one. – Dekkadeci Jun 19 '17 at 7:57
I think you have a very cursory understanding of what a Time Signature is. The beats of time signatures can have any number of notes in them. Compound vs Simple is much more an issue of what beats consist of.
Compound time has dotted note values for beats where Simple time simply has notes without dots for beats, this has very little to do with what particular notes a beat can or should consist of.
A compound time signature can have two notes in a bar trough the use of duplets, just in the same way as a simple time signature can have three notes in a beat trough the use of triplets
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Can we use any note without caring much on rule of particular meter as long as note value fits in the meter?. As we know there are two main common meter :- simple and compound. We all know that simple meter divides a beat into two and compound meter into three. We also know that a measure holding two beat is duple, three beat triple and four beat quadruple so on. Now my question rise here. I saw some where in internet a music sheet using 2/4 meter which seems "simple duple". 2 beat in each measure "duple" and it would have been normal if a quarter note or eighth note use those beat space because they can be divided into two which make sense to be a simple. But a triplet? I found they using two triplet of eight note which we know 1-triplet = 1 beat so, 2-triplet = 2 beat. It fitted fine with the 2/4 time signature in the view of "duple" I think we can say two dotted quarter note made a triplet (if i'm wrong at this point tell me guys). But in view of "simple" a triplet seems to be divided in three part of a note then how is it fitted on "simple" ? If triplet is a note divided in three then isn't it suitable to be in compound? Is 2/4 meter really a compound duple or simple duple? Please help me to clear my misunderstanding.. • Could I ask you to work on the formatting of this question. It is very hard to read as is and that may very easily affect the quality of the answers. – Neil Meyer Jun 19 '17 at 9:58. • I'm not that great on english never mind if you can edit my question keeping in care what I'm trying to say that will be great @NeilMeyer – SOuřaan Gřg Jun 19 '17 at 10:45. • @SOuřaanGřg I think your English is off to a good start; I understood your question no problem! – Richard Jun 19 '17 at 13:20. If ALL the notes relate to triplets, then you will have six quaver triplets per bar, and it would be better written in 6/8 (compound duple). If only some bars have triplets, and others contain the normal 4 quavers, 2 crotchets, or a combination, the time sig will probably stay as simple duple, 2/4. So, it depends on the writer and the tune itself. One will always be a better fit than the other.. • I have less reputation for now but submitted a up vote for your kind help! When we have a dot with any note value is the dot left uncount ? I mean doesn't the dot take effect on the number of beat per measure of a meter? A two "quarter dotted" note is on 2/4 meter which seem only "quarter" note been counted for beat per measure but "dot" doesn't seem to be counted? – SOuřaan Gřg Jun 19 '17 at 2:17.
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• A dot after a note increases its value or length by an extra half of its original. BUT - triplets are actually three identical notes that are played in the time of two of those same notes. A kind of opposite to what the dot does! – Tim Jun 19 '17 at 6:16. 2/4 is a simple duple meter. It doesn't matter how many triplets, quintuplets, septuplets, or any other tuplets are in the music--it's still a simple duple meter.. Now, if you notate all eighth-note triplets and no other tuplets (including bog-standard duplets) in the entire 2/4 piece, people may argue that your music should be in 6/8 (a compound duple meter) instead.. (I'm answering based on my best understanding of your question. I hope I'm interpreting your words right...). • I have less reputation for now but submitted a up vote for your kind help! It mean as long as a note value fit in the meter we can add it without worrying about "simple or compound" ? Just keeping in mind with duple,triple,quadruple etc.. ? – SOuřaan Gřg Jun 19 '17 at 2:00. • You can put in any notes you want as long as they fit in the meter (such as a 16th-note quintuplet and an 8th-note triplet in the same 2/4 bar). Consistently put triplets instead of any other type of tuplet in a piece, though, and people will wonder why the piece is written in a simple meter instead of a compound one. – Dekkadeci Jun 19 '17 at 7:57. I think you have a very cursory understanding of what a Time Signature is. The beats of time signatures can have any number of notes in them. Compound vs Simple is much more an issue of what beats consist of.. Compound time has dotted note values for beats where Simple time simply has notes without dots for beats, this has very little to do with what particular notes a beat can or should consist of.. A compound time signature can have two notes in a bar trough the use of duplets, just in the same way as a simple time signature can have three notes in a beat trough the use of triplets.
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## #1 2006-04-11 07:11:33
Chemist
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### Line Integral
I just can't figure out what's the difference b/w the following two methods for solving line integrals.
In the book, they say for evaluating a line integral for a function f(x,y,z) = x - 3y2 + z over the line segment C joining the origin and the point (1,1,1) , we must 1st determine the parametric equations in terms of a parameter t.
r = t i + t j + t k 0 ≤ t ≤ 1 ( smooth parameter , 1st derivative is cont and never zero )
they give the line integral form : ∫ f(x,y,z)ds = ∫ f(t,t,t) |v(t)| dt 0 ≤ t ≤ 1
In class, however, this method is never used.
We start from ∫ F.dr ( both vectors ) where F = vector field = M i + N j + O k
∫ F.Tds ( T is the tangent unit vector, F.T----> dot product ) = ∫ F.dr = ∫ F.drdt /dt = ∫ ( M dx/dt + N dy/dt + O dz/dt ) dt
We've used this form to determine the line integral in all the problems we have encountered so far. If I try the 1st way, the answer is not always the same for the 2nd, actually, rarely is it the same. I tried to see the common point between the two:
∫ F.Tds = ∫ F.dr = ∫ F.dr dt / dr = ∫ F.V(t) dt , here obviously V is a velocity vector.
so, ∫ F.V dt = ∫ F |V(t)| cosθ dt ( θ b/w F and V )
In the book , they're using ∫ f(t,t,t) |V(t)| dt , so perhaps they assume that V is parallel to F ? Why would they do that, and what's the significance of either method?
"Fundamentally one will never be able to renounce abstraction."
Werner Heisenberg
## #2 2006-04-11 08:35:58
Ricky
Moderator
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### Re: Line Integral
The main point of below is that you must do different things when your function goes from R³⇒R and R³⇒ R³
For a Path integral, you have the integral over a path c = (x(t), y(t), z(t)) of a function f(x, y, z), where f goes from R^3 -> R.
To integrate this, you integrate from a to b (the start and end points of your path) of f(c(t)) ||c'(t)|| dt, where ||c'(t)|| is the length of the derivative of c(t).
But a Line integral is completely different (sort of...). Here, F goes from R^3 -> R^3, not to just R. Then you need to do:
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
## #3 2006-04-11 08:38:05
Ricky
Moderator
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### Re: Line Integral
If you tried to use both formulas on the same problem, then you either:
Tried to integrate a vector
or
Tried to take the dot product of a scalar.
Of course, neither makes sense.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
## #4 2006-04-11 09:21:14
Chemist
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### Re: Line Integral
Thanks a lot Ricky, I know that a line integral is used to compute the work or circulation depending on what the vector field is, but what's the path integral used for?
Tried to integrate a vector
or
Tried to take the dot product of a scalar.
Of course, neither makes sense.
Nah , I actually changed the scalar function to a vector field ... why can't we do that? Can't every scalar function have its vector components too?
"Fundamentally one will never be able to renounce abstraction."
Werner Heisenberg
## #5 2006-04-11 11:07:04
Chemist
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### Re: Line Integral
If we assume that the total differential form of a vector field is exact, the line integral is always dependent on the path, can we relate this to a conservative field? That is, a line integral exists only for conservative vector fields, which means if a vector field has a line integral over a certain path ( does it make a difference if it's differentiable or not here? ) , it must have a scalar potential function. Does this make any sense? I'm just wondering ...
"Fundamentally one will never be able to renounce abstraction."
Werner Heisenberg
## #6 2006-04-11 11:12:25
Ricky
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### Re: Line Integral
Nah , I actually changed the scalar function to a vector field ... why can't we do that? Can't every scalar function have its vector components too?
Well, take the example: f(x, y, z) = x + y + z
That's from R³⇒R. You could define it from R³⇒R³ by:
f(x, y, z) = (x + y + z, 0, 0)
But these are two completely different things with completely different properties. They look the same, but they aren't.
Is this what you are talking about?
Thanks a lot Ricky, I know that a line integral is used to compute the work or circulation depending on what the vector field is, but what's the path integral used for?
They are pretty much the same exact thing. That is, the both "sum up" the value of a function along a certain path. But one sums up vector values, such as a force, and one sums up constant values, such as heat or density.
If you have f(x, y, z) as the density of a wire at a point in 3d space, and c(t) be the path of that wire. Then the path integral will mass of that wire.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
## #7 2006-04-11 15:02:10
George,Y
Super Member
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### Re: Line Integral
The second integral you mentioned is mainly for computing work done by conservative force. As you can see, the integrated is a dot product of work on a small piece F.dr, where F and dr are both vectors.
The first integral can be used to integrate f(t,t,t) |v(t)|, f(t,t,t) is displacement from origin, and if v(t) stands for velocity, then|v(t)| stands for speed, or the magnitude of velocity. i cannot figure out what the product means.
So you'd better use the second defaultly.
So far, i haven't encountered the situation to use your first integral.
X'(y-Xβ)=0
## #8 2006-04-11 19:34:48
Chemist
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### Re: Line Integral
Well, take the example: f(x, y, z) = x + y + z
That's from R³⇒R. You could define it from R³⇒R³ by:
f(x, y, z) = (x + y + z, 0, 0)
I'm not sure what this is, what's the 0,0 for?
But these are two completely different things with completely different properties. They look the same, but they aren't.
Is this what you are talking about?
No , I was simply saying that if we have a scalar function f(x,y,z) = x +y +z , we can change it to a vector field or to its total differential form. This is what I did in calculating the path integral of a function f, I changed it to its vector form and calculated the scalar product.
They are pretty much the same exact thing. That is, the both "sum up" the value of a function along a certain path. But one sums up vector values, such as a force, and one sums up constant values, such as heat or density.
If you have f(x, y, z) as the density of a wire at a point in 3d space, and c(t) be the path of that wire. Then the path integral will mass of that wire.
Thanks for the explanation.
If we assume that the total differential form of a vector field is exact, the line integral is always dependent on the path, can we relate this to a conservative field? That is, a line integral exists only for conservative vector fields, which means if a vector field has a line integral over a certain path ( does it make a difference if it's differentiable or not here? ) , it must have a scalar potential function. Does this make any sense? I'm just wondering ...
I think I may be right here. I was just solving some problems, it turns out that whenever the total diff form of a vector field is exact ( rotor = 0 ), the line integral is independent of the path, and I can simply compute it F(B) - F(A) if we assume the line integral is from A to B across an arc. This really makes the computation easy for a line integral.
But if the total diff form is exact, continuous and its 1st derivative is continuous too, the line integral in a closed domain would be zero. Why's that?
And can anyone prove Green-Rieman's formula? I have an idea but I think it's incorrect.
"Fundamentally one will never be able to renounce abstraction."
Werner Heisenberg
## #9 2006-04-11 19:36:37
Chemist
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### Re: Line Integral
#### George,Y wrote:
The second integral you mentioned is mainly for computing work done by conservative force. As you can see, the integrated is a dot product of work on a small piece F.dr, where F and dr are both vectors.
The first integral can be used to integrate f(t,t,t) |v(t)|, f(t,t,t) is displacement from origin, and if v(t) stands for velocity, then|v(t)| stands for speed, or the magnitude of velocity. i cannot figure out what the product means.
So you'd better use the second defaultly.
So far, i haven't encountered the situation to use your first integral.
Thanks for the info , I think I got the difference between the two. And yes |V(t)| is the velocity magnitude or speed.
"Fundamentally one will never be able to renounce abstraction."
Werner Heisenberg
## #10 2006-04-12 07:54:48
Chemist
Member
Offline
### Re: Line Integral
We can obtain Green's theorem area formula by going backward :
A = ∫ ∫ dydx = ∫ ∫ ( 1/2 + 1/2 ) dydx = 1/2 ∫ ( xdy - ydx )
, ofcourse the domain is closed.
But it seems sometimes we can compute the area directly from A = ∫ xdy
Example:
Determine the area b/w y=x2 and y = x across C+
A = 1/2 ( ∫ xdy - ydx )
I divided the line integral into arc OA and a straight line OA where A (1,1), the point of intersection of the parabola and the straight line, and O is the origin ofcourse.
I chose the parametirc equations,
For [OA] : y = x , x=x 0 ≤ x ≤ 1
For arc OA : x=y2 , y=y 1 ≤ y ≤ 0
I1 = 1/2 ∫ (x-x)dx = 0 ----> for [OA]
I2 = 1/2 ∫ (y2 - 2y2)dy = 1/6 ---> for arc OA
A = I1 + I2 = 1/6
Now notice that the area could have been calculated directly from:
A = ∫ xdy ( line int across C+ ) = ∫ xdy ( across [OA] ) + ∫ xdy ( across arc OA )
Using the same variables, A = 1/6
The second way is quicker, but I can't seem to figure out when to use it.
"Fundamentally one will never be able to renounce abstraction."
Werner Heisenberg
## #11 2006-04-12 12:45:06
George,Y
Super Member
Offline
### Re: Line Integral
to compute the area enclosed by an ellipse.
X'(y-Xβ)=0
## #12 2006-04-12 23:40:24
Chemist
Member
Offline
### Re: Line Integral
You mean the area A = 1/2 ∫ ( xdy - ydx ) can be simplified to A = ∫ xdy in the case of an ellipse? It was correct for the closed region between x=y2 and y=x. There was no ellipse there ...
"Fundamentally one will never be able to renounce abstraction."
Werner Heisenberg
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °. You are not logged in.. ## #1 2006-04-11 07:11:33. Chemist. Member. Offline. ### Line Integral. I just can't figure out what's the difference b/w the following two methods for solving line integrals.. In the book, they say for evaluating a line integral for a function f(x,y,z) = x - 3y2 + z over the line segment C joining the origin and the point (1,1,1) , we must 1st determine the parametric equations in terms of a parameter t.. r = t i + t j + t k 0 ≤ t ≤ 1 ( smooth parameter , 1st derivative is cont and never zero ). they give the line integral form : ∫ f(x,y,z)ds = ∫ f(t,t,t) |v(t)| dt 0 ≤ t ≤ 1. In class, however, this method is never used.. We start from ∫ F.dr ( both vectors ) where F = vector field = M i + N j + O k. ∫ F.Tds ( T is the tangent unit vector, F.T----> dot product ) = ∫ F.dr = ∫ F.drdt /dt = ∫ ( M dx/dt + N dy/dt + O dz/dt ) dt. We've used this form to determine the line integral in all the problems we have encountered so far. If I try the 1st way, the answer is not always the same for the 2nd, actually, rarely is it the same. I tried to see the common point between the two:. ∫ F.Tds = ∫ F.dr = ∫ F.dr dt / dr = ∫ F.V(t) dt , here obviously V is a velocity vector.. so, ∫ F.V dt = ∫ F |V(t)| cosθ dt ( θ b/w F and V ). In the book , they're using ∫ f(t,t,t) |V(t)| dt , so perhaps they assume that V is parallel to F ? Why would they do that, and what's the significance of either method?. "Fundamentally one will never be able to renounce abstraction.". Werner Heisenberg. ## #2 2006-04-11 08:35:58. Ricky. Moderator. Offline. ### Re: Line Integral. The main point of below is that you must do different things when your function goes from R³⇒R and R³⇒ R³. For a Path integral, you have the integral over a path c = (x(t), y(t), z(t)) of a function f(x, y, z), where f goes from R^3 -> R.. To integrate this, you integrate from a to b (the start and end points of your path) of f(c(t)) ||c'(t)|| dt, where ||c'(t)|| is the length of the derivative of c(t).. But a Line integral is completely different (sort of...). Here, F goes from R^3 -> R^3, not to just R. Then you need to do:. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now...". ## #3 2006-04-11 08:38:05. Ricky. Moderator. Offline. ### Re: Line Integral. If you tried to use both formulas on the same problem, then you either:. Tried to integrate a vector. or. Tried to take the dot product of a scalar.. Of course, neither makes sense.. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now...". ## #4 2006-04-11 09:21:14. Chemist. Member. Offline. ### Re: Line Integral. Thanks a lot Ricky, I know that a line integral is used to compute the work or circulation depending on what the vector field is, but what's the path integral used for?. Tried to integrate a vector. or. Tried to take the dot product of a scalar.. Of course, neither makes sense.. Nah , I actually changed the scalar function to a vector field ... why can't we do that? Can't every scalar function have its vector components too?. "Fundamentally one will never be able to renounce abstraction.". Werner Heisenberg. ## #5 2006-04-11 11:07:04. Chemist. Member. Offline. ### Re: Line Integral. If we assume that the total differential form of a vector field is exact, the line integral is always dependent on the path, can we relate this to a conservative field? That is, a line integral exists only for conservative vector fields, which means if a vector field has a line integral over a certain path ( does it make a difference if it's differentiable or not here? ) , it must have a scalar potential function. Does this make any sense? I'm just wondering .... "Fundamentally one will never be able to renounce abstraction.". Werner Heisenberg. ## #6 2006-04-11 11:12:25. Ricky. Moderator. Offline. ### Re: Line Integral. Nah , I actually changed the scalar function to a vector field ... why can't we do that? Can't every scalar function have its vector components too?. Well, take the example: f(x, y, z) = x + y + z. That's from R³⇒R. You could define it from R³⇒R³ by:. f(x, y, z) = (x + y + z, 0, 0). But these are two completely different things with completely different properties. They look the same, but they aren't.. Is this what you are talking about?. Thanks a lot Ricky, I know that a line integral is used to compute the work or circulation depending on what the vector field is, but what's the path integral used for?. They are pretty much the same exact thing. That is, the both "sum up" the value of a function along a certain path. But one sums up vector values, such as a force, and one sums up constant values, such as heat or density.. If you have f(x, y, z) as the density of a wire at a point in 3d space, and c(t) be the path of that wire. Then the path integral will mass of that wire.. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now...". ## #7 2006-04-11 15:02:10. George,Y. Super Member. Offline. ### Re: Line Integral. The second integral you mentioned is mainly for computing work done by conservative force. As you can see, the integrated is a dot product of work on a small piece F.dr, where F and dr are both vectors.. The first integral can be used to integrate f(t,t,t) |v(t)|, f(t,t,t) is displacement from origin, and if v(t) stands for velocity, then|v(t)| stands for speed, or the magnitude of velocity.
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i cannot figure out what the product means.. So you'd better use the second defaultly.. So far, i haven't encountered the situation to use your first integral.. X'(y-Xβ)=0. ## #8 2006-04-11 19:34:48. Chemist. Member. Offline. ### Re: Line Integral. Well, take the example: f(x, y, z) = x + y + z. That's from R³⇒R. You could define it from R³⇒R³ by:. f(x, y, z) = (x + y + z, 0, 0). I'm not sure what this is, what's the 0,0 for?. But these are two completely different things with completely different properties. They look the same, but they aren't.. Is this what you are talking about?. No , I was simply saying that if we have a scalar function f(x,y,z) = x +y +z , we can change it to a vector field or to its total differential form. This is what I did in calculating the path integral of a function f, I changed it to its vector form and calculated the scalar product.. They are pretty much the same exact thing. That is, the both "sum up" the value of a function along a certain path. But one sums up vector values, such as a force, and one sums up constant values, such as heat or density.. If you have f(x, y, z) as the density of a wire at a point in 3d space, and c(t) be the path of that wire. Then the path integral will mass of that wire.. Thanks for the explanation.. If we assume that the total differential form of a vector field is exact, the line integral is always dependent on the path, can we relate this to a conservative field? That is, a line integral exists only for conservative vector fields, which means if a vector field has a line integral over a certain path ( does it make a difference if it's differentiable or not here? ) , it must have a scalar potential function. Does this make any sense? I'm just wondering .... I think I may be right here. I was just solving some problems, it turns out that whenever the total diff form of a vector field is exact ( rotor = 0 ), the line integral is independent of the path, and I can simply compute it F(B) - F(A) if we assume the line integral is from A to B across an arc. This really makes the computation easy for a line integral.. But if the total diff form is exact, continuous and its 1st derivative is continuous too, the line integral in a closed domain would be zero. Why's that?. And can anyone prove Green-Rieman's formula? I have an idea but I think it's incorrect.. "Fundamentally one will never be able to renounce abstraction.". Werner Heisenberg. ## #9 2006-04-11 19:36:37. Chemist. Member. Offline. ### Re: Line Integral. #### George,Y wrote:. The second integral you mentioned is mainly for computing work done by conservative force. As you can see, the integrated is a dot product of work on a small piece F.dr, where F and dr are both vectors.. The first integral can be used to integrate f(t,t,t) |v(t)|, f(t,t,t) is displacement from origin, and if v(t) stands for velocity, then|v(t)| stands for speed, or the magnitude of velocity. i cannot figure out what the product means.. So you'd better use the second defaultly.. So far, i haven't encountered the situation to use your first integral.. Thanks for the info , I think I got the difference between the two. And yes |V(t)| is the velocity magnitude or speed.. "Fundamentally one will never be able to renounce abstraction.". Werner Heisenberg. ## #10 2006-04-12 07:54:48. Chemist. Member. Offline. ### Re: Line Integral. We can obtain Green's theorem area formula by going backward :. A = ∫ ∫ dydx = ∫ ∫ ( 1/2 + 1/2 ) dydx = 1/2 ∫ ( xdy - ydx ). , ofcourse the domain is closed.. But it seems sometimes we can compute the area directly from A = ∫ xdy. Example:. Determine the area b/w y=x2 and y = x across C+. A = 1/2 ( ∫ xdy - ydx ). I divided the line integral into arc OA and a straight line OA where A (1,1), the point of intersection of the parabola and the straight line, and O is the origin ofcourse.. I chose the parametirc equations,. For [OA] : y = x , x=x 0 ≤ x ≤ 1. For arc OA : x=y2 , y=y 1 ≤ y ≤ 0. I1 = 1/2 ∫ (x-x)dx = 0 ----> for [OA]. I2 = 1/2 ∫ (y2 - 2y2)dy = 1/6 ---> for arc OA. A = I1 + I2 = 1/6. Now notice that the area could have been calculated directly from:. A = ∫ xdy ( line int across C+ ) = ∫ xdy ( across [OA] ) + ∫ xdy ( across arc OA ). Using the same variables, A = 1/6. The second way is quicker, but I can't seem to figure out when to use it.. "Fundamentally one will never be able to renounce abstraction.". Werner Heisenberg. ## #11 2006-04-12 12:45:06. George,Y. Super Member. Offline. ### Re: Line Integral. to compute the area enclosed by an ellipse.. X'(y-Xβ)=0. ## #12 2006-04-12 23:40:24. Chemist. Member. Offline. ### Re: Line Integral. You mean the area A = 1/2 ∫ ( xdy - ydx ) can be simplified to A = ∫ xdy in the case of an ellipse? It was correct for the closed region between x=y2 and y=x. There was no ellipse there .... "Fundamentally one will never be able to renounce abstraction.". Werner Heisenberg.
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Pre-Calc Homework Solutions 92
# Pre-Calc Homework Solutions 92 - 92 Section 3.6 d sin 2x dx...
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28. Continuous: Note that g (0) 5 lim x 0 1 g ( x ) 5 lim x 0 1 cos x 5 cos (0) 5 1, and lim x 0 2 g ( x ) 5 lim x 0 2 ( x 1 b ) 5 b . We require lim x 0 2 g ( x ) 5 g (0), so b 5 1. The function is continuous if b 5 1. Differentiable: For b 5 1, the left-hand derivative is 1 and the right-hand derivative is 2 sin (0) 5 0, so the function is not differentiable. For other values of b , the function is discontinuous at x 5 0 and there is no left-hand derivative. So, there is no value of b that will make the function differentiable at x 5 0. 29. Observe the pattern: } d d x } cos x 5 2 sin x } d d x 5 5 } cos x 5 2 sin x } d d x 2 2 } cos x 5 2 cos x } d d x 6 6 } cos x 5 2 cos x } d d x 3 3 } cos x 5 sin x } d d x 7 7 } cos x 5 sin x } d d x 4 4 } cos x 5 cos x } d d x 8 8 } cos x 5 cos x Continuing the pattern, we see that } d d x n n } cos x 5 sin x when n 5 4 k 1 3 for any whole number k . Since 999 5 4(249) 1 3, } d d x 9 9 9 9 9 9 } cos x 5 sin x . 30. Observe the pattern: } d d x } sin x 5 cos x } d d x 5 5 } sin x 5 cos x } d d x 2 2 } sin x 5 2 sin x } d d x 6 6 } sin x 5 2 sin x } d
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{[ promptMessage ]}. Bookmark it. {[ promptMessage ]}. Pre-Calc Homework Solutions 92. # Pre-Calc Homework Solutions 92 - 92 Section 3.6 d sin 2x dx.... This preview shows page 1. Sign up to view the full content.. 28. Continuous: Note that g (0) 5 lim x 0 1 g ( x ) 5 lim x 0 1 cos x 5 cos (0) 5 1, and lim x 0 2 g ( x ) 5 lim x 0 2 ( x 1 b ) 5 b . We require lim x 0 2 g ( x ) 5 g (0), so b 5 1. The function is continuous if b 5 1. Differentiable: For b 5 1, the left-hand derivative is 1 and the right-hand derivative is 2 sin (0) 5 0, so the function is not differentiable.
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For other values of b , the function is discontinuous at x 5 0 and there is no left-hand derivative. So, there is no value of b that will make the function differentiable at x 5 0. 29. Observe the pattern: } d d x } cos x 5 2 sin x } d d x 5 5 } cos x 5 2 sin x } d d x 2 2 } cos x 5 2 cos x } d d x 6 6 } cos x 5 2 cos x } d d x 3 3 } cos x 5 sin x } d d x 7 7 } cos x 5 sin x } d d x 4 4 } cos x 5 cos x } d d x 8 8 } cos x 5 cos x Continuing the pattern, we see that } d d x n n } cos x 5 sin x when n 5 4 k 1 3 for any whole number k . Since 999 5 4(249) 1 3, } d d x 9 9 9 9 9 9 } cos x 5 sin x . 30. Observe the pattern: } d d x } sin x 5 cos x } d d x 5 5 } sin x 5 cos x } d d x 2 2 } sin x 5 2 sin x } d d x 6 6 } sin x 5 2 sin x } d. This is the end of the preview. Sign up to access the rest of the document.. {[ snackBarMessage ]}. Ask a homework question - tutors are online.
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# Multi-step Problems
## Solve multi-step problems in contexts, deciding which operations and methods to use and why.
##### Menu Level 1Level 2Level 3More Arithmetic
This is level 2; Multiplication and division multi-step problems. You can earn a trophy if you get at least 7 questions correct.
1. I have 6 crayons which are each 17cm long. I also have a ruler which is 31cm long. How far would they reach if I arranged them on the floor in a long line? Working: cm 2. The angles inside an irregular hexagon add up to 720°. If three of the angles are 119° and two of them are 132° what is the size of the other angle? Working: ° 3. Frank has two boxes of screws. The first box contains 58 screws and the second box contains 170 screws. If he needs four screws for every model he makes, how many models can he make with the screws he has? Working: 4. Using the most efficient method you can think of multiply together the numbers from 3 to 8 then divide the product by seven? Working: 5. Farmer Brown has two fields of sheep. The first field contains 17 sheep and the second field contains 64 sheep. He can only transport the sheep to the shearing station nine at a time. What is the minimum number of journeys he needs to make? Working: 6. A 50p coin weighs 8 grams. What is the value in pounds of two kilograms of 50p coins? Working: £ 7. If four soldiers can stand on one square metre. How many soldiers could stand on a ractanguler parade ground measuting 50 metres long and 20 metres wide? Working: 8. Six people share the cost of a meal in a restaurant. The total cost of the starters was £32, The total cost of the main courses was £34 and the total cost of the deserts was £78. How much does each person pay? Working: £ 9. Ayden makes Christmas crackers. Each cracker contains a toy, 2 sweets and a party hat. If each toy costs 46p, each sweet costs 8p and each hat costs 25p, how much change will he have from £50 if he makes 12 crackers? Working: £ 10. Six security guards are deciding on the times they should be on duty so that one of them is guarding the bank at every moment of every day of the week. They multiply the number of days in a week by the number hours in a day then divide the result by six. How many hours will each guard have to work so that between them the whole week is covered? Working: hours
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# Multi-step Problems. ## Solve multi-step problems in contexts, deciding which operations and methods to use and why.. ##### Menu Level 1Level 2Level 3More Arithmetic. This is level 2; Multiplication and division multi-step problems. You can earn a trophy if you get at least 7 questions correct.. 1. I have 6 crayons which are each 17cm long. I also have a ruler which is 31cm long. How far would they reach if I arranged them on the floor in a long line? Working: cm 2. The angles inside an irregular hexagon add up to 720°. If three of the angles are 119° and two of them are 132° what is the size of the other angle? Working: ° 3. Frank has two boxes of screws. The first box contains 58 screws and the second box contains 170 screws. If he needs four screws for every model he makes, how many models can he make with the screws he has? Working: 4. Using the most efficient method you can think of multiply together the numbers from 3 to 8 then divide the product by seven? Working: 5. Farmer Brown has two fields of sheep. The first field contains 17 sheep and the second field contains 64 sheep. He can only transport the sheep to the shearing station nine at a time. What is the minimum number of journeys he needs to make? Working: 6. A 50p coin weighs 8 grams. What is the value in pounds of two kilograms of 50p coins? Working: £ 7. If four soldiers can stand on one square metre. How many soldiers could stand on a ractanguler parade ground measuting 50 metres long and 20 metres wide? Working: 8. Six people share the cost of a meal in a restaurant. The total cost of the starters was £32, The total cost of the main courses was £34 and the total cost of the deserts was £78. How much does each person pay? Working: £ 9. Ayden makes Christmas crackers. Each cracker contains a toy, 2 sweets and a party hat. If each toy costs 46p, each sweet costs 8p and each hat costs 25p, how much change will he have from £50 if he makes 12 crackers? Working: £ 10. Six security guards are deciding on the times they should be on duty so that one of them is guarding the bank at every moment of every day of the week. They multiply the number of days in a week by the number hours in a day then divide the result by six. How many hours will each guard have to work so that between them the whole week is covered? Working: hours. Check. This is Multi-step Problems level 2. You can also try:. Level 1 Level 3. ## Instructions. Try your best to answer the questions above. Type your answers into the boxes provided leaving no spaces. As you work through the exercise regularly click the "check" button. If you have any wrong answers, do your best to do corrections but if there is anything you don't understand, please ask your teacher for help.. When you have got all of the questions correct you may want to print out this page and paste it into your exercise book. If you keep your work in an ePortfolio you could take a screen shot of your answers and paste that into your Maths file.. ## More Activities:. Mathematicians are not the people who find Maths easy; they are the people who enjoy how mystifying, puzzling and hard it is. Are you a mathematician?. Comment recorded on the 14 October 'Starter of the Day' page by Inger Kisby, Herts and Essex High School:. "Just a quick note to say that we use a lot of your starters. It is lovely to have so many different ideas to start a lesson with.
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Thank you very much and keep up the good work.". Comment recorded on the 28 September 'Starter of the Day' page by Malcolm P, Dorset:. "A set of real life savers!!. Keep it up and thank you!". Each month a newsletter is published containing details of the new additions to the Transum website and a new puzzle of the month.. The newsletter is then duplicated as a podcast which is available on the major delivery networks. You can listen to the podcast while you are commuting, exercising or relaxing.. Transum breaking news is available on Twitter @Transum and if that's not enough there is also a Transum Facebook page.. #### Beat The Clock. It is a race against the clock to answer 30 mental arithmetic questions. There are nine levels to choose from to suit pupils of different abilities.. There are answers to this exercise but they are available in this space to teachers, tutors and parents who have logged in to their Transum subscription on this computer.. A Transum subscription unlocks the answers to the online exercises, quizzes and puzzles. It also provides the teacher with access to quality external links on each of the Transum Topic pages and the facility to add to the collection themselves.. Subscribers can manage class lists, lesson plans and assessment data in the Class Admin application and have access to reports of the Transum Trophies earned by class members.. If you would like to enjoy ad-free access to the thousands of Transum resources, receive our monthly newsletter, unlock the printable worksheets and see our Maths Lesson Finishers then sign up for a subscription now:. Subscribe. ## Go Maths. Learning and understanding Mathematics, at every level, requires learner engagement. Mathematics is not a spectator sport. Sometimes traditional teaching fails to actively involve students. One way to address the problem is through the use of interactive activities and this web site provides many of those. The Go Maths page is an alphabetical list of free activities designed for students in Secondary/High school.. ## Maths Map. Are you looking for something specific? An exercise to supplement the topic you are studying at school at the moment perhaps. Navigate using our Maths Map to find exercises, puzzles and Maths lesson starters grouped by topic.. ## Teachers. If you found this activity useful don't forget to record it in your scheme of work or learning management system. The short URL, ready to be copied and pasted, is as follows:. Alternatively, if you use Google Classroom, all you have to do is click on the green icon below in order to add this activity to one of your classes.. It may be worth remembering that if Transum.org should go offline for whatever reason, there is a mirror site at Transum.info that contains most of the resources that are available here on Transum.org.. When planning to use technology in your lesson always have a plan B!. Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for those learning Mathematics anywhere in the world. Click here to enter your comments.. For Students:. For All:. ## Description of Levels. Close. Level 1 - Addition and subtraction multi-step problems. Level 2 - Multiplication and division multi-step problems. Level 3 - Mixed multi-step problems. More Arithmetic including lesson Starters, visual aids, investigations and self-marking exercises.. Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.. ## Curriculum Reference. See the National Curriculum page for links to related online activities and resources.
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PROBLEM 09 – 0359: A uniform circular disk of radius 25 cm is pivoted at a point
20 cm from its center and allowed to oscillate in a vertical
plane under the action of gravity. What is the period of its
small oscillations? Take g as π2 m.s−2.
Solution: The moment of inertia of a uniform circular disk of radius R and
mass M about an axis through its center perpendicular to its plane is
1/2 MR2. By the parallel-axes theorem, the moment of inertia about a
parallel axis a distance h from the first is
I = 1/2 MR2 + Mh2.
The disk is acting as a physical pendulum, and hence its period for
small oscillations is given by
T = 2π √(1/Mgh)
where I is the moment of inertia of the physical pendulum about its axis of
suspension. Hence,
T = 2π √[(1/2 MR2 + Mh2)/Mgh]
= 2π √[(1/2 × 0.252 m2 + 0.022 m2)/(π2 m∙s−2 × 0.2 m)]
= 2 √(0.35625 s)
= 1.193 s.
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PROBLEM 09 – 0359: A uniform circular disk of radius 25 cm is pivoted at a point. 20 cm from its center and allowed to oscillate in a vertical. plane under the action of gravity. What is the period of its. small oscillations? Take g as π2 m.s−2.. Solution: The moment of inertia of a uniform circular disk of radius R and. mass M about an axis through its center perpendicular to its plane is. 1/2 MR2. By the parallel-axes theorem, the moment of inertia about a. parallel axis a distance h from the first is. I = 1/2 MR2 + Mh2.
|
The disk is acting as a physical pendulum, and hence its period for. small oscillations is given by. T = 2π √(1/Mgh). where I is the moment of inertia of the physical pendulum about its axis of. suspension. Hence,. T = 2π √[(1/2 MR2 + Mh2)/Mgh]. = 2π √[(1/2 × 0.252 m2 + 0.022 m2)/(π2 m∙s−2 × 0.2 m)]. = 2 √(0.35625 s). = 1.193 s.
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https://astronomy.stackexchange.com/questions/53152/how-much-of-the-variance-of-matter-is-due-to-scales
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# How much of the variance of matter is due to scales
Imagine a Universe in which the matter power spectrum behaves as $$\mathcal{P}\sim k^{0.8}$$
How much of the variance (not variance squared!) of matter is due to scales around $$k=k_1=10^{2.9}\mathrm{Mpc}^{-1}$$, versus scales around $$k=k_2=10^{-3.0}\mathrm{Mpc}^{-1}$$
To be explicit, how much is: $$\frac{\sigma(k\sim k_1)}{\sigma(k\sim k_2)}?$$
• I'd recommend formatting this with latex, currently, with images cutting through the text, it looks very confusing.. Mar 14, 2023 at 17:07
• Please explain some of the magic numbers here. Why 0.8? Why 2.9? Why -3.0? Mar 14, 2023 at 18:08
• They are random values assigned, nothing theoretical just to obtain certain value.
– Alba
Mar 14, 2023 at 18:38
I'll assume $$\mathcal{P}(k)$$ is the "dimensionless" power spectrum (also known as $$\Delta^2(k)$$), i.e. $$\mathcal{P}(k)\equiv k^3/(2\pi^2) P(k)$$. The density variance is $$\sigma^2=\int_0^\infty \frac{\mathrm{d}k}{k}\mathcal{P}(k).\tag{1}\label{1}$$
"How much of $$\sigma^2$$ is contributed by scales around $$k$$" is ambiguous, but the most natural interpretation (to me) is "how much of $$\sigma^2$$ is contributed by a logarithmic interval around $$k$$", in which case the answer is just (proportional to) the integrand, $$\mathcal{P}(k)$$. To see this explicitly, we can define $$\sigma^2(k)=\int_0^k\frac{\mathrm{d}k^\prime}{k^\prime}\mathcal{P}(k^\prime)\tag{2}\label{2}$$ as the variance contributed by wavenumbers smaller than $$k$$ (i.e. scales larger than $$k$$). Then the differential variance contributed by scales near $$k$$, per logarithmic interval in $$k$$, is $$\frac{\mathrm{d}}{\mathrm{d}\log k}\sigma^2(k)=\mathcal{P}(k).\tag{3}\label{3}$$
Now, you are not asking about $$\sigma^2$$ but about $$\sigma$$. In this case there is further ambiguity to the question, because $$\sigma$$ is no longer simply a sum over the power contributed at different scales and is instead a nonlinear function thereof. In general, the answer to "how much rms variance is contributed by scales near $$k$$" depends on the order in which you want to add up the power at different scales. If we add up power in order from the largest scales (smallest $$k$$) to the smallest scales (largest $$k$$), as in equation \eqref{2}, then the answer is $$\frac{\mathrm{d}}{\mathrm{d}\log k}\sigma(k) =\frac{\mathrm{d}}{\mathrm{d}\log k}[\sigma^2(k)]^{1/2} =\frac{1}{2}[\sigma^2(k)]^{-1/2}\frac{\mathrm{d}\sigma^2(k)}{\mathrm{d}\log k} =\frac{\mathcal{P}(k)}{2\sigma(k)}.\tag{4}\label{4}$$ To add up the scales in a different order, just reorder the integral in equation \eqref{2}; equation \eqref{4} remains accurate with the redefined $$\sigma(k)$$.
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# How much of the variance of matter is due to scales. Imagine a Universe in which the matter power spectrum behaves as $$\mathcal{P}\sim k^{0.8}$$. How much of the variance (not variance squared!) of matter is due to scales around $$k=k_1=10^{2.9}\mathrm{Mpc}^{-1}$$, versus scales around $$k=k_2=10^{-3.0}\mathrm{Mpc}^{-1}$$. To be explicit, how much is: $$\frac{\sigma(k\sim k_1)}{\sigma(k\sim k_2)}?$$. • I'd recommend formatting this with latex, currently, with images cutting through the text, it looks very confusing.. Mar 14, 2023 at 17:07. • Please explain some of the magic numbers here. Why 0.8? Why 2.9? Why -3.0? Mar 14, 2023 at 18:08. • They are random values assigned, nothing theoretical just to obtain certain value.. – Alba.
|
Mar 14, 2023 at 18:38. I'll assume $$\mathcal{P}(k)$$ is the "dimensionless" power spectrum (also known as $$\Delta^2(k)$$), i.e. $$\mathcal{P}(k)\equiv k^3/(2\pi^2) P(k)$$. The density variance is $$\sigma^2=\int_0^\infty \frac{\mathrm{d}k}{k}\mathcal{P}(k).\tag{1}\label{1}$$. "How much of $$\sigma^2$$ is contributed by scales around $$k$$" is ambiguous, but the most natural interpretation (to me) is "how much of $$\sigma^2$$ is contributed by a logarithmic interval around $$k$$", in which case the answer is just (proportional to) the integrand, $$\mathcal{P}(k)$$. To see this explicitly, we can define $$\sigma^2(k)=\int_0^k\frac{\mathrm{d}k^\prime}{k^\prime}\mathcal{P}(k^\prime)\tag{2}\label{2}$$ as the variance contributed by wavenumbers smaller than $$k$$ (i.e. scales larger than $$k$$). Then the differential variance contributed by scales near $$k$$, per logarithmic interval in $$k$$, is $$\frac{\mathrm{d}}{\mathrm{d}\log k}\sigma^2(k)=\mathcal{P}(k).\tag{3}\label{3}$$. Now, you are not asking about $$\sigma^2$$ but about $$\sigma$$. In this case there is further ambiguity to the question, because $$\sigma$$ is no longer simply a sum over the power contributed at different scales and is instead a nonlinear function thereof. In general, the answer to "how much rms variance is contributed by scales near $$k$$" depends on the order in which you want to add up the power at different scales. If we add up power in order from the largest scales (smallest $$k$$) to the smallest scales (largest $$k$$), as in equation \eqref{2}, then the answer is $$\frac{\mathrm{d}}{\mathrm{d}\log k}\sigma(k) =\frac{\mathrm{d}}{\mathrm{d}\log k}[\sigma^2(k)]^{1/2} =\frac{1}{2}[\sigma^2(k)]^{-1/2}\frac{\mathrm{d}\sigma^2(k)}{\mathrm{d}\log k} =\frac{\mathcal{P}(k)}{2\sigma(k)}.\tag{4}\label{4}$$ To add up the scales in a different order, just reorder the integral in equation \eqref{2}; equation \eqref{4} remains accurate with the redefined $$\sigma(k)$$.
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https://calculus.subwiki.org/w/index.php?title=Sinc_function&oldid=832
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# Sinc function
This article is about a particular function from a subset of the real numbers to the real numbers. Information about the function, including its domain, range, and key data relating to graphing, differentiation, and integration, is presented in the article.
View a complete list of particular functions on this wiki
For functions involving angles (trigonometric functions, inverse trigonometric functions, etc.) we follow the convention that all angles are measured in radians. Thus, for instance, the angle of $90\,^\circ$ is measured as $\pi/2$.
## Definition
This function, denoted $\operatorname{sinc}$, is defined as follows:
$\operatorname{sinc} x := \left\lbrace \begin{array}{rl} 1, & x = 0 \\ \frac{\sin x}{x} & x \ne 0\end{array}\right.$
## Key data
Item Value
default domain all real numbers, i.e., all of $\R$.
range the closed interval $[\alpha,1]$ where $\alpha$ is approximately $-2/(3\pi)$.
period none; the function is not periodic
horizontal asymptotes $y = 0$, i.e., the $x$-axis. This is because $\lim_{x \to \pm \infty} \frac{\sin x}{x} = 0$, which in turn can be deduced from the fact that the numerator is bounded while the magnitude of the denominator approaches $\infty$.
local maximum values and points of attainment The local maxima occur at points $x$ satisfying $\tan x = x$ and $x \in [2n\pi,(2n + 1)\pi]$ or $x \in [-(2n + 1)\pi,-2n\pi]$ for $n$ a positive integer.
There is an anomalous local maximum at $x = 0$ with value 1. Apart from that, the other local maxima occur at points of the form $\pm(2n\pi + \alpha_n)$ where $\alpha_n$ is fairly close to $\pi/2$ for all $n > 0$. The local maximum value at this point is slightly more than $1/(2n\pi + \pi/2)$.
local minimum values and points of attainment The local minima occur at points $x$ satisfying $\tan x = -x$ and $x \in [(2n - 1)\pi,2n\pi]$ or $x \in [-2n\pi,-(2n - 1)\pi]$ for $n$ a positive integer.
The local minima occur at points of the form $\pm(2n\pi - \alpha_n)$ where $\alpha_n$ is fairly close to $\pi/2$ for all $n > 0$. The local minimum value at this point is slightly less than $-1/(2n\pi - \pi/2)$.
points of inflection Fill this in later
important symmetries The function is an even function, i.e., the graph has symmetry about the $y$-axis.
derivative $\operatorname{sinc}'x = \left\lbrace \begin{array}{rl} 0, & x = 0 \\ \frac{x \cos x - \sin x}{x^2}, & x \ne 0\\\end{array}\right.$
antiderivative the sine integral (this is defined as the antiderivative of the sinc function that takes the value 0 at 0)
power series and Taylor series The power series about 0 (which is also the Taylor series) is
$\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k + 1)!} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \dots$
The power series converges globally to the function.
## Graph
Below is a graph of the function for the domain restricted to $[-3\pi,3\pi]$:
The picture is a little unclear, so we consider an alternative depiction of the graph where the $x$-axis and $y$-axis are scaled differently to make it clearer:
## Differentation
### First derivative
Computation at $x = 0$:
WARNING ON ERRONEOUS DIFFERENTIATION APPROACH: Suppose a function is given a separate definition at an isolated point from the definition at points surrounding it (on its immediate left or immediate right). To compute the derivative, it is not correct to simply differentiate the definition at the point and arrive at the derivative. Rather, we need to compute the derivative from first principles as a limit of a difference quotient, where the function value at the point is taken as the specified value and the function value at nearby points is given by the expression used to define the function around the point.
Here, we need to compute the derivative using first principles, as a limit of a difference quotient:
$\operatorname{sinc}' 0 := \lim_{x \to 0} \frac{\operatorname{sinc} x - \operatorname{sinc} 0}{x - 0} = \lim_{x \to 0} \frac{(\sin x)/x - 1}{x} = \lim_{x \to 0} \frac{\sin x - x}{x^2}$
This limit can be computed in many ways. For instance, it can be computed using the L'Hopital rule:
$\lim_{x \to 0} \frac{\sin x - x}{x^2} \stackrel{*}{=} \lim_{x \to 0} \frac{\cos x - 1}{2x} \stackrel{*}{=} \lim_{x \to 0} \frac{-\sin x}{2} = \frac{-\sin 0}{2} = 0$
DERIVATIVE OF EVEN FUNCTION AT ZERO: For an even function, if the function is differentiable at zero, then the derivative at zero is zero. However, it is possible for an even function to not be differentiable at zero, so we cannot directly conclude merely from the function being even that its derivative at zero is zero.
Computation at $x \ne 0$: At any such point, we know that the function looks like $(\sin x)/x$ in an open interval about the point, so we can use the quotient rule for differentiation, which states that:
$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}$
In this case, $f(x) = \sin x$ and $g(x) = x$, so we get:
$\frac{d}{dx}(\operatorname{sinc}\ x) = \frac{x \cos x - \sin x}{x^2}$
Combining the two computations, we get:
$\operatorname{sinc}'(x) = \left\lbrace \begin{array}{rl} 0, & x = 0 \\ \frac{x \cos x - \sin x}{x^2}, & x \ne 0 \\\end{array}\right.$
### Second derivative
Computation at $x = 0$:
WARNING ON ERRONEOUS DIFFERENTIATION APPROACH: Suppose a function is given a separate definition at an isolated point from the definition at points surrounding it (on its immediate left or immediate right). To compute the derivative, it is not correct to simply differentiate the definition at the point and arrive at the derivative. Rather, we need to compute the derivative from first principles as a limit of a difference quotient, where the function value at the point is taken as the specified value and the function value at nearby points is given by the expression used to define the function around the point.
Here, we need to compute the derivative using first principles, as a limit of a difference quotient:
$\operatorname{sinc}'' 0 := \lim_{x \to 0} \frac{\operatorname{sinc}' x - \operatorname{sinc}' 0}{x - 0} = \lim_{x \to 0} \frac{(x \cos x - \sin x)/x^2 - 0}{x} = \lim_{x \to 0} \frac{x \cos x - \sin x}{x^3}$
The latter limit can be computed in many ways. One way is to use the L'Hopital rule:
$\lim_{x \to 0} \frac{x \cos x - \sin x}{x^3} \stackrel{*}{=} \lim_{x \to 0} \frac{-x \sin x + \cos x - \cos x}{3x^2} = \lim_{x \to 0} \frac{-\sin x}{3x} \stackrel{*}{=} \lim_{x \to 0} \frac{-\cos x}{3} = \frac{-1}{3}$
Thus, $\operatorname{sinc}'' 0 = -1/3$.
The computation at other points can be done using the usual method for computing derivatives, i.e., with the quotient rule for differentiation combined with other rules. Overall, we get:
$\operatorname{sinc}'(x) = \left\lbrace \begin{array}{rl} \frac{-1}{3}, & x = 0 \\ \frac{(2 - x^2)\sin x - 2x\cos x}{x^3}, & x \ne 0 \\\end{array}\right.$
## Taylor series and power series
### Computation of power series
We use the power series for the sine function (see sine function#Computation of power series):
$\! \sin x := x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots = \sum_{k=0}^\infty \frac{(-1)^kx^{2k+1}}{(2k + 1)!}$
Dividing both sides by $x$ (valid when $x \ne 0$), we get:
$\! \frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \dots = \sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k+1)!}$
We note that the power series also works at $x = 0$ (because $\operatorname{sinc} \ 0 = 1$), hence it works globally, and is the power series for the sinc function.
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# Sinc function. This article is about a particular function from a subset of the real numbers to the real numbers. Information about the function, including its domain, range, and key data relating to graphing, differentiation, and integration, is presented in the article.. View a complete list of particular functions on this wiki. For functions involving angles (trigonometric functions, inverse trigonometric functions, etc.) we follow the convention that all angles are measured in radians. Thus, for instance, the angle of $90\,^\circ$ is measured as $\pi/2$.. ## Definition. This function, denoted $\operatorname{sinc}$, is defined as follows:. $\operatorname{sinc} x := \left\lbrace \begin{array}{rl} 1, & x = 0 \\ \frac{\sin x}{x} & x \ne 0\end{array}\right.$. ## Key data. Item Value. default domain all real numbers, i.e., all of $\R$.. range the closed interval $[\alpha,1]$ where $\alpha$ is approximately $-2/(3\pi)$.. period none; the function is not periodic. horizontal asymptotes $y = 0$, i.e., the $x$-axis. This is because $\lim_{x \to \pm \infty} \frac{\sin x}{x} = 0$, which in turn can be deduced from the fact that the numerator is bounded while the magnitude of the denominator approaches $\infty$.. local maximum values and points of attainment The local maxima occur at points $x$ satisfying $\tan x = x$ and $x \in [2n\pi,(2n + 1)\pi]$ or $x \in [-(2n + 1)\pi,-2n\pi]$ for $n$ a positive integer.. There is an anomalous local maximum at $x = 0$ with value 1. Apart from that, the other local maxima occur at points of the form $\pm(2n\pi + \alpha_n)$ where $\alpha_n$ is fairly close to $\pi/2$ for all $n > 0$. The local maximum value at this point is slightly more than $1/(2n\pi + \pi/2)$.. local minimum values and points of attainment The local minima occur at points $x$ satisfying $\tan x = -x$ and $x \in [(2n - 1)\pi,2n\pi]$ or $x \in [-2n\pi,-(2n - 1)\pi]$ for $n$ a positive integer.. The local minima occur at points of the form $\pm(2n\pi - \alpha_n)$ where $\alpha_n$ is fairly close to $\pi/2$ for all $n > 0$. The local minimum value at this point is slightly less than $-1/(2n\pi - \pi/2)$.. points of inflection Fill this in later. important symmetries The function is an even function, i.e., the graph has symmetry about the $y$-axis.. derivative $\operatorname{sinc}'x = \left\lbrace \begin{array}{rl} 0, & x = 0 \\ \frac{x \cos x - \sin x}{x^2}, & x \ne 0\\\end{array}\right.$. antiderivative the sine integral (this is defined as the antiderivative of the sinc function that takes the value 0 at 0). power series and Taylor series The power series about 0 (which is also the Taylor series) is. $\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k + 1)!} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \dots$. The power series converges globally to the function.. ## Graph. Below is a graph of the function for the domain restricted to $[-3\pi,3\pi]$:. The picture is a little unclear, so we consider an alternative depiction of the graph where the $x$-axis and $y$-axis are scaled differently to make it clearer:. ## Differentation. ### First derivative. Computation at $x = 0$:. WARNING ON ERRONEOUS DIFFERENTIATION APPROACH: Suppose a function is given a separate definition at an isolated point from the definition at points surrounding it (on its immediate left or immediate right). To compute the derivative, it is not correct to simply differentiate the definition at the point and arrive at the derivative.
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Rather, we need to compute the derivative from first principles as a limit of a difference quotient, where the function value at the point is taken as the specified value and the function value at nearby points is given by the expression used to define the function around the point.. Here, we need to compute the derivative using first principles, as a limit of a difference quotient:. $\operatorname{sinc}' 0 := \lim_{x \to 0} \frac{\operatorname{sinc} x - \operatorname{sinc} 0}{x - 0} = \lim_{x \to 0} \frac{(\sin x)/x - 1}{x} = \lim_{x \to 0} \frac{\sin x - x}{x^2}$. This limit can be computed in many ways. For instance, it can be computed using the L'Hopital rule:. $\lim_{x \to 0} \frac{\sin x - x}{x^2} \stackrel{*}{=} \lim_{x \to 0} \frac{\cos x - 1}{2x} \stackrel{*}{=} \lim_{x \to 0} \frac{-\sin x}{2} = \frac{-\sin 0}{2} = 0$. DERIVATIVE OF EVEN FUNCTION AT ZERO: For an even function, if the function is differentiable at zero, then the derivative at zero is zero. However, it is possible for an even function to not be differentiable at zero, so we cannot directly conclude merely from the function being even that its derivative at zero is zero.. Computation at $x \ne 0$: At any such point, we know that the function looks like $(\sin x)/x$ in an open interval about the point, so we can use the quotient rule for differentiation, which states that:. $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}$. In this case, $f(x) = \sin x$ and $g(x) = x$, so we get:. $\frac{d}{dx}(\operatorname{sinc}\ x) = \frac{x \cos x - \sin x}{x^2}$. Combining the two computations, we get:. $\operatorname{sinc}'(x) = \left\lbrace \begin{array}{rl} 0, & x = 0 \\ \frac{x \cos x - \sin x}{x^2}, & x \ne 0 \\\end{array}\right.$. ### Second derivative. Computation at $x = 0$:. WARNING ON ERRONEOUS DIFFERENTIATION APPROACH: Suppose a function is given a separate definition at an isolated point from the definition at points surrounding it (on its immediate left or immediate right). To compute the derivative, it is not correct to simply differentiate the definition at the point and arrive at the derivative. Rather, we need to compute the derivative from first principles as a limit of a difference quotient, where the function value at the point is taken as the specified value and the function value at nearby points is given by the expression used to define the function around the point.. Here, we need to compute the derivative using first principles, as a limit of a difference quotient:. $\operatorname{sinc}'' 0 := \lim_{x \to 0} \frac{\operatorname{sinc}' x - \operatorname{sinc}' 0}{x - 0} = \lim_{x \to 0} \frac{(x \cos x - \sin x)/x^2 - 0}{x} = \lim_{x \to 0} \frac{x \cos x - \sin x}{x^3}$. The latter limit can be computed in many ways. One way is to use the L'Hopital rule:. $\lim_{x \to 0} \frac{x \cos x - \sin x}{x^3} \stackrel{*}{=} \lim_{x \to 0} \frac{-x \sin x + \cos x - \cos x}{3x^2} = \lim_{x \to 0} \frac{-\sin x}{3x} \stackrel{*}{=} \lim_{x \to 0} \frac{-\cos x}{3} = \frac{-1}{3}$. Thus, $\operatorname{sinc}'' 0 = -1/3$.. The computation at other points can be done using the usual method for computing derivatives, i.e., with the quotient rule for differentiation combined with other rules. Overall, we get:. $\operatorname{sinc}'(x) = \left\lbrace \begin{array}{rl} \frac{-1}{3}, & x = 0 \\ \frac{(2 - x^2)\sin x - 2x\cos x}{x^3}, & x \ne 0 \\\end{array}\right.$. ## Taylor series and power series. ### Computation of power series. We use the power series for the sine function (see sine function#Computation of power series):. $\! \sin x := x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots = \sum_{k=0}^\infty \frac{(-1)^kx^{2k+1}}{(2k + 1)!}$. Dividing both sides by $x$ (valid when $x \ne 0$), we get:. $\! \frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \dots = \sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k+1)!}$. We note that the power series also works at $x = 0$ (because $\operatorname{sinc} \ 0 = 1$), hence it works globally, and is the power series for the sinc function.
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https://mathematica.stackexchange.com/questions/185596/finding-the-best-representation-of-a-numerically-inverted-function-via-interpola?noredirect=1
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# Finding the best representation of a numerically-inverted function via InterpolatingPolynomial and/or variations
Below is the routine I am using to sort of represent the numerically-inverted function TP. Basically I am finding a necessary interpolating polynomial TPint that fits with the data points given by TPtab, so that I can represent TP as TPint and use in my other calculations. I just wonder, is this by far the best and proper way to find a polynomial that fits my data points to properly represent my numerically-inverted function? The output looks like a gross. Is there other routine wherein the output is compact and yet the accuracy of the function will not be compromised?
Thank you.
b = 1;
q = -1;
rat = 10^-30;
rho[r_,b_,q_]:=(2b/(1-q))(1-(b/r)^(1-q))^(1/2)Hypergeometric2F1[1/2,1-1/(q-1),3/2,1-(b/r)^(1-q)]
TP=Rationalize[InverseFunction[Function[{r,b,q},rho[r,b,q]],1,3]];
TPtab=Table[{ρ,TP[ρ,b,q]},{ρ,0,1,.1}]
TPint=Rationalize[InterpolatingPolynomial[TPtab,ρ],rat]//Simplify//Expand
Why not use NDSolveValue to compute an interpolating function of the inverse? Basically, given $$y = f(x)$$ we want to find $$x = f^{-1}(y)$$. To do this differentiate $$y = f(x(y))$$ with respect to $$y$$ to obtain $$1 = f'(x(y)) x'(y)$$
Hence, the ODE to be solved is: $$x'(y) = 1/f'(x(y))$$
and the following NDSolveValue call should produce your desired result:
if = NDSolveValue[
{x'[y] == 1/Derivative[1,0,0][rho][x[y], b, q], x[1] == Sqrt[2]},
x,
{y, 0, 1}
];
(where I used rho[Sqrt[2], b, q] == 1)
Visualization:
GraphicsRow[{Plot[if[ρ], {ρ, 0, 1}], Plot[TPint, {ρ, 0, 1}]}]
• That's fine also. But my problem is that further in my calculations, i am to use inverse function which is an interpolating function. So how would I able to execute algebraic calculations involving an interpolating function? That's basically my problem, that's why i need to find for an interpolating polynomial to best represent my inverse function. Commented Nov 9, 2018 at 0:26
• Notice that the output that TPint gives is a polynomial of certain degree. Is there other way to obtain a reasonable fit with corresponding polynomial to the inverse function? Commented Nov 9, 2018 at 0:36
• @user583893 What calculations are you doing that need a polynomial approximation to the inverse? Perhaps there's a way to modify those calculations to work with an interpolation function instead. Also, note that approximations using InterpolatingPolynomial can be very bad, see the Possible issues section in the documentation for InterpolatingPolynomial. Commented Nov 9, 2018 at 0:40
Interpolation over a regular grid has known problems. Over the Chebyshev extreme grid, however, there no such problems in interpolating smooth functions, or even continuous functions although convergence can be slow. A Chebyshev interpolation will be near minimax (best possible with respect to the infinity norm) for a smooth function. With machine-precision input, a near machine-precision approximation is possible. With higher precision input, higher precision approximations are possible.
I was introduced to this approach here and one can use adaptiveChebSeries to automatically determine the degree of the Chebyshev interpolation needed to approximate a function within a given error. For more, see Trefethen, Approximation Theory and Approximation Practice and Boyd, Solving Transcendental Equations. There are two methods of implementation, barycentric interpolation (see Berrut & Trefethen (2004) for its advantages) and Chebyshev series. Both methods are generally numerically better than an interpolation based on a power basis expansion. In this case, the difference is negligible.
ClearAll[rho, TP];
b = 1;
q = -1;
rat = 10^-30;
rho[r_, b_,
q_] := (2 b/(1 - q)) (1 - (b/r)^(1 - q))^(1/2) Hypergeometric2F1[1/2,
1 - 1/(q - 1), 3/2, 1 - (b/r)^(1 - q)];
TP[ρ_?NumericQ, b_?NumericQ, q_?NumericQ] :=
Block[{r}, With[{ρ0 = SetPrecision[ρ, 32]},
r /. FindRoot[rho[r, b, q] == ρ0, {r, 1001/1000},
WorkingPrecision -> 32] // Re]];
(* Chebyshev nodes and points *)
deg = 24;
chebnodes = N[Rescale[Sin[Pi/2 Range[-deg, deg, 2]/deg]], 32];
TPtab = Table[{ρ, TP[ρ, b, q]}, {ρ, chebnodes}];
(* Barycentric interpolation *)
rif = StatisticsLibraryBarycentricInterpolation[N@TPtab[[All, 1]],
N@TPtab[[All, 2]],
"Weights" ->
ReplacePart[
Table[(-1)^k, {k, 0, Length@TPtab - 1}], {1 -> 1/2, -1 -> 1/2}]];
Plot[rho[rif[ρ], b, q] - ρ // RealExponent, {ρ, 0, 1}]
(* Power basis interpolation *)
(* To diminish round-off error, increase precision *)
TPint = SetPrecision[InterpolatingPolynomial[TPtab, ρ], 50] // Expand // N
Plot[rho[N@TPint, b, q] - ρ // RealExponent, {ρ, 0, 1}]
(* Chebyshev series coefficients via FFT/DCT *)
cc = Sqrt[2/(Length@TPtab - 1)] FourierDCT[Reverse@TPtab[[All, 2]], 1];
cc[[{1, -1}]] /= 2;
rCS[ρ_] := cc.Cos[Range[0, Length@cc - 1] ArcCos[2 ρ - 1]];
Plot[rho[rCS[ρ], b, q] - ρ // RealExponent, {ρ, 0, 1}]
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# Finding the best representation of a numerically-inverted function via InterpolatingPolynomial and/or variations. Below is the routine I am using to sort of represent the numerically-inverted function TP. Basically I am finding a necessary interpolating polynomial TPint that fits with the data points given by TPtab, so that I can represent TP as TPint and use in my other calculations. I just wonder, is this by far the best and proper way to find a polynomial that fits my data points to properly represent my numerically-inverted function? The output looks like a gross. Is there other routine wherein the output is compact and yet the accuracy of the function will not be compromised?. Thank you.. b = 1;. q = -1;. rat = 10^-30;. rho[r_,b_,q_]:=(2b/(1-q))(1-(b/r)^(1-q))^(1/2)Hypergeometric2F1[1/2,1-1/(q-1),3/2,1-(b/r)^(1-q)]. TP=Rationalize[InverseFunction[Function[{r,b,q},rho[r,b,q]],1,3]];. TPtab=Table[{ρ,TP[ρ,b,q]},{ρ,0,1,.1}]. TPint=Rationalize[InterpolatingPolynomial[TPtab,ρ],rat]//Simplify//Expand. Why not use NDSolveValue to compute an interpolating function of the inverse? Basically, given $$y = f(x)$$ we want to find $$x = f^{-1}(y)$$. To do this differentiate $$y = f(x(y))$$ with respect to $$y$$ to obtain $$1 = f'(x(y)) x'(y)$$. Hence, the ODE to be solved is: $$x'(y) = 1/f'(x(y))$$. and the following NDSolveValue call should produce your desired result:. if = NDSolveValue[. {x'[y] == 1/Derivative[1,0,0][rho][x[y], b, q], x[1] == Sqrt[2]},. x,. {y, 0, 1}. ];. (where I used rho[Sqrt[2], b, q] == 1). Visualization:. GraphicsRow[{Plot[if[ρ], {ρ, 0, 1}], Plot[TPint, {ρ, 0, 1}]}]. • That's fine also. But my problem is that further in my calculations, i am to use inverse function which is an interpolating function. So how would I able to execute algebraic calculations involving an interpolating function? That's basically my problem, that's why i need to find for an interpolating polynomial to best represent my inverse function. Commented Nov 9, 2018 at 0:26. • Notice that the output that TPint gives is a polynomial of certain degree. Is there other way to obtain a reasonable fit with corresponding polynomial to the inverse function? Commented Nov 9, 2018 at 0:36. • @user583893 What calculations are you doing that need a polynomial approximation to the inverse? Perhaps there's a way to modify those calculations to work with an interpolation function instead. Also, note that approximations using InterpolatingPolynomial can be very bad, see the Possible issues section in the documentation for InterpolatingPolynomial. Commented Nov 9, 2018 at 0:40. Interpolation over a regular grid has known problems. Over the Chebyshev extreme grid, however, there no such problems in interpolating smooth functions, or even continuous functions although convergence can be slow. A Chebyshev interpolation will be near minimax (best possible with respect to the infinity norm) for a smooth function. With machine-precision input, a near machine-precision approximation is possible. With higher precision input, higher precision approximations are possible.
|
I was introduced to this approach here and one can use adaptiveChebSeries to automatically determine the degree of the Chebyshev interpolation needed to approximate a function within a given error. For more, see Trefethen, Approximation Theory and Approximation Practice and Boyd, Solving Transcendental Equations. There are two methods of implementation, barycentric interpolation (see Berrut & Trefethen (2004) for its advantages) and Chebyshev series. Both methods are generally numerically better than an interpolation based on a power basis expansion. In this case, the difference is negligible.. ClearAll[rho, TP];. b = 1;. q = -1;. rat = 10^-30;. rho[r_, b_,. q_] := (2 b/(1 - q)) (1 - (b/r)^(1 - q))^(1/2) Hypergeometric2F1[1/2,. 1 - 1/(q - 1), 3/2, 1 - (b/r)^(1 - q)];. TP[ρ_?NumericQ, b_?NumericQ, q_?NumericQ] :=. Block[{r}, With[{ρ0 = SetPrecision[ρ, 32]},. r /. FindRoot[rho[r, b, q] == ρ0, {r, 1001/1000},. WorkingPrecision -> 32] // Re]];. (* Chebyshev nodes and points *). deg = 24;. chebnodes = N[Rescale[Sin[Pi/2 Range[-deg, deg, 2]/deg]], 32];. TPtab = Table[{ρ, TP[ρ, b, q]}, {ρ, chebnodes}];. (* Barycentric interpolation *). rif = StatisticsLibraryBarycentricInterpolation[N@TPtab[[All, 1]],. N@TPtab[[All, 2]],. "Weights" ->. ReplacePart[. Table[(-1)^k, {k, 0, Length@TPtab - 1}], {1 -> 1/2, -1 -> 1/2}]];. Plot[rho[rif[ρ], b, q] - ρ // RealExponent, {ρ, 0, 1}]. (* Power basis interpolation *). (* To diminish round-off error, increase precision *). TPint = SetPrecision[InterpolatingPolynomial[TPtab, ρ], 50] // Expand // N. Plot[rho[N@TPint, b, q] - ρ // RealExponent, {ρ, 0, 1}]. (* Chebyshev series coefficients via FFT/DCT *). cc = Sqrt[2/(Length@TPtab - 1)] FourierDCT[Reverse@TPtab[[All, 2]], 1];. cc[[{1, -1}]] /= 2;. rCS[ρ_] := cc.Cos[Range[0, Length@cc - 1] ArcCos[2 ρ - 1]];. Plot[rho[rCS[ρ], b, q] - ρ // RealExponent, {ρ, 0, 1}].
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https://doubtnut.com/question-answer/let-a1-a2-a3-an-1an-be-an-ap-statement-1-a1-a2-a3-ann-2a1-an-statement-2-ak-an-k-1a1-an-for-k123-n-53796475
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or
# Let a_(1)+a_(2)+a_(3), . . . ,a_(n-1),a_(n) be an A.P. <br> Statement -1: a_(1)+a_(2)+a_(3)+ . . . +a_(n)=(n)/(2)(a_(1)+a_(n)) <br> Statement -2 a_(k)+a_(n-k+1)=a_(1)+a_(n)" for "k=1,2,3, . . . , n
Question from Class 11 Chapter Sequences And Series
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Statement -1 is true, Statement -2 is True, Statement -2 is a correct explanation for Statement for Statement -1.Statement -1 is true, Statement -2 is True, Statement -2 is not a correct explanation for Statement for Statement -1.Statement -1 is true, Statement -2 is False.Statement -1 is False, Statement -2 is True.
A
Solution :
Let `S_(n)=a_(1)+a_(2)+a_(3)+ . . . +a_(n-1)+a_(n)`. Then, <br> `S_(n)=a_(n)+a_(n-1)+a_(n-2)+ . . . +a_(2)+a_(1)` <br> `rArr" "2S_(n)=n(a_(1)+a_(n))" [Using statement-2]"` <br> `rArr" "S_(n)=(n)/(2)(a_(1)+a_(2))` <br> So, statement -2 correct explanation for statement -1.
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or. # Let a_(1)+a_(2)+a_(3), . . . ,a_(n-1),a_(n) be an A.P. <br> Statement -1: a_(1)+a_(2)+a_(3)+ . . . +a_(n)=(n)/(2)(a_(1)+a_(n)) <br> Statement -2 a_(k)+a_(n-k+1)=a_(1)+a_(n)" for "k=1,2,3, . . . , n. Question from Class 11 Chapter Sequences And Series. Apne doubts clear karein ab Whatsapp par bhi. Try it now.. Watch 1000+ concepts & tricky questions explained!. 300+ views | 23.6 K+ people like this. Share. Share. Statement -1 is true, Statement -2 is True, Statement -2 is a correct explanation for Statement for Statement -1.Statement -1 is true, Statement -2 is True, Statement -2 is not a correct explanation for Statement for Statement -1.Statement -1 is true, Statement -2 is False.Statement -1 is False, Statement -2 is True.. A. Solution :. Let `S_(n)=a_(1)+a_(2)+a_(3)+ . . . +a_(n-1)+a_(n)`. Then, <br> `S_(n)=a_(n)+a_(n-1)+a_(n-2)+ . . . +a_(2)+a_(1)` <br> `rArr" "2S_(n)=n(a_(1)+a_(n))" [Using statement-2]"` <br> `rArr" "S_(n)=(n)/(2)(a_(1)+a_(2))` <br> So, statement -2 correct explanation for statement -1.. Related Video.
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# How to Solve Trig Ratios of General Angles? (+FREE Worksheet!)
Learn trigonometric ratios of general angles and how to solve math problems related to trig ratios by the following step-by-step guide.
## Step by step guide to solve Trig Ratios of General Angles
• Learn common trigonometric functions:
### Trig Ratios of General Angles – Example 1:
Find the trigonometric function: $$cos$$ $$120^\circ$$
Solution:
$$cos$$ $$120^{\circ}$$
Use the following property: $$cos$$$$(x)=$$ $$sin$$$$(90^{\circ}-x)$$
$$cos$$ $$120^{\circ} =$$ $$sin$$ $$( 90^{\circ} -120^{\circ})=$$ $$sin ( -30^{\circ})$$
Now use the following property: $$sin (-x)$$$$=- sin (x)$$
Then: $$sin ( -30^{\circ})=-sin (30^{\circ}$$)$$=-\frac{1}{2 }$$
### Trig Ratios of General Angles – Example 2:
Find the trigonometric function: $$sin$$ $$135^\circ$$
Solution:
Use the following property: $$sin$$$$(x)=$$ $$cos$$$$(90^\circ-x)$$
$$sin$$ $$135^\circ=$$ $$cos$$$$(90^\circ-135^\circ)=$$ $$cos$$$$(-45^\circ)$$
Now use the following property: $$cos$$$$(-x)=cos x$$
Then: $$cos$$$$(-45^\circ)=$$ $$cos$$$$(45^\circ)=\frac{\sqrt{2}}{2 }$$
### Trig Ratios of General Angles – Example 3:
Find the trigonometric function: $$sin$$ $$-120^\circ$$
Solution:
Use the following property: $$sin$$$$(-x)=-$$ $$sin$$$$(x)$$
$$sin$$$$-120^\circ=-$$ $$sin$$ $$120^\circ$$ , $$sin$$$$120^\circ=\frac{\sqrt{3}}{2}$$
Then: $$sin$$$$-120^\circ=-\frac{\sqrt{3}}{2}$$
### Trig Ratios of General Angles – Example 4:
Find the trigonometric function: $$cos$$ $$150^\circ$$
Solution:
$$cos$$ $$150^{\circ}$$
Use the following property: $$cos$$$$(x)=$$ $$sin$$$$(90^{\circ}-x)$$
$$cos$$ $$150^{\circ} =$$ $$sin$$ $$( 90^{\circ} -150^{\circ})=$$ $$sin ( -60^{\circ})$$
Now use the following property: $$sin (-x)$$$$=- sin (x)$$
Then: $$sin ( -60^{\circ})=-sin (60^{\circ}$$)$$= -\frac{\sqrt{3}}{2}$$
## Exercises
### Use a calculator to find each. Round your answers to the nearest ten–thousandth.
• $$\color{blue}{sin \ – 120^\circ}$$
• $$\color{blue}{sin \ 150^\circ}$$
• $$\color{blue}{cos \ 315^\circ}$$
• $$\color{blue}{cos \ 180^\circ}$$
• $$\color{blue}{sin \ 120^\circ}$$
• $$\color{blue}{sin \ – 330^\circ }$$
### Download Trig Ratios of General Angles Worksheet
• $$\color{blue}{-\frac{\sqrt{3}}{2}}$$
• $$\color{blue}{\frac{1}{2}}$$
• $$\color{blue}{\frac{\sqrt{2}}{2}}$$
• $$\color{blue}{-1}$$
• $$\color{blue}{\frac{\sqrt{3}}{2}}$$
• $$\color{blue}{\frac{1}{2}}$$
.
### What people say about "How to Solve Trig Ratios of General Angles? (+FREE Worksheet!)"?
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# How to Solve Trig Ratios of General Angles? (+FREE Worksheet!). Learn trigonometric ratios of general angles and how to solve math problems related to trig ratios by the following step-by-step guide.. ## Step by step guide to solve Trig Ratios of General Angles. • Learn common trigonometric functions:. ### Trig Ratios of General Angles – Example 1:. Find the trigonometric function: $$cos$$ $$120^\circ$$. Solution:. $$cos$$ $$120^{\circ}$$. Use the following property: $$cos$$$$(x)=$$ $$sin$$$$(90^{\circ}-x)$$. $$cos$$ $$120^{\circ} =$$ $$sin$$ $$( 90^{\circ} -120^{\circ})=$$ $$sin ( -30^{\circ})$$. Now use the following property: $$sin (-x)$$$$=- sin (x)$$. Then: $$sin ( -30^{\circ})=-sin (30^{\circ}$$)$$=-\frac{1}{2 }$$. ### Trig Ratios of General Angles – Example 2:. Find the trigonometric function: $$sin$$ $$135^\circ$$. Solution:. Use the following property: $$sin$$$$(x)=$$ $$cos$$$$(90^\circ-x)$$. $$sin$$ $$135^\circ=$$ $$cos$$$$(90^\circ-135^\circ)=$$ $$cos$$$$(-45^\circ)$$. Now use the following property: $$cos$$$$(-x)=cos x$$. Then: $$cos$$$$(-45^\circ)=$$ $$cos$$$$(45^\circ)=\frac{\sqrt{2}}{2 }$$. ### Trig Ratios of General Angles – Example 3:. Find the trigonometric function: $$sin$$ $$-120^\circ$$. Solution:. Use the following property: $$sin$$$$(-x)=-$$ $$sin$$$$(x)$$. $$sin$$$$-120^\circ=-$$ $$sin$$ $$120^\circ$$ , $$sin$$$$120^\circ=\frac{\sqrt{3}}{2}$$. Then: $$sin$$$$-120^\circ=-\frac{\sqrt{3}}{2}$$. ### Trig Ratios of General Angles – Example 4:. Find the trigonometric function: $$cos$$ $$150^\circ$$. Solution:. $$cos$$ $$150^{\circ}$$.
|
Use the following property: $$cos$$$$(x)=$$ $$sin$$$$(90^{\circ}-x)$$. $$cos$$ $$150^{\circ} =$$ $$sin$$ $$( 90^{\circ} -150^{\circ})=$$ $$sin ( -60^{\circ})$$. Now use the following property: $$sin (-x)$$$$=- sin (x)$$. Then: $$sin ( -60^{\circ})=-sin (60^{\circ}$$)$$= -\frac{\sqrt{3}}{2}$$. ## Exercises. ### Use a calculator to find each. Round your answers to the nearest ten–thousandth.. • $$\color{blue}{sin \ – 120^\circ}$$. • $$\color{blue}{sin \ 150^\circ}$$. • $$\color{blue}{cos \ 315^\circ}$$. • $$\color{blue}{cos \ 180^\circ}$$. • $$\color{blue}{sin \ 120^\circ}$$. • $$\color{blue}{sin \ – 330^\circ }$$. ### Download Trig Ratios of General Angles Worksheet. • $$\color{blue}{-\frac{\sqrt{3}}{2}}$$. • $$\color{blue}{\frac{1}{2}}$$. • $$\color{blue}{\frac{\sqrt{2}}{2}}$$. • $$\color{blue}{-1}$$. • $$\color{blue}{\frac{\sqrt{3}}{2}}$$. • $$\color{blue}{\frac{1}{2}}$$. .. ### What people say about "How to Solve Trig Ratios of General Angles? (+FREE Worksheet!)"?. No one replied yet.. X. 30% OFF. Limited time only!. Save Over 30%. SAVE $5 It was$16.99 now it is \$11.99.
|
http://www.numbersaplenty.com/50358
| 1,597,276,783,000,000,000
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text/html
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crawl-data/CC-MAIN-2020-34/segments/1596439738950.31/warc/CC-MAIN-20200812225607-20200813015607-00400.warc.gz
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Search a number
50358 = 23711109
BaseRepresentation
bin1100010010110110
32120002010
430102312
53102413
61025050
7266550
oct142266
976063
1050358
1134920
1225186
1319bc9
14144d0
15edc3
hexc4b6
50358 has 32 divisors (see below), whose sum is σ = 126720. Its totient is φ = 12960.
The previous prime is 50341. The next prime is 50359. The reversal of 50358 is 85305.
It can be divided in two parts, 503 and 58, that added together give a triangular number (561 = T33).
50358 is digitally balanced in base 2 and base 4, because in such bases it contains all the possibile digits an equal number of times.
It is a Harshad number since it is a multiple of its sum of digits (21).
It is a nialpdrome in base 15.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (50359) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 408 + ... + 516.
It is an arithmetic number, because the mean of its divisors is an integer number (3960).
250358 is an apocalyptic number.
It is a practical number, because each smaller number is the sum of distinct divisors of 50358, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (63360).
50358 is an abundant number, since it is smaller than the sum of its proper divisors (76362).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
50358 is a wasteful number, since it uses less digits than its factorization.
50358 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 132.
The product of its (nonzero) digits is 600, while the sum is 21.
The square root of 50358 is about 224.4058822758. The cubic root of 50358 is about 36.9280315200.
The spelling of 50358 in words is "fifty thousand, three hundred fifty-eight".
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Search a number. 50358 = 23711109. BaseRepresentation. bin1100010010110110. 32120002010. 430102312. 53102413. 61025050. 7266550. oct142266. 976063. 1050358. 1134920. 1225186. 1319bc9. 14144d0. 15edc3. hexc4b6. 50358 has 32 divisors (see below), whose sum is σ = 126720. Its totient is φ = 12960.. The previous prime is 50341. The next prime is 50359.
|
The reversal of 50358 is 85305.. It can be divided in two parts, 503 and 58, that added together give a triangular number (561 = T33).. 50358 is digitally balanced in base 2 and base 4, because in such bases it contains all the possibile digits an equal number of times.. It is a Harshad number since it is a multiple of its sum of digits (21).. It is a nialpdrome in base 15.. It is a congruent number.. It is not an unprimeable number, because it can be changed into a prime (50359) by changing a digit.. It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 408 + ... + 516.. It is an arithmetic number, because the mean of its divisors is an integer number (3960).. 250358 is an apocalyptic number.. It is a practical number, because each smaller number is the sum of distinct divisors of 50358, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (63360).. 50358 is an abundant number, since it is smaller than the sum of its proper divisors (76362).. It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.. 50358 is a wasteful number, since it uses less digits than its factorization.. 50358 is an evil number, because the sum of its binary digits is even.. The sum of its prime factors is 132.. The product of its (nonzero) digits is 600, while the sum is 21.. The square root of 50358 is about 224.4058822758. The cubic root of 50358 is about 36.9280315200.. The spelling of 50358 in words is "fifty thousand, three hundred fifty-eight".
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https://math.answers.com/Q/What_is_42_out_of_50_in_percent
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0
# What is 42 out of 50 in percent?
Updated: 9/23/2023
Wiki User
8y ago
84%
Wiki User
8y ago
Earn +20 pts
Q: What is 42 out of 50 in percent?
Submit
Still have questions?
Related questions
### What is 50 percent of 84?
50 percent of 84 = .50 x 84 = 42
### What number is 42 percent of 50?
What is 42% of 50:= 42% * 50= 0.42 * 50= 21
### What is 42 percent percent written as a fraction?
42 percent written as a fraction is 42/100, which simplifies to 21/50.
### What whole number is 50 42 percent of?
50 is 42% of= 50 / 42= 50 / 0.42= 119
### What is 42 out of 50 as a percent?
42 out of 50 can be expressed as the fraction 42/50. 42 divided by 50 = 0.84To convert 0.84 to percent multiply by 100: 0.84 × 100 = 84 %
50% of 84 is 42
### What is the percentage of 50 percent of 42?
If you mean 50% of 42 then it is 21
### What is the fraction for 42 percent?
42% := 42/100= 21/50
42% = 21/50
42% = 21/50
It is 42%
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0. # What is 42 out of 50 in percent?. Updated: 9/23/2023. Wiki User. 8y ago. 84%. Wiki User. 8y ago. Earn +20 pts. Q: What is 42 out of 50 in percent?. Submit. Still have questions?. Related questions. ### What is 50 percent of 84?. 50 percent of 84 = .50 x 84 = 42. ### What number is 42 percent of 50?. What is 42% of 50:= 42% * 50= 0.42 * 50= 21.
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### What is 42 percent percent written as a fraction?. 42 percent written as a fraction is 42/100, which simplifies to 21/50.. ### What whole number is 50 42 percent of?. 50 is 42% of= 50 / 42= 50 / 0.42= 119. ### What is 42 out of 50 as a percent?. 42 out of 50 can be expressed as the fraction 42/50. 42 divided by 50 = 0.84To convert 0.84 to percent multiply by 100: 0.84 × 100 = 84 %. 50% of 84 is 42. ### What is the percentage of 50 percent of 42?. If you mean 50% of 42 then it is 21. ### What is the fraction for 42 percent?. 42% := 42/100= 21/50. 42% = 21/50. 42% = 21/50. It is 42%.
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https://math.answers.com/questions/If_f_of_x_equals_2x_squared_what_is_the_positive_value_of_x_for_which_f_of_x_equals_136_With_the_explanation_please_Thanks
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0
If f of x equals 2x squared what is the positive value of x for which f of x equals 136 With the explanation please Thanks?
Wiki User
2015-04-25 12:04:58
f(x) = 2x²
f(x) = 136
→ 2x² = 136
→ x² = 68
→ x = ±√68
As the positive value is required, x = √68 ≈ 8.246
Wiki User
2015-04-25 12:04:58
Study guides
20 cards
A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
➡️
See all cards
3.74
1190 Reviews
Wiki User
2015-01-29 11:34:48
f(x) = 2*x^2 = 136that is x^2 = 136/2 = 68
and so x = + or - sqrt(68).
But x is said to be the positive value so x = +sqrt(68).
Earn +20 pts
Q: If f of x equals 2x squared what is the positive value of x for which f of x equals 136 With the explanation please Thanks?
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0. If f of x equals 2x squared what is the positive value of x for which f of x equals 136 With the explanation please Thanks?. Wiki User. 2015-04-25 12:04:58. f(x) = 2x². f(x) = 136. → 2x² = 136. → x² = 68. → x = ±√68. As the positive value is required, x = √68 ≈ 8.246. Wiki User. 2015-04-25 12:04:58. Study guides. 20 cards.
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A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials. ➡️. See all cards. 3.74. 1190 Reviews. Wiki User. 2015-01-29 11:34:48. f(x) = 2*x^2 = 136that is x^2 = 136/2 = 68. and so x = + or - sqrt(68).. But x is said to be the positive value so x = +sqrt(68).. Earn +20 pts. Q: If f of x equals 2x squared what is the positive value of x for which f of x equals 136 With the explanation please Thanks?.
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Associated Topics || Dr. Math Home || Search Dr. Math
### An Ordinary Differential Equation in Everyday Life?
```Date: 03/26/2011 at 05:33:53
From: Hassan
Subject: ODE's
I just want to ask about ordinary differential equations. Where do these
find practical applications?
```
```
Date: 03/26/2011 at 20:45:22
From: Doctor Jordan
Subject: Re: ODE's
Hi Hassan,
Let's say I have a small cube of metal of mass m sitting on a table. Say
the cube of metal is attached to a spring in such a way that the cube will
move back and forth on the table.
Suppose we want to know the position of this cube of metal at time t. How
can we obtain this?
This is the kind of problem that call for ordinary differential equations.
Methods for solving ODE let us write down an equation using data
that we have, and then the solution to the ODE will give us the
information that we want -- in this case, the position of the cube at
time t.
To get the differential equation for this mechanical system, we need to
know the forces acting on the object at any time -- namely, the restoring
force of the spring, which is approximately -kx, where x is the distance
from the rest position of the cube and k is an experimentally determined
number that depends on the spring's material composition and
manufacturing; and the damping force, which is approximately cx', where c
is called the viscous damping coefficient. Here, x' is the derivative of x
with respect to t.
Then by Newton's second law here, the force cx' - kx is equal to mx'',
where x'' is the second derivative of x with respect to t -- in other
words, the acceleration of the metal cube. So here the differential
equation is
mx'' = cx' - kx
It is called a differential equation because it involves derivatives. It
says how the system is moving at each point in time.
We have mathematical techniques for solving ordinary differential
equations. We apply these techniques to this equation and we get a
solution, x(t). But what is x(t)? x(t) is the position of the metal cube
at time t. Therefore, solving this differential equation gives us a
function which tells us where the metal cube is at any point in time.
We started knowing only the forces acting on the cube at each point in
time and put this together into a differential equation. By applying
mathematical techniques, we ended up with a function that told us the
exact arrangement of our system at each point in time.
- Doctor Jordan, The Math Forum
http://mathforum.org/dr.math/
```
```
Date: 03/27/2011 at 00:27:38
From: Hassan
Subject: Thank you (ODE's)
Nice job; very grateful.
May Allah bless you more.
Jazak Allah.
```
Associated Topics:
High School Calculus
Search the Dr. Math Library:
Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words
Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search
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Associated Topics || Dr. Math Home || Search Dr. Math. ### An Ordinary Differential Equation in Everyday Life?. ```Date: 03/26/2011 at 05:33:53. From: Hassan. Subject: ODE's. I just want to ask about ordinary differential equations. Where do these. find practical applications?. ```. ```. Date: 03/26/2011 at 20:45:22. From: Doctor Jordan. Subject: Re: ODE's. Hi Hassan,. Let's say I have a small cube of metal of mass m sitting on a table. Say. the cube of metal is attached to a spring in such a way that the cube will. move back and forth on the table.. Suppose we want to know the position of this cube of metal at time t. How. can we obtain this?. This is the kind of problem that call for ordinary differential equations.. Methods for solving ODE let us write down an equation using data. that we have, and then the solution to the ODE will give us the. information that we want -- in this case, the position of the cube at. time t.. To get the differential equation for this mechanical system, we need to. know the forces acting on the object at any time -- namely, the restoring. force of the spring, which is approximately -kx, where x is the distance. from the rest position of the cube and k is an experimentally determined. number that depends on the spring's material composition and. manufacturing; and the damping force, which is approximately cx', where c. is called the viscous damping coefficient. Here, x' is the derivative of x. with respect to t.. Then by Newton's second law here, the force cx' - kx is equal to mx'',. where x'' is the second derivative of x with respect to t -- in other.
|
words, the acceleration of the metal cube. So here the differential. equation is. mx'' = cx' - kx. It is called a differential equation because it involves derivatives. It. says how the system is moving at each point in time.. We have mathematical techniques for solving ordinary differential. equations. We apply these techniques to this equation and we get a. solution, x(t). But what is x(t)? x(t) is the position of the metal cube. at time t. Therefore, solving this differential equation gives us a. function which tells us where the metal cube is at any point in time.. We started knowing only the forces acting on the cube at each point in. time and put this together into a differential equation. By applying. mathematical techniques, we ended up with a function that told us the. exact arrangement of our system at each point in time.. - Doctor Jordan, The Math Forum. http://mathforum.org/dr.math/. ```. ```. Date: 03/27/2011 at 00:27:38. From: Hassan. Subject: Thank you (ODE's). Nice job; very grateful.. May Allah bless you more.. Jazak Allah.. ```. Associated Topics:. High School Calculus. Search the Dr. Math Library:. Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words. Submit your own question to Dr. Math. Math Forum Home || Math Library || Quick Reference || Math Forum Search.
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It is currently Tue Oct 16, 2018 5:58 pm
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Post subject: GMAT Geometry (Data Sufficiency)Posted: Fri Aug 20, 2010 4:09 am
Joined: Sun May 30, 2010 3:15 am
Posts: 424
In the figure above, three segments are drawn to connect opposite vertices of a hexagon, forming six triangles. All three of these segments intersect at point A. What is the area of the hexagon?
(1) One of the triangles has an area of 12.
(2) All sides of the hexagon are of equal length.
A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
D. Either statement BY ITSELF is sufficient to answer the question.
E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.
(E) Statement (1) is not sufficient because we do not know whether the hexagon is a regular hexagon (meaning that all 6 sides
have the same length). We could only calculate the area if the hexagon could be divided into six congruent triangles.
Statement (2) is also insufficient because we are not given any numbers. So, it is impossible to calculate the area. Combined, the two statements are still insufficient. A hexagon could have six equal sides, yet not be a regular hexagon. Imagine pushing opposite sides of a regular hexagon closer to each other: the
hexagon would flatten, while all the sides would remain the same length. One of the triangles within this flattened hexagon could have an area of 12, but not all of the triangles would have the same area. Therefore, we can't calculate the area, even by combining the statements, and the correct answer is choice (E).
-------------
In my opinion the the answer should be (A).
I will attempt to give a detailed explanation below:
Since it is a Hexagon, sum of internal angles is 720 using the formula: (n – 2) × 180, where 'n' is the number of sides of the polygon.
Therefore each internal angle is 120 degree. From this I say each triangle formed inside is an equilateral triangle. From (1), the area is 12 for each triangle. This is SUFFICIENT to conclude the hexagon having six triangles having an area of 12 each, has an area of
72.
Please confirm
.
Attachments: hexagon1.gif [3.85 KiB] Not downloaded yet
Top
Post subject: Re: math (test 1, question 25): geometry, data sufficiencyPosted: Fri Aug 20, 2010 4:48 am
Joined: Sun May 30, 2010 2:23 am
Posts: 498
When you solve any data sufficiency geometry question you should avoid graphics distorting your reasoning.
In this case the hexagon seems to be regular (all sides and angles are equal) and it affects the reasoning. Therefore you may conclude that all angles are equal (120⁰) but we have no facts that this conclusion can be based on.
The question statement gives us just the fact that we have a hexagon. It is just some arbitrary hexagon and can look in many different ways:
or
and many other different shapes.
That gives us a clear idea why statement (1) is insufficient.
Statement (2) also doesn't imply that hexagon is regular (regular hexagon must have all equal sides AND angles). If we know that all sides are equal then hexagon can still have diverse shapes:
or
and many other variants.
So be careful with geometry graphics and always check if what you see is based on facts.
Attachments: hexagon5.gif [4.27 KiB] Not downloaded yet hexagon4.gif [3.52 KiB] Not downloaded yet hexagon3.gif [4.3 KiB] Not downloaded yet hexagon2.gif [3.82 KiB] Not downloaded yet
Top
Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 7:59 am
Joined: Sun May 30, 2010 3:15 am
Posts: 424
Hi,
As per my understanding, every polygon with equal sides is always cyclic in nature.
Could you, please, correct if I am wrong.
Best regards.
Top
Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 7:59 am
Joined: Sun May 30, 2010 2:23 am
Posts: 498
Quote:
As per my understanding, every polygon with equal sides is always cyclic in nature.
A polygon with equal sides is cyclic if all its angles are equal. In other words such polygon is a regular polygon (all sides are equal and all angles are equal).
Here is an example of a polygon with equal sides but not with equal angles. It is not cyclic.
Top
Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 8:00 am
Joined: Sun May 30, 2010 3:15 am
Posts: 424
On the gmat, will it specifically state whether the hexagon is a regular
hexagon or not? I assumed that when you mentioned hexagon it was a regular
hexagon.
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Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 8:01 am
Joined: Sun May 30, 2010 2:23 am
Posts: 498
With the problem at hand you don't need to know if the hexagon is regular or not. All of the necessary information is provided. However, on the GMAT when you actually encounter a problem that involves a geometric figure that is regular, and, it is required that you know the latter in order to solve the problem, then yes, the GMAT question will provide you with that information.
Take care.
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Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 8:01 am
Joined: Sun May 30, 2010 3:15 am
Posts: 424
I understand the question, but I think "three segments connecting opposite vertices of a hexagon intersect at a point" means that the hexagon is a regular hexagon!
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Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 8:02 am
Joined: Sun May 30, 2010 2:23 am
Posts: 498
questioner wrote:
I understand the question, but I think "three segments connecting opposite vertices of a hexagon intersect at a point" means that the hexagon is a regular hexagon!
No. You can see that if you draw various cases of 3 arbitrary segments that intersect in one point, then you just need to connect the ends of the segments to create various hexagons.
Attachments: arbitrary_hexagon.gif [4.44 KiB] Not downloaded yet
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Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Thu Jun 20, 2013 4:03 pm
Joined: Sun May 30, 2010 3:15 am
Posts: 424
If all sides are equal, its a regular hexagon. An area of a regular hexagon can be considered 6 × area of the equilateral triangles. Thus using both the statements we could arrive at the solution . Hence option C is correct.
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Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Thu Jun 20, 2013 4:08 pm
Joined: Sun May 30, 2010 2:23 am
Posts: 498
questioner wrote:
If all sides are equal, its a regular hexagon.
A hexagon is regular if all sides are equal AND all interior angles are equal.
Think of 6 equal sticks connected in form of a hexagon and imagine moving those sticks. Here is an examples of a non-regular hexagon, which sides are all equal:
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It is currently Tue Oct 16, 2018 5:58 pm. All times are UTC - 5 hours [ DST ]. Page 1 of 1 [ 10 posts ]. Print view Previous topic | Next topic. Author Message. Post subject: GMAT Geometry (Data Sufficiency)Posted: Fri Aug 20, 2010 4:09 am. Joined: Sun May 30, 2010 3:15 am. Posts: 424. In the figure above, three segments are drawn to connect opposite vertices of a hexagon, forming six triangles. All three of these segments intersect at point A. What is the area of the hexagon?. (1) One of the triangles has an area of 12.. (2) All sides of the hexagon are of equal length.. A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.. B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.. C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.. D. Either statement BY ITSELF is sufficient to answer the question.. E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.. (E) Statement (1) is not sufficient because we do not know whether the hexagon is a regular hexagon (meaning that all 6 sides. have the same length). We could only calculate the area if the hexagon could be divided into six congruent triangles.. Statement (2) is also insufficient because we are not given any numbers. So, it is impossible to calculate the area. Combined, the two statements are still insufficient. A hexagon could have six equal sides, yet not be a regular hexagon. Imagine pushing opposite sides of a regular hexagon closer to each other: the. hexagon would flatten, while all the sides would remain the same length. One of the triangles within this flattened hexagon could have an area of 12, but not all of the triangles would have the same area. Therefore, we can't calculate the area, even by combining the statements, and the correct answer is choice (E).. -------------. In my opinion the the answer should be (A).. I will attempt to give a detailed explanation below:. Since it is a Hexagon, sum of internal angles is 720 using the formula: (n – 2) × 180, where 'n' is the number of sides of the polygon.. Therefore each internal angle is 120 degree. From this I say each triangle formed inside is an equilateral triangle. From (1), the area is 12 for each triangle. This is SUFFICIENT to conclude the hexagon having six triangles having an area of 12 each, has an area of. 72.. Please confirm. .. Attachments: hexagon1.gif [3.85 KiB] Not downloaded yet. Top. Post subject: Re: math (test 1, question 25): geometry, data sufficiencyPosted: Fri Aug 20, 2010 4:48 am. Joined: Sun May 30, 2010 2:23 am. Posts: 498. When you solve any data sufficiency geometry question you should avoid graphics distorting your reasoning.. In this case the hexagon seems to be regular (all sides and angles are equal) and it affects the reasoning. Therefore you may conclude that all angles are equal (120⁰) but we have no facts that this conclusion can be based on.. The question statement gives us just the fact that we have a hexagon. It is just some arbitrary hexagon and can look in many different ways:. or. and many other different shapes.. That gives us a clear idea why statement (1) is insufficient.. Statement (2) also doesn't imply that hexagon is regular (regular hexagon must have all equal sides AND angles). If we know that all sides are equal then hexagon can still have diverse shapes:. or. and many other variants.. So be careful with geometry graphics and always check if what you see is based on facts.. Attachments: hexagon5.gif [4.27 KiB] Not downloaded yet hexagon4.gif [3.52 KiB] Not downloaded yet hexagon3.gif [4.3 KiB] Not downloaded yet hexagon2.gif [3.82 KiB] Not downloaded yet. Top. Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 7:59 am. Joined: Sun May 30, 2010 3:15 am. Posts: 424. Hi,. As per my understanding, every polygon with equal sides is always cyclic in nature.
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Could you, please, correct if I am wrong.. Best regards.. Top. Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 7:59 am. Joined: Sun May 30, 2010 2:23 am. Posts: 498. Quote:. As per my understanding, every polygon with equal sides is always cyclic in nature.. A polygon with equal sides is cyclic if all its angles are equal. In other words such polygon is a regular polygon (all sides are equal and all angles are equal).. Here is an example of a polygon with equal sides but not with equal angles. It is not cyclic.. Top. Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 8:00 am. Joined: Sun May 30, 2010 3:15 am. Posts: 424. On the gmat, will it specifically state whether the hexagon is a regular. hexagon or not? I assumed that when you mentioned hexagon it was a regular. hexagon.. Top. Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 8:01 am. Joined: Sun May 30, 2010 2:23 am. Posts: 498. With the problem at hand you don't need to know if the hexagon is regular or not. All of the necessary information is provided. However, on the GMAT when you actually encounter a problem that involves a geometric figure that is regular, and, it is required that you know the latter in order to solve the problem, then yes, the GMAT question will provide you with that information.. Take care.. Top. Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 8:01 am. Joined: Sun May 30, 2010 3:15 am. Posts: 424. I understand the question, but I think "three segments connecting opposite vertices of a hexagon intersect at a point" means that the hexagon is a regular hexagon!. Top. Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 8:02 am. Joined: Sun May 30, 2010 2:23 am. Posts: 498. questioner wrote:. I understand the question, but I think "three segments connecting opposite vertices of a hexagon intersect at a point" means that the hexagon is a regular hexagon!. No. You can see that if you draw various cases of 3 arbitrary segments that intersect in one point, then you just need to connect the ends of the segments to create various hexagons.. Attachments: arbitrary_hexagon.gif [4.44 KiB] Not downloaded yet. Top. Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Thu Jun 20, 2013 4:03 pm. Joined: Sun May 30, 2010 3:15 am. Posts: 424. If all sides are equal, its a regular hexagon. An area of a regular hexagon can be considered 6 × area of the equilateral triangles. Thus using both the statements we could arrive at the solution . Hence option C is correct.. Top. Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Thu Jun 20, 2013 4:08 pm. Joined: Sun May 30, 2010 2:23 am. Posts: 498. questioner wrote:. If all sides are equal, its a regular hexagon.. A hexagon is regular if all sides are equal AND all interior angles are equal.. Think of 6 equal sticks connected in form of a hexagon and imagine moving those sticks. Here is an examples of a non-regular hexagon, which sides are all equal:. Top. Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending. Page 1 of 1 [ 10 posts ]. All times are UTC - 5 hours [ DST ]. #### Who is online. Users browsing this forum: No registered users and 4 guests. You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot post attachments in this forum. Search for:. Jump to: Select a forum ------------------ GMAT GMAT: Quantitative Section (Math) GMAT: Verbal Section GMAT: Integrated Reasoning GMAT: General Questions GRE GRE: Quantitative Reasoning (Math) GRE: Verbal Reasoning GRE: General Questions General questions Other questions. Powered by phpBB © 2000, 2002, 2005, 2007 phpBB Group.
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https://opencourser.com/course/h1dxgd/khanacademy-introduction-to-polynomials
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# Polynomials
This course is a part of Algebra I, a 17-course Topic series from Khan Academy.
Learn how to add, subtract, and multiply polynomial expressions. For example, write (2x+3)(x-1) as 2x²+x-3.
This course contains 9 segments:
Intro to polynomials
Learn about polynomial expressions: What are they? How are they constructed? What can we do with them?
Learn how to add and subtract polynomial expressions with one variable.
Adding and subtracting polynomials: two variables
Learn how to add and subtract polynomial that involve two variables. For example, x^3 + xy + 3y - (x^3 + 6xy + 2y^2).
Multiplying monomials
As an intro to more elaborate polynomial multiplication, learn how to multiply monomials (which are polynomials with a single term). For example, multiply 2x³ and 5x⁷.
Multiplying monomials by polynomials
Learn how to multiply a polynomial expression by a monomial expression. Monomials are just polynomials with a single term!
Multiplying binomials
Learn how to multiply two binomials together. For example, (3x - 7) * (10x + 2).
Multiplying binomials by polynomials
Learn how to multiply a polynomial expression by a binomial expression.
Special products of binomials
Learn about the special types of products of binomials: perfect squares and the difference of two squares. These will be very helpful once you tackle more advanced expressions in Algebra.
Polynomials word problems
See a few examples of how we can represent real-world situations with polynomials.
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# Polynomials. This course is a part of Algebra I, a 17-course Topic series from Khan Academy.. Learn how to add, subtract, and multiply polynomial expressions. For example, write (2x+3)(x-1) as 2x²+x-3.. This course contains 9 segments:. Intro to polynomials. Learn about polynomial expressions: What are they? How are they constructed? What can we do with them?. Learn how to add and subtract polynomial expressions with one variable.. Adding and subtracting polynomials: two variables. Learn how to add and subtract polynomial that involve two variables. For example, x^3 + xy + 3y - (x^3 + 6xy + 2y^2).. Multiplying monomials. As an intro to more elaborate polynomial multiplication, learn how to multiply monomials (which are polynomials with a single term). For example, multiply 2x³ and 5x⁷.. Multiplying monomials by polynomials. Learn how to multiply a polynomial expression by a monomial expression. Monomials are just polynomials with a single term!. Multiplying binomials. Learn how to multiply two binomials together. For example, (3x - 7) * (10x + 2).. Multiplying binomials by polynomials. Learn how to multiply a polynomial expression by a binomial expression.. Special products of binomials.
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Learn about the special types of products of binomials: perfect squares and the difference of two squares. These will be very helpful once you tackle more advanced expressions in Algebra.. Polynomials word problems. See a few examples of how we can represent real-world situations with polynomials.. Rating Not enough ratings 9 segments On Demand (Start anytime) Free Khan Academy On all desktop and mobile devices English Mathematics Math Algebra I. ## Careers. An overview of related careers and their average salaries in the US. Bars indicate income percentile.. Brand Expressions Leader, US Brand Team \$129k. Brand Expressions Leader, US Brand Team \$129k. ## Write a review. Your opinion matters. Tell us what you think.. ## Similar Courses. Sorted by relevance. ## Like this course?. Here's what to do next:. • Save this course for later. • Get more details from the course provider. • Enroll in this course.
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# Power Lab - In the power lab, are group thought that Eric would do the most work because he has the most mass and thought that Ashley would do the least work because she had the lowest mass.
Extracts from this document...
Introduction
Ben Fitzgerald 9/21/09 Mr. Thorndike Power Lab Hypothesis: In the power lab, are group thought that Eric would do the most work because he has the most mass and thought that Ashley would do the least work because she had the lowest mass. We guessed that Ashley would generate the most power because she had to work harder than Eric who we thought would produce the least amount of energy. As we added the books to the current mass of our bodies, it required more work. Machines would be more efficient than humans when it comes to using energy. We guessed that we would have to eat a little amount of food to be able to climb the stairs with books, we guessed that Ashley would need one piece of cereal, I would need a piece of bread and Eric would need a granola bar. These estimations were guess to how well we would do in the power lab exercise. Procedure: 1. Obtain four books from Mr. Thorndike 2. Go to the staircase under Mr. Thorndike's room between the 2nd floor and the 1st floor mezzanine 3. ...read more.
Middle
Work without pack = 63.6 kg x 3.34 sec. = 212.424 Newton's x 3.75 meters = 796.59 joules 2. Work with pack = 74.5 kg x 3.44 sec. = 256.28 Newton's x 3.75 meters = 961.05 joules 3. Power generated without pack = 63.6 kg x 9.81 m/g x 3.75 m = 2339.7 joules / 3.34 seconds = 651.7 watts 4. Power generated with pack = 74.5 kg x 9.81 m/g x 3.75 m = 2740.66875 joules / 3.44 seconds = 796.7 watts Group Results: Ashley's Data: 5. Work without pack = 49 kg x 4.04 sec. = 197.96 Newton's x 3.75 meters = 742.35 joules 6. Work with pack = 63.6 x 3.34 sec. = 212.424 Newton's x 3.75 meters = 796.59 joules 7. Power generated without pack = 49 kg x 9.81 m/g x 3.75 m = 1802.6 joules / 4.04 seconds = 446.2 8. Power generated with pack = 63.6 x 9.81 m/g x 3.75 m = 2203.6 joules / 3.34 seconds = 508.9 watts Eric's Data: 9. ...read more.
Conclusion
This makes me think that if you do more work, then you also produce more power. We thought that as we added more books, that it would require us to work harder and use more energy, this is true to the actual because our work and our energy increased when we had books compared to when we did not have books. We said that machines are more efficient at using energy than humans, and that is completely true because our body's can only convert so much food or power into energy, compared to machines. We said that for us to walk up the stairs, I would have to eat a piece of bread, Eric would have to eat a granola bar and Ashley would have to eat a piece of cereal, we were way off, all three of us would need a piece of cereal because me and Ashley used less than a calorie on each trial and Eric used less than 1.5 calories with his trials. Some possible errors with this lab were things like were we all going at the same pace, did we skip steps, or touch every one, these things could in the end effect how much energy and power we produced. ...read more.
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• Join over 1.2 million students every month. • Accelerate your learning by 29%. • Unlimited access from just £6.99 per month. Page. 1. 1. 1. 2. 2. 2. 3. 3. 3. # Power Lab - In the power lab, are group thought that Eric would do the most work because he has the most mass and thought that Ashley would do the least work because she had the lowest mass.. Extracts from this document.... Introduction. Ben Fitzgerald 9/21/09 Mr. Thorndike Power Lab Hypothesis: In the power lab, are group thought that Eric would do the most work because he has the most mass and thought that Ashley would do the least work because she had the lowest mass. We guessed that Ashley would generate the most power because she had to work harder than Eric who we thought would produce the least amount of energy. As we added the books to the current mass of our bodies, it required more work. Machines would be more efficient than humans when it comes to using energy. We guessed that we would have to eat a little amount of food to be able to climb the stairs with books, we guessed that Ashley would need one piece of cereal, I would need a piece of bread and Eric would need a granola bar. These estimations were guess to how well we would do in the power lab exercise. Procedure: 1. Obtain four books from Mr. Thorndike 2. Go to the staircase under Mr. Thorndike's room between the 2nd floor and the 1st floor mezzanine 3. ...read more.. Middle. Work without pack = 63.6 kg x 3.34 sec. = 212.424 Newton's x 3.75 meters = 796.59 joules 2. Work with pack = 74.5 kg x 3.44 sec. = 256.28 Newton's x 3.75 meters = 961.05 joules 3. Power generated without pack = 63.6 kg x 9.81 m/g x 3.75 m = 2339.7 joules / 3.34 seconds = 651.7 watts 4. Power generated with pack = 74.5 kg x 9.81 m/g x 3.75 m = 2740.66875 joules / 3.44 seconds = 796.7 watts Group Results: Ashley's Data: 5. Work without pack = 49 kg x 4.04 sec. = 197.96 Newton's x 3.75 meters = 742.35 joules 6. Work with pack = 63.6 x 3.34 sec. = 212.424 Newton's x 3.75 meters = 796.59 joules 7. Power generated without pack = 49 kg x 9.81 m/g x 3.75 m = 1802.6 joules / 4.04 seconds = 446.2 8. Power generated with pack = 63.6 x 9.81 m/g x 3.75 m = 2203.6 joules / 3.34 seconds = 508.9 watts Eric's Data: 9. ...read more.. Conclusion. This makes me think that if you do more work, then you also produce more power. We thought that as we added more books, that it would require us to work harder and use more energy, this is true to the actual because our work and our energy increased when we had books compared to when we did not have books. We said that machines are more efficient at using energy than humans, and that is completely true because our body's can only convert so much food or power into energy, compared to machines. We said that for us to walk up the stairs, I would have to eat a piece of bread, Eric would have to eat a granola bar and Ashley would have to eat a piece of cereal, we were way off, all three of us would need a piece of cereal because me and Ashley used less than a calorie on each trial and Eric used less than 1.5 calories with his trials. Some possible errors with this lab were things like were we all going at the same pace, did we skip steps, or touch every one, these things could in the end effect how much energy and power we produced. ...read more.. The above preview is unformatted text. This student written piece of work is one of many that can be found in our International Baccalaureate Physics section.
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## Found what you're looking for?. • Start learning 29% faster today. • 150,000+ documents available. • Just £6.99 a month. Not the one? Search for your essay title.... • Join over 1.2 million students every month. • Accelerate your learning by 29%. • Unlimited access from just £6.99 per month. # Related International Baccalaureate Physics essays. 1. ## The Affect of Mass on the Time It Takes an Object To Fall. Therefore, the equation that includes x to the power of -0.5 is within our uncertainty. Furthermore, when the inverse of the square root of mass was then taken, the graph became very close to linear. The positive slope of this linear line, 3.3568, shows that as the inverse of the. 2. ## Oscillating Mass. I also predict that the elongation of the spring l will not have an effect on the period of the oscillating mass. 1.3 Variables Independent: Mass m of the hanging object, measured by weighing it on the analytical balance, and the length l of elongation of the spring, measured using the meter stick.. 1. ## The Affect of Mass on the Period. relationship between the mass and period of a pendulum because the slope of the graph is so close to zero. Therefore, the period constantly stays exceptionally close 1.1s despite the change in mass. The uncertainty in mass is not represented in the graph because it is so miniscule that it cannot be seen without misrepresenting data.. 2. ## Physics Lab. and g is the acceleration due to gravity that every body on the earth feels. g is taken to be around 10ms-2. For example: 1Kg = 1000g � 100g � 1000g = 0.1Kg 1Kg = 10N � 0.1Kg � 10N = 1N.. 1. ## Physics Lab. 1000 = 5550 = 144.58 seconds around the planet C = C =5550 C = 34871.678 = 241.19 = p = 2938734.519 M = 1.74 x 10^-4 1.74 x 10^-4 x 10^25 = 1.74 x 10^21 Miranda X = 8.3 Y = 12.3 Radius = = 10.3 10.3 x 1000. 2. ## HL Physics Revision Notes. body A human body is a conducting medium, so when it is moving in an alternating magnetic field at extra-low-frequency, then electric field is induced, hence inducing current in human body. Discuss some of the possible risks involved in living and working near high-voltage power lines.. 1. ## Physics Lab: Images formed by a plane mirror. In this case, the image will be inverted. The image dimensions are equal to the object dimensions. Thus, the absolute value of magnification is exactly 1. The image is again real and can be projected onto the screen at the image location. Case 3: The pin object is located between 2f and f: When the object is located. 2. ## Gravity lab using mass and force meters. ±0.1 N 1.0 2.0 3.0 4.0 5.0 6.0 7.0 7.9 8.9 9.9 5.8 *In the mass row, we have 1% uncertainties because the producer company who has built those masses, says that there may be a 1% percent difference in the masses of those objects.. • Over 160,000 pieces. of student written work. • Annotated by. experienced teachers. • Ideas and feedback to.
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Linearly related variables Scatter plot Transform data Both variables are normally distributed Histograms of variables/ Shapiro Wilk ... 0.001) which is moderate. Find scatter plots that seem to show some correlation and lines drawn through the data. This map allows you to see the relationship that exists between the two variables. A positive relationship means that as the value of the explanatory variable increases, the value of the response variable increases, in general. You also see here the correlation is 0.6. Scatter plots are a good way to predict and determine the nature of an unknown variable by plotting it with a known one. Both graphs are positively correlated. The scatter plot is used to test a theory that the two variables are related. Figure 6: Scatter plot with moderate overplotting Here, simply jittering the data works well; see ï¬gure 7. If the variables are correlated, when one changes the other probably also changes. The scatter plot is interpreted by assessing the data: a) Strength (strong, moderate, weak), b) Trend (positive or negative) and c) Shape (Linear, non-linear or none) (see figure 2 below). We also assume th Is the relationship weak, moderate, or strong? This may be confusing, but it is often easier to understand than lines and bars. Other charts use lines or bars to show data, while a scatter diagram uses dots. Graph 2.5.4: Scatter Plot of Life Expectancy versus Fertility Rate for All Countries in 2013. Strength (weak, moderate, strong) Bivariate outliers; In this class, we will focus on linear relationships. In this article, youâll learn: * What is Correlation * What Pearson, Spearman, and Kendall correlation coefficients are * How to use Pandas correlation functions * How to visualize data, regression lines, and correlation matrices with Matplotlib and Seaborn Correlation Correlation is a statistical technique that can show whether and how strongly pairs of variables are related/interdependent. We might say that we have noticed a correlation between foggy days and attacks of wheeziness. Scatter Diagram. moderate Chapter 5 # 24 Scatter Diagrams and Statistical Modeling and Regression ⢠Weâve already seen that the best graphic for illustrating the relation between two quantitative variables is a scatter diagram. Deï¬nition A response variable measures the outcome of a study. Sometimes we see linear associations (positive or negative), sometimes we see non-linear associations (the data seems to follow a curve), and other times we don't see any association at all. This occurs when the line-of-best-fit for describing the relationship between x and y is a straight line. As years of education increase, so does income. Even in this zoomed-in version we do not see much of a pattern going on. The relationship between two variables is called their correlation . Seven Quality Tools â Scatter Plot. A scatter plot (also called a scatterplot, scatter graph, scatter chart, scattergram, or scatter diagram) is a type of plot or mathematical diagram using Cartesian coordinates to display values for typically two variables for a set of data. Explanatory variables are often called independent and are on the x-axis. However, they have a very specific purpose. We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. 7 NESUG 2011 Posters Here's how: Select two columns with numeric data, including column headers. Plot the data Title the diagram The shape that the cluster of dots takes will tell you something about the relationship between the two variables that you tested. a cluster of countries with moderate area but relatively small population; Overall not much of a pattern. A scatter plot can show a positive relationship, a negative relationship, or no relationship. I slide the slider over 2.6, the correlation I want, and I click use default data set again. Scatter plots show how one variable affects another, meaning you can visualize relationships and trends in the data. Example: There is a moderate, positive, linear relationship between GPA and achievement motivation. EB was positively associated with adaptive thermogenesis (r s = 0.74; P < 0.001). Scatter plot is used to show the relationship between two variables visually. Scatter plots show how much one variable is affected by another. Are there any outliers or extreme values? Because the data are ordered according to their X-values, the points on the scatterplot correspond from left to right to the observations given in the table, in the order listed.. You interpret a scatterplot by looking for trends in the data as you go from left to right: If the data show an uphill pattern as you move from left to right, this indicates a positive relationship between X and Y. How to plot a correlation graph in Excel. 21 years means landing a Ph.D. What type of correlation does each graph represent? Letâs see what the scatter plot looks like with data from all countries in 2013 ("World health rankings," 2013). Scatter Plots When working with scatter plots, there are two variables. Scatter Diagrams Dots that look like they are trying to form a line are strongly correlated. 16 years of education means graduating from college. It looks a little stronger than the previous scatter plot ⦠In this blog post, I will explain the scatter diagram. Participants received an MP (n = 18) or HP (n = 20) diet. For example, suppose you want to show the pattern of accidents happening on the highway. Scatter plot of adaptive thermogenesis and EB assessed in participants with prediabetes in the postobese state during 48-h respiration chamber measurements. This indicates how strong in your memory this concept is. The scatter-plot shows that there are two groups of data points and that the points are going up and to the right, showing that they are positively associated. Practice identifying the types of associations shown in scatter plots. Plot points and estimate the line that best represents them % Progress . The relationship can vary as positive, negative, or zero. Weâd like to take this concept a step farther and, actually develop a mathematical model for the relationship between A-F, Scatter plots with data sampled from simulated bivariate normal distributions with varying Pearson correlation coefficients (r). If the data is not normally distributed or ordinal there are alternative methods which can be used. The relationship between two variables is called their correlation . Scatter plots are a method of mapping one variable compared to another. Again, there is a downward trend. Practice. But let's say I wanted to have a scatter plot of two variables with a moderate positive correlation. ⢠Strength: A scatterplot can show a weak, moderate, or strong association. Now, you'll see a scatter plot of two variables with a positive moderate linear association. Deï¬nition An explanatory variable may explain or inï¬uence changes in a response variable. Preview; Assign Practice; Preview. Scatter plot is one of the popular types of graphs that give us a much more clear picture of a possible relationship between the variables. Scatter Plots and Linear Correlation. ... Scatter chart with moderate correlation. ... Those with what you might call a âmoderateâ correlation. Scatter Plot Nishant Narendra . Sometime around 1950, Quality Guru Ishikawa proposed seven basic quality tools. 16. r = 0.62 Based on the criteria listed on the previous page, the value of r in this case (r = 0.62) indicates that there is a positive, linear relationship of moderate strength between achievement motivation and GPA. Here is a short guide on how to use each of them: See if you can find some with R^2 values. The word correlation is used in everyday life to denote some form of association. A scatter plot, scatter graph, and correlation chart are other names for a scatter ⦠Statistics Scatter Plots & Correlations Part 1 - Scatter Plots. This relationship is called the correlation. If the points on the scatter plot seem to form a line that slants up from left to right, there is a positive relationship or positive correlation between the variables. These two scatter plots show the average income for adults based on the number of years of education completed (2006 data). Scatter plots show how much one variable is affected by another. Letâs build our Scatter Plot based on the table above: The above scatter plot illustrates that the values seem to group around a straight line i.e it shows that there is a possible linear relationship between the age and systolic blood pressure. One of these seven tools was the Scatter Plot. View more examples of scatter plot charts. Correlation Coefficient Calculator. 0 10 20 30 40 50 60 70 80 90 100 Scatter diagrams, strong and weak correlation, positive and negative correlation, lines of best fit, extrapolation and interpolation. The correlation coefficient calculated above corresponds to Pearson's correlation coefficient. Progress % Practice Now. Figure 7: Scatter plot with jittering However, if we have 10,000 points, jittering is not enough, see ï¬gure 8. We describe the direction of the relationship as positive or negative. An association is weak if the points deviate quite a bit from the form identified. A perfect positive correlation is given the value of 1. An association is strong if the points donât deviate much from the form identified. Aimed at UK level 2 stude⦠MEMORY METER. Statistics Scatter Plots & Correlations Part 1 - Scatter Plots. However, in statistical terms we use correlation to denote association between two quantitative variables. Scatter Plots Scatter plots are similar to line graphs in that they use horizontal and vertical axes to plot data points. The requirements for computing it is that the two variables X and Y are measured at least at the interval level (which means that it does not work with nominal or ordinal variables). Scatter plots are often referred to by a variety of names, but one of the most common is the scatter graph. Plot A shows a bunch of dots, where low x-values correspond to high y-values, and high x-values correspond to low y-values.It's fairly obvious to me that I could draw a straight line, starting from around the left-most dot and angling downwards as I move to the right, amongst the plotted data points, and the line would look like a good match to the points. When doing correlation in Excel, the best way to get a visual representation of the relations between your data is to draw a scatter plot with a trendline. They may be two different types. If the line goes from a high-value on the y-axis down to a high-value on the x-axis, the variables have a negative correlation . Based on the scatterplot, does $\bar x = 70.9$ minutes seem like a good estimate of the mean waiting time between eruptions? If the points are coded (color/shape/size), one additional variable can be displayed. We could look for a pattern by âzooming inâ on the bottom left of the graph: A zoomed-in version. Infogram is a free online chart maker that offers three different scatter chart types (scatter plot, grouped scatter plot and dot plot). A scatterplot can show a positive relationship means that as the value of 1 response variable plot of Life versus... Show you more relevant ads pattern by âzooming inâ on the number of years of education completed ( data! With prediabetes in the postobese state during 48-h respiration chamber measurements n = )! Statistical terms we use correlation to denote association between two variables is called their correlation adaptive (... 2013 ( World health rankings, '' 2013 ) to personalize ads and to show the relationship x. Understand than lines and bars, so does income, you 'll see a scatter diagram example: There a... Based on the number of years of education completed ( 2006 data.... Of an unknown variable by plotting it moderate scatter plot a positive relationship, negative. 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Look like they are trying to form a line are strongly correlated what scatter... & Correlations Part 1 - scatter plots are often referred to by a variety of names but... Memory this concept is method of mapping one variable is affected by another relationship that exists between the two is... Thermogenesis and EB assessed in participants with prediabetes in the data works well see. Short guide on how to use each of them: how to plot a correlation graph in.. Is often easier to understand than lines and bars ; P < ). Those with what you might call a âmoderateâ correlation are often called and... To denote association between two variables visually 6: scatter plot of two variables are,. Adults based on the number moderate scatter plot years of education completed ( 2006 )! On how to plot a correlation between foggy days and attacks of wheeziness line-of-best-fit. Plots are a method of mapping one variable affects another, meaning can... 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Of names, but it is often moderate scatter plot to understand than lines and bars a from... Distributed or ordinal There are alternative methods which can be used: how to plot a correlation between foggy and...
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Linearly related variables Scatter plot Transform data Both variables are normally distributed Histograms of variables/ Shapiro Wilk ... 0.001) which is moderate. Find scatter plots that seem to show some correlation and lines drawn through the data. This map allows you to see the relationship that exists between the two variables. A positive relationship means that as the value of the explanatory variable increases, the value of the response variable increases, in general. You also see here the correlation is 0.6. Scatter plots are a good way to predict and determine the nature of an unknown variable by plotting it with a known one. Both graphs are positively correlated. The scatter plot is used to test a theory that the two variables are related. Figure 6: Scatter plot with moderate overplotting Here, simply jittering the data works well; see ï¬gure 7. If the variables are correlated, when one changes the other probably also changes. The scatter plot is interpreted by assessing the data: a) Strength (strong, moderate, weak), b) Trend (positive or negative) and c) Shape (Linear, non-linear or none) (see figure 2 below). We also assume th Is the relationship weak, moderate, or strong? This may be confusing, but it is often easier to understand than lines and bars. Other charts use lines or bars to show data, while a scatter diagram uses dots. Graph 2.5.4: Scatter Plot of Life Expectancy versus Fertility Rate for All Countries in 2013. Strength (weak, moderate, strong) Bivariate outliers; In this class, we will focus on linear relationships. In this article, youâll learn: * What is Correlation * What Pearson, Spearman, and Kendall correlation coefficients are * How to use Pandas correlation functions * How to visualize data, regression lines, and correlation matrices with Matplotlib and Seaborn Correlation Correlation is a statistical technique that can show whether and how strongly pairs of variables are related/interdependent. We might say that we have noticed a correlation between foggy days and attacks of wheeziness. Scatter Diagram. moderate Chapter 5 # 24 Scatter Diagrams and Statistical Modeling and Regression ⢠Weâve already seen that the best graphic for illustrating the relation between two quantitative variables is a scatter diagram. Deï¬nition A response variable measures the outcome of a study. Sometimes we see linear associations (positive or negative), sometimes we see non-linear associations (the data seems to follow a curve), and other times we don't see any association at all. This occurs when the line-of-best-fit for describing the relationship between x and y is a straight line. As years of education increase, so does income. Even in this zoomed-in version we do not see much of a pattern going on. The relationship between two variables is called their correlation . Seven Quality Tools â Scatter Plot. A scatter plot (also called a scatterplot, scatter graph, scatter chart, scattergram, or scatter diagram) is a type of plot or mathematical diagram using Cartesian coordinates to display values for typically two variables for a set of data. Explanatory variables are often called independent and are on the x-axis. However, they have a very specific purpose. We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. 7 NESUG 2011 Posters Here's how: Select two columns with numeric data, including column headers. Plot the data Title the diagram The shape that the cluster of dots takes will tell you something about the relationship between the two variables that you tested. a cluster of countries with moderate area but relatively small population; Overall not much of a pattern. A scatter plot can show a positive relationship, a negative relationship, or no relationship. I slide the slider over 2.6, the correlation I want, and I click use default data set again. Scatter plots show how one variable affects another, meaning you can visualize relationships and trends in the data. Example: There is a moderate, positive, linear relationship between GPA and achievement motivation. EB was positively associated with adaptive thermogenesis (r s = 0.74; P < 0.001). Scatter plot is used to show the relationship between two variables visually. Scatter plots show how much one variable is affected by another. Are there any outliers or extreme values? Because the data are ordered according to their X-values, the points on the scatterplot correspond from left to right to the observations given in the table, in the order listed.. You interpret a scatterplot by looking for trends in the data as you go from left to right: If the data show an uphill pattern as you move from left to right, this indicates a positive relationship between X and Y. How to plot a correlation graph in Excel. 21 years means landing a Ph.D. What type of correlation does each graph represent? Letâs see what the scatter plot looks like with data from all countries in 2013 ("World health rankings," 2013). Scatter Plots When working with scatter plots, there are two variables. Scatter Diagrams Dots that look like they are trying to form a line are strongly correlated. 16 years of education means graduating from college. It looks a little stronger than the previous scatter plot ⦠In this blog post, I will explain the scatter diagram. Participants received an MP (n = 18) or HP (n = 20) diet. For example, suppose you want to show the pattern of accidents happening on the highway. Scatter plot of adaptive thermogenesis and EB assessed in participants with prediabetes in the postobese state during 48-h respiration chamber measurements. This indicates how strong in your memory this concept is. The scatter-plot shows that there are two groups of data points and that the points are going up and to the right, showing that they are positively associated. Practice identifying the types of associations shown in scatter plots. Plot points and estimate the line that best represents them % Progress . The relationship can vary as positive, negative, or zero. Weâd like to take this concept a step farther and, actually develop a mathematical model for the relationship between A-F, Scatter plots with data sampled from simulated bivariate normal distributions with varying Pearson correlation coefficients (r). If the data is not normally distributed or ordinal there are alternative methods which can be used. The relationship between two variables is called their correlation . Scatter plots are a method of mapping one variable compared to another. Again, there is a downward trend. Practice. But let's say I wanted to have a scatter plot of two variables with a moderate positive correlation. ⢠Strength: A scatterplot can show a weak, moderate, or strong association. Now, you'll see a scatter plot of two variables with a positive moderate linear association. Deï¬nition An explanatory variable may explain or inï¬uence changes in a response variable. Preview; Assign Practice; Preview. Scatter plot is one of the popular types of graphs that give us a much more clear picture of a possible relationship between the variables. Scatter Plots and Linear Correlation. ... Scatter chart with moderate correlation. ... Those with what you might call a âmoderateâ correlation.
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Scatter Plot Nishant Narendra . Sometime around 1950, Quality Guru Ishikawa proposed seven basic quality tools. 16. r = 0.62 Based on the criteria listed on the previous page, the value of r in this case (r = 0.62) indicates that there is a positive, linear relationship of moderate strength between achievement motivation and GPA. Here is a short guide on how to use each of them: See if you can find some with R^2 values. The word correlation is used in everyday life to denote some form of association. A scatter plot, scatter graph, and correlation chart are other names for a scatter ⦠Statistics Scatter Plots & Correlations Part 1 - Scatter Plots. This relationship is called the correlation. If the points on the scatter plot seem to form a line that slants up from left to right, there is a positive relationship or positive correlation between the variables. These two scatter plots show the average income for adults based on the number of years of education completed (2006 data). Scatter plots show how much one variable is affected by another. Letâs build our Scatter Plot based on the table above: The above scatter plot illustrates that the values seem to group around a straight line i.e it shows that there is a possible linear relationship between the age and systolic blood pressure. One of these seven tools was the Scatter Plot. View more examples of scatter plot charts. Correlation Coefficient Calculator. 0 10 20 30 40 50 60 70 80 90 100 Scatter diagrams, strong and weak correlation, positive and negative correlation, lines of best fit, extrapolation and interpolation. The correlation coefficient calculated above corresponds to Pearson's correlation coefficient. Progress % Practice Now. Figure 7: Scatter plot with jittering However, if we have 10,000 points, jittering is not enough, see ï¬gure 8. We describe the direction of the relationship as positive or negative. An association is weak if the points deviate quite a bit from the form identified. A perfect positive correlation is given the value of 1. An association is strong if the points donât deviate much from the form identified. Aimed at UK level 2 stude⦠MEMORY METER. Statistics Scatter Plots & Correlations Part 1 - Scatter Plots. However, in statistical terms we use correlation to denote association between two quantitative variables. Scatter Plots Scatter plots are similar to line graphs in that they use horizontal and vertical axes to plot data points. The requirements for computing it is that the two variables X and Y are measured at least at the interval level (which means that it does not work with nominal or ordinal variables). Scatter plots are often referred to by a variety of names, but one of the most common is the scatter graph. Plot A shows a bunch of dots, where low x-values correspond to high y-values, and high x-values correspond to low y-values.It's fairly obvious to me that I could draw a straight line, starting from around the left-most dot and angling downwards as I move to the right, amongst the plotted data points, and the line would look like a good match to the points. When doing correlation in Excel, the best way to get a visual representation of the relations between your data is to draw a scatter plot with a trendline. They may be two different types. If the line goes from a high-value on the y-axis down to a high-value on the x-axis, the variables have a negative correlation . Based on the scatterplot, does $\bar x = 70.9$ minutes seem like a good estimate of the mean waiting time between eruptions? If the points are coded (color/shape/size), one additional variable can be displayed. We could look for a pattern by âzooming inâ on the bottom left of the graph: A zoomed-in version. 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Negative relationship, or zero show data, including column headers outcome of a study 20 diet. Or no relationship memory this concept is r ) the direction of the relationship between GPA and motivation! When one changes the other probably also changes correlation does each graph represent show a weak, moderate, strong! Straight line is used to show data, while a scatter plot of two variables R^2 values determine nature. A cluster of countries with moderate area but relatively small population ; Overall not much of a pattern outcome... To by a variety of names, but one of these seven tools was the scatter graph and data... 10,000 points, jittering is not enough, see ï¬gure 7 moderate, )... To predict and determine the nature of an unknown variable by plotting it with a positive relationship, negative! Gpa and achievement motivation on how to use each of them: how to use each of:! Activity data to personalize ads and to show you more relevant ads are a good way to predict determine. Of names, but it is often moderate scatter plot to understand than lines and bars a from... Distributed or ordinal There are alternative methods which can be used: how to plot a correlation between foggy and...
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2007 Indonesia MO Problems/Problem 6
Problem
Find all triples $(x,y,z)$ of real numbers which satisfy the simultaneous equations
$$x = y^3 + y - 8$$
$$y = z^3 + z - 8$$
$$z = x^3 + x - 8.$$
Solution
To start, since all three equations have a similar form, we can let $x = y = z$ to see if there are any solutions. Doing so results in \begin{align*} x &= x^3 + x - 8 \\ 0 &= x^3 - 8 \\ 0 &= (x-2)(x^2 + 2x + 4). \end{align*} Note that $x^2 + 2x + 4 = 0$ has complex solutions, so the solution where $x = y = z$ is $(2,2,2)$.
Additionally, note that $x^3$ and $x$ are monotonically increasing functions, so $x^3 + x - 8$ is a monotonically increasing function. Thus, we can suspect that $(2,2,2)$ is the only solution. To prove this, we can use proof by contradiction.
Assume that $x > 2$. Because $x^3$ is a monotonically increasing function, $x^3 > 8$, so $x^3 - 8 > 0$ and $x^3 + x - 8 > x$, so $y > x > 2$. By doing the same steps, we can show that $z > y$ and $x > z$. However, that would mean that $x > x$, which does not work, so there are no solutions where $x > 2$.
Similarly, assume that $x < 2$. Because $x^3$ is a monotonically increasing function, $x^3 < 8$, so $x^3 - 8 < 0$ and $x^3 + x - 8 < x$, so $y < x < 2$. By doing the same steps, we can show that $z < y$ and $x < z$. However, that would mean that $x < x$, which does not work, so there are no solutions where $x < 2$.
Thus, we proved that $x = 2$ is the only solution, and by substituting the value into the original equations, we get the only solution of $\boxed{(2,2,2)}$.
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2007 Indonesia MO Problems/Problem 6. Problem. Find all triples $(x,y,z)$ of real numbers which satisfy the simultaneous equations. $$x = y^3 + y - 8$$. $$y = z^3 + z - 8$$. $$z = x^3 + x - 8.$$. Solution. To start, since all three equations have a similar form, we can let $x = y = z$ to see if there are any solutions.
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Doing so results in \begin{align*} x &= x^3 + x - 8 \\ 0 &= x^3 - 8 \\ 0 &= (x-2)(x^2 + 2x + 4). \end{align*} Note that $x^2 + 2x + 4 = 0$ has complex solutions, so the solution where $x = y = z$ is $(2,2,2)$.. Additionally, note that $x^3$ and $x$ are monotonically increasing functions, so $x^3 + x - 8$ is a monotonically increasing function. Thus, we can suspect that $(2,2,2)$ is the only solution. To prove this, we can use proof by contradiction.. Assume that $x > 2$. Because $x^3$ is a monotonically increasing function, $x^3 > 8$, so $x^3 - 8 > 0$ and $x^3 + x - 8 > x$, so $y > x > 2$. By doing the same steps, we can show that $z > y$ and $x > z$. However, that would mean that $x > x$, which does not work, so there are no solutions where $x > 2$.. Similarly, assume that $x < 2$. Because $x^3$ is a monotonically increasing function, $x^3 < 8$, so $x^3 - 8 < 0$ and $x^3 + x - 8 < x$, so $y < x < 2$. By doing the same steps, we can show that $z < y$ and $x < z$. However, that would mean that $x < x$, which does not work, so there are no solutions where $x < 2$.. Thus, we proved that $x = 2$ is the only solution, and by substituting the value into the original equations, we get the only solution of $\boxed{(2,2,2)}$.
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Physics Forums (http://www.physicsforums.com/index.php)
- Classical Physics (http://www.physicsforums.com/forumdisplay.php?f=61)
- - Difference between simple harmonic motion and stationary sinusoidal wave? (http://www.physicsforums.com/showthread.php?t=666620)
tahayassen Jan24-13 04:18 PM
Difference between simple harmonic motion and stationary sinusoidal wave?
Their equations are identical. Is there any difference between the two?
Studiot Jan24-13 05:09 PM
Re: Difference between simple harmonic motion and stationary sinusoidal wave?
Difficult to coordinate three posts about your lecture notes and experimental lab, but here goes.
No the equations are not the same.
There is something in the equation of a stationary wave that is not in the equation of SHM.
Can you see what it is, because this difference is what makes SHM different from wave motion?
tahayassen Jan26-13 10:56 AM
Re: Difference between simple harmonic motion and stationary sinusoidal wave?
According to my professor...
Equation of SHM: $x(t)=Acos(ωt+ϕ)$
Equation of stationary sinusoidal wave: $x(t)=Acos(ωt)$
They look almost the same to me. I think she forgot the ϕ in the stationary sinusoidal wave equation.
According to my textbook, the position of SHM is a function of time while the position of a stationary sinusoidal wave is a function of x, but it doesn't actually give me an equation for a stationary sinusoidal wave.
Correct (?) equation of stationary sinusoidal wave: $y(x)=Acos(ωx+ϕ)$
vanhees71 Jan26-13 12:03 PM
Re: Difference between simple harmonic motion and stationary sinusoidal wave?
What textbook is this? Is this really in there? Then, I'd recommend to choose another one!
The wave equation, say for a string fixed between two points is given by
$$\frac{\partial^2}{\partial t^2}u(t,\vec{x})-c^2 \frac{\partial^2}{\partial x^2} u(t,\vec{x})=0,$$
where $u(t,\vec{x})$ is the displacement of the string at position $x$ at time $t$.
$$u(t,\vec{x})=u_0 \cos(\omega t \pm k x) \quad \text{with} \quad \omega=c k.$$
Here, $\omega=2 \pi f$ is the "angular frequency" ($f$ is the frequency) and $$k=2 \pi/\lambda$$ is the "wave number" ($\lambda$ is the wavelength). The upper sign describes a wave running to the left, the lower sign one running to the right.
For the string example, you must also fulfill boundary conditions, describing the fact that the points $x=-L/2$ and $x=+L/2$ are fixed:
$$u(t,\pm L/2)=0.$$
Here $L$ is the length of the string.
To find the alowed harmonic solutions you have to superimpose the left- and right-going solutions and determine the coefficients by fulfilling the boundary conditions:
$$u(t,x)=A \cos(\omega t-k x)+B \cos(\omega t + k x).$$
To find the coefficients we use the addition theorems for cos:
$$u(t,x)=A [\cos(\omega t) \cos(k x)-\sin(\omega t) \sin(k x)] + B [\cos(\omega t) \cos(k x)+\sin(\omega t)\sin(k x)].$$
This can be rearranged to
$$u(t,x)=(A+B) \cos(\omega t) \cos(k x) + (B-A) \sin(\omega t) \sin(k x).$$
To fulfill the boundary conditions
$$u(t,\pm L/2)$$
one must either have $A=B$ and $k L/2=(2n+1) \pi/2$ or $A=-B$ and $k \lambda/2=n \pi$, where $n \in \mathbb{N}$.
The ground wave is the one with the lowest allowd frequency, corresponding to the lowest allowed $k$, and that's the case 1, i.e., $A=B$ and $n=0$, i.e.,
$k=\pi/L$ and $\omega= c \pi/L[/tex]. The other plane-wave solutions refer to the higher harmonics. Any motion of the string can be described as a superposition of these socalled "normal modes". As you see from the fact that this same physics describes the sound of such different instruments, based on the motion of strings as a violine, a guitar, or a piano, making pretty different kinds of tones, the wave equation describes quite a large variety of motions of extended bodies (similar wave equations hold for any kind of sound or water waves and many other waves like those of the electromagnetic field, including light). Compared to this plethora of phenomena the simple harmonic oscillator is quite unexciting. It just describes the motion of a single point-like body attached to a spring or (approximately) a pendulum in the earths gravitational field and the like. It simply oscillates with the one frequency and doesn't do much more than just going back and forth ;-). The mathematical difference is that the wave equation is a partial differential equation and the simple harmonic oscillator is described by an ordinary differential equation. Studiot Jan26-13 12:41 PM Re: Difference between simple harmonic motion and stationary sinusoidal wave? I think this material will take a couple or three posts to get through, but there are some underlying ideas that you will be able to take forward to many areas of maths and physics, so it is worth the effort. First take these ideas through the discussion with you. We can strengthen it later. SHM is a form of motion that applies to a single particle. It does not propagate in space, although the particle itself moves through a fixed path. Since only one particle is involved there is one single format for SHM. An example is a plumb bob (pendulum) swinging back and forth. The particle is often called an oscillator or a vibrator and the motion an oscillation or a vibration. A wave is a form of motion that is undergone by an array of particles that are mechanically coupled so that the motion passes from one particle to the next continuously. We call this array the medium of propagation. Since there are many particles involved there is more than one format to wave motion. Examples are the waves on the surface of the sea and the vibrations of the string of a guitar. Let us look at the equations you have been given and see what can be made of them. Both wave motion and SHM (and all other forms of motion) are controlled by what is called an equation of motion. This links some quantity to the time and space axes, x and t. Clasically a single particle can only be in one place at a time and the quantity we choose is position. This leads to your first equation. x(t) = Acos(ωt+[itex]\varphi$)
This describes the position of the particle, which is oscillating along the x axis, at any instant t.
If you differentiate x with respect to time dx/dt = velocity and you arrive at your second equation.
v(t) = -Aωsin(ωt+$\varphi$)
Given any value of t, this descibes the velocity of the particle at that instant. Remember that the particle is at only one point x at that instant.
Now a wave extends over a region of space. The quantity we choose on the left hand side of the equation of motion will vary from point to point in space in a systematic manner. This quantity is not position (like with SHM). We call it the wave variable. It may be vertical displacement as in the case of the guitar string, or it may be a more complicated property like pressure in a sound wave.
Sticking with the guitar string, the vertical displacement, y is given by your equation for a wave.
Your equation is not correct, which is why I asked if you could see what was missing. It needs to contain the wave variable (Y in the case of our string) on the LHS and the independent variables position (x) and time (t) in the RHS.
Then we can say Y = Y(t,x) ie Y is a function of position and time.
This leads to the equation for a travelling wave
Y(x,t) = Acos (ωt-x)
How are we doing so far?
Do you want to ask questions at this stage or continue to the next stage which is to move on to the stationary wave?
All times are GMT -5. The time now is 04:36 AM.
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Physics Forums (http://www.physicsforums.com/index.php). - Classical Physics (http://www.physicsforums.com/forumdisplay.php?f=61). - - Difference between simple harmonic motion and stationary sinusoidal wave? (http://www.physicsforums.com/showthread.php?t=666620). tahayassen Jan24-13 04:18 PM. Difference between simple harmonic motion and stationary sinusoidal wave?. Their equations are identical. Is there any difference between the two?. Studiot Jan24-13 05:09 PM. Re: Difference between simple harmonic motion and stationary sinusoidal wave?. Difficult to coordinate three posts about your lecture notes and experimental lab, but here goes.. No the equations are not the same.. There is something in the equation of a stationary wave that is not in the equation of SHM.. Can you see what it is, because this difference is what makes SHM different from wave motion?. tahayassen Jan26-13 10:56 AM. Re: Difference between simple harmonic motion and stationary sinusoidal wave?. According to my professor.... Equation of SHM: $x(t)=Acos(ωt+ϕ)$. Equation of stationary sinusoidal wave: $x(t)=Acos(ωt)$. They look almost the same to me. I think she forgot the ϕ in the stationary sinusoidal wave equation.. According to my textbook, the position of SHM is a function of time while the position of a stationary sinusoidal wave is a function of x, but it doesn't actually give me an equation for a stationary sinusoidal wave.. Correct (?) equation of stationary sinusoidal wave: $y(x)=Acos(ωx+ϕ)$. vanhees71 Jan26-13 12:03 PM. Re: Difference between simple harmonic motion and stationary sinusoidal wave?. What textbook is this? Is this really in there? Then, I'd recommend to choose another one!. The wave equation, say for a string fixed between two points is given by. $$\frac{\partial^2}{\partial t^2}u(t,\vec{x})-c^2 \frac{\partial^2}{\partial x^2} u(t,\vec{x})=0,$$. where $u(t,\vec{x})$ is the displacement of the string at position $x$ at time $t$.. $$u(t,\vec{x})=u_0 \cos(\omega t \pm k x) \quad \text{with} \quad \omega=c k.$$. Here, $\omega=2 \pi f$ is the "angular frequency" ($f$ is the frequency) and $$k=2 \pi/\lambda$$ is the "wave number" ($\lambda$ is the wavelength). The upper sign describes a wave running to the left, the lower sign one running to the right.. For the string example, you must also fulfill boundary conditions, describing the fact that the points $x=-L/2$ and $x=+L/2$ are fixed:. $$u(t,\pm L/2)=0.$$. Here $L$ is the length of the string.. To find the alowed harmonic solutions you have to superimpose the left- and right-going solutions and determine the coefficients by fulfilling the boundary conditions:. $$u(t,x)=A \cos(\omega t-k x)+B \cos(\omega t + k x).$$. To find the coefficients we use the addition theorems for cos:. $$u(t,x)=A [\cos(\omega t) \cos(k x)-\sin(\omega t) \sin(k x)] + B [\cos(\omega t) \cos(k x)+\sin(\omega t)\sin(k x)].$$. This can be rearranged to. $$u(t,x)=(A+B) \cos(\omega t) \cos(k x) + (B-A) \sin(\omega t) \sin(k x).$$. To fulfill the boundary conditions. $$u(t,\pm L/2)$$. one must either have $A=B$ and $k L/2=(2n+1) \pi/2$ or $A=-B$ and $k \lambda/2=n \pi$, where $n \in \mathbb{N}$.. The ground wave is the one with the lowest allowd frequency, corresponding to the lowest allowed $k$, and that's the case 1, i.e., $A=B$ and $n=0$, i.e.,.
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$k=\pi/L$ and $\omega= c \pi/L[/tex]. The other plane-wave solutions refer to the higher harmonics. Any motion of the string can be described as a superposition of these socalled "normal modes". As you see from the fact that this same physics describes the sound of such different instruments, based on the motion of strings as a violine, a guitar, or a piano, making pretty different kinds of tones, the wave equation describes quite a large variety of motions of extended bodies (similar wave equations hold for any kind of sound or water waves and many other waves like those of the electromagnetic field, including light). Compared to this plethora of phenomena the simple harmonic oscillator is quite unexciting. It just describes the motion of a single point-like body attached to a spring or (approximately) a pendulum in the earths gravitational field and the like. It simply oscillates with the one frequency and doesn't do much more than just going back and forth ;-). The mathematical difference is that the wave equation is a partial differential equation and the simple harmonic oscillator is described by an ordinary differential equation. Studiot Jan26-13 12:41 PM Re: Difference between simple harmonic motion and stationary sinusoidal wave? I think this material will take a couple or three posts to get through, but there are some underlying ideas that you will be able to take forward to many areas of maths and physics, so it is worth the effort. First take these ideas through the discussion with you. We can strengthen it later. SHM is a form of motion that applies to a single particle. It does not propagate in space, although the particle itself moves through a fixed path. Since only one particle is involved there is one single format for SHM. An example is a plumb bob (pendulum) swinging back and forth. The particle is often called an oscillator or a vibrator and the motion an oscillation or a vibration. A wave is a form of motion that is undergone by an array of particles that are mechanically coupled so that the motion passes from one particle to the next continuously. We call this array the medium of propagation. Since there are many particles involved there is more than one format to wave motion. Examples are the waves on the surface of the sea and the vibrations of the string of a guitar. Let us look at the equations you have been given and see what can be made of them. Both wave motion and SHM (and all other forms of motion) are controlled by what is called an equation of motion. This links some quantity to the time and space axes, x and t. Clasically a single particle can only be in one place at a time and the quantity we choose is position. This leads to your first equation. x(t) = Acos(ωt+[itex]\varphi$). This describes the position of the particle, which is oscillating along the x axis, at any instant t.. If you differentiate x with respect to time dx/dt = velocity and you arrive at your second equation.. v(t) = -Aωsin(ωt+$\varphi$). Given any value of t, this descibes the velocity of the particle at that instant. Remember that the particle is at only one point x at that instant.. Now a wave extends over a region of space. The quantity we choose on the left hand side of the equation of motion will vary from point to point in space in a systematic manner. This quantity is not position (like with SHM). We call it the wave variable. It may be vertical displacement as in the case of the guitar string, or it may be a more complicated property like pressure in a sound wave.. Sticking with the guitar string, the vertical displacement, y is given by your equation for a wave.. Your equation is not correct, which is why I asked if you could see what was missing. It needs to contain the wave variable (Y in the case of our string) on the LHS and the independent variables position (x) and time (t) in the RHS.. Then we can say Y = Y(t,x) ie Y is a function of position and time.. This leads to the equation for a travelling wave. Y(x,t) = Acos (ωt-x). How are we doing so far?. Do you want to ask questions at this stage or continue to the next stage which is to move on to the stationary wave?. All times are GMT -5. The time now is 04:36 AM.
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# How many 15
How many six-eights are there in 18? Which of the following would you choose to solve this question?
18 x 6/8 or 18 : 6/8
n = 24
### Step-by-step explanation:
$n=18\mathrm{/}\frac{6}{8}=24$
Did you find an error or inaccuracy? Feel free to write us. Thank you!
Tips to related online calculators
Need help to calculate sum, simplify or multiply fractions? Try our fraction calculator.
## Related math problems and questions:
• Simplest form of a fraction
Which one of the following fraction after reducing in simplest form is not equal to 3/2? a) 15/20 b) 12/8 c) 27/18 d) 6/4
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The ratio of the number of giraffes to the number of monkeys in a zoo is 2 to 5. Which statement about the giraffes and monkeys could be true? A. For every 10 monkeys in the zoo, there are 4 giraffes. B. For every giraffe in the zoo, there are 3 monkeys.
• Six-eights
Six-eights of the one hundred pupils joined the Math Glee club. If the Math Glee club members were grouped into three, how many members were in each group?
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Of the 28 golf balls, 1/7 are yellow. How many golf balls are yellow? Use the model to help you. Enter your answer in the box.
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# How many 15. How many six-eights are there in 18? Which of the following would you choose to solve this question?. 18 x 6/8 or 18 : 6/8. n = 24. ### Step-by-step explanation:. $n=18\mathrm{/}\frac{6}{8}=24$. Did you find an error or inaccuracy? Feel free to write us. Thank you!. Tips to related online calculators. Need help to calculate sum, simplify or multiply fractions? Try our fraction calculator.. ## Related math problems and questions:. • Simplest form of a fraction. Which one of the following fraction after reducing in simplest form is not equal to 3/2? a) 15/20 b) 12/8 c) 27/18 d) 6/4. • Dividing walnuts into crates. There are 8 and 2 over 3 pounds of walnuts in a container, which will be divided equally into containers that hold 1 and 1 over 5 pounds. This would fill n and 4 over 18 containers. What is n?. • Shirts. In a classroom,1/6 of the students are wearing blue shirts, and 2/3 are wearing white shirts. There are 18 students in the classroom. How many students are wearing shirts other than blue shirts or white shirts?. • Jonah. Jonah has an 8-pound bag of potting soil. He divides it evenly among 5 flowerpots. How much soil is in each pot? Show your answer as a fraction or mixed number. Solve this problem any way you choose.. • A rope. A rope can be cut into equal length with no rope left over. The lengths can be 15cm,18cm or 25cm. What is the shortest possible length of the rope?. • Digits. How many odd four-digit numbers can we create from digits: 0, 3, 5, 6, 7? (a) the figures may be repeated (b) the digits may not be repeated. The size of two internal angles of a triangle ABC are α=6/18π and β=7/18π.
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Calculate the size of the third angle.. • To improper fraction. Change mixed number to improper fraction a) 1 2/15 b) -2 15/17. • Giraffes to monkeys. The ratio of the number of giraffes to the number of monkeys in a zoo is 2 to 5. Which statement about the giraffes and monkeys could be true? A. For every 10 monkeys in the zoo, there are 4 giraffes. B. For every giraffe in the zoo, there are 3 monkeys.. • Six-eights. Six-eights of the one hundred pupils joined the Math Glee club. If the Math Glee club members were grouped into three, how many members were in each group?. • Golf balls. Of the 28 golf balls, 1/7 are yellow. How many golf balls are yellow? Use the model to help you. Enter your answer in the box.. • Sally. Sally is making 15 individual blueberry cheesecakes. How many pounds of blueberries will be in each cheesecake if 5 1/4 pounds of blueberries are divided equally among them?. • Equation 15. Solve equation with variables on both sides:. • Plums. In the bowl are plums. How many would be there if we can divide it equally among 8, 10 and 11 children?. • Denissa. Denissa has 18 9/10 kilos of bananas to sell if she was able to sell 15 ⅗. How many kilos of bananas were left?. • Four pavers. Four pavers would pave the square in 18 days. How many pavers do you need to add to done work in 12 days?. • How many 16. How many three-tenths are there in two and one-fourths?.
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# Thread: C number operator question
1. ## C number operator question
here it is I am currently learning C by reading "Beginning C" by Ivor Horton and there is an example that is not explained as detailed as I would hope:
#include <stdio.h>
void main()
{
int number = 0;
int rebmun = 0;
int temp = 0;
printf("\nEnter a positive integer: ");
scanf("%d", &number);
temp = number;
do
{
rebmun = 10 * rebmun + temp % 10;
temp = temp/10;
} while(temp);
printf("\nThe number %d reversed is %d rebmun ehT\n", number, rebmun);
}
that which is bolded is what I have a question on, why does C claculate the modulus before multiplying. I have a chart that explains it but it has only:
* / % From left to right
now from what I learned back in Algebra that meant that they were all equal and would be performed in order they appeared except that the modulus operator is located after the multiplication. Some one please explain the logic behind this.
Sean
2. Actually, it goes like this:
multiplication, then division, then addition, then subtraction.
Thus, % is basicly division, so it is after multiplication.
In reality, if you want to be _sure_ of what you're doing, it is a good idea to use parenthesis (Just like in algebra) for clarification.
Quzah.
3. ## yes,but...
That I do understand but in the book it tells me that first it calculates the modulus and it is equal to 3 and that is before it multiplies rebmun by 10. Why would it not be in reverse order?
I know that with parenthesis it would make sense but without I am jsut confused about the example.
4. I don't think that it matters in your example since the plus is the last operation.
5. It does the modulus after the multiplication.
This is a three step problem:
1) rebmun is multiplied by 10.
2) temp is taken to the modulus of 10.
3) The results from 1) and 2) are added.
6. ## thanx
ok i think i got it tell me if i am wrong:
let's say we input 43
-first it multiply 10 by zero(since rembun is 0)
-then it found the remainder you get while dividing 43 by 10
which is 3
-then it took 43(what I inputed) and divided it by 10 and got 4
-then it went back and took rebmun(which is now equal to 3)
and multiplied it by ten and added 30 and 4 to equal 34
then some how temp became 0 and it printed out the opposite of 43 as 34
7. Correct -- the "somehow" of how temp becomes 0 is that in integer math, 4/10 = 0
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# Thread: C number operator question. 1. ## C number operator question. here it is I am currently learning C by reading "Beginning C" by Ivor Horton and there is an example that is not explained as detailed as I would hope:. #include <stdio.h>. void main(). {. int number = 0;. int rebmun = 0;. int temp = 0;. printf("\nEnter a positive integer: ");. scanf("%d", &number);. temp = number;. do. {. rebmun = 10 * rebmun + temp % 10;. temp = temp/10;. } while(temp);. printf("\nThe number %d reversed is %d rebmun ehT\n", number, rebmun);. }. that which is bolded is what I have a question on, why does C claculate the modulus before multiplying. I have a chart that explains it but it has only:. * / % From left to right. now from what I learned back in Algebra that meant that they were all equal and would be performed in order they appeared except that the modulus operator is located after the multiplication. Some one please explain the logic behind this.. Sean. 2. Actually, it goes like this:. multiplication, then division, then addition, then subtraction.. Thus, % is basicly division, so it is after multiplication.
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In reality, if you want to be _sure_ of what you're doing, it is a good idea to use parenthesis (Just like in algebra) for clarification.. Quzah.. 3. ## yes,but.... That I do understand but in the book it tells me that first it calculates the modulus and it is equal to 3 and that is before it multiplies rebmun by 10. Why would it not be in reverse order?. I know that with parenthesis it would make sense but without I am jsut confused about the example.. 4. I don't think that it matters in your example since the plus is the last operation.. 5. It does the modulus after the multiplication.. This is a three step problem:. 1) rebmun is multiplied by 10.. 2) temp is taken to the modulus of 10.. 3) The results from 1) and 2) are added.. 6. ## thanx. ok i think i got it tell me if i am wrong:. let's say we input 43. -first it multiply 10 by zero(since rembun is 0). -then it found the remainder you get while dividing 43 by 10. which is 3. -then it took 43(what I inputed) and divided it by 10 and got 4. -then it went back and took rebmun(which is now equal to 3). and multiplied it by ten and added 30 and 4 to equal 34. then some how temp became 0 and it printed out the opposite of 43 as 34. 7. Correct -- the "somehow" of how temp becomes 0 is that in integer math, 4/10 = 0.
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Friday
May 6, 2016
# Homework Help: Algebra 2
Posted by Devon on Tuesday, February 6, 2007 at 4:00pm.
What is the determinant of -2 -3
5 0
15
How you get 15
Use the definition of a determinant. From the first column you take take the element that is in the position of a permutation of 1, from the second you take the element that is in the position given by the permutation of 2. Multiply the result by the sign of the permutation and then sum over all possible permutations.
In this case the identity permutation yields zero, while the permutation that interchanges the numbers 1 and 2 yields -15. The sign of that permutation is minus one, so the result is 15.
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Friday. May 6, 2016. # Homework Help: Algebra 2. Posted by Devon on Tuesday, February 6, 2007 at 4:00pm.. What is the determinant of -2 -3. 5 0. 15.
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How you get 15. Use the definition of a determinant. From the first column you take take the element that is in the position of a permutation of 1, from the second you take the element that is in the position given by the permutation of 2. Multiply the result by the sign of the permutation and then sum over all possible permutations.. In this case the identity permutation yields zero, while the permutation that interchanges the numbers 1 and 2 yields -15. The sign of that permutation is minus one, so the result is 15.
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You are currently browsing the tag archive for the ‘The Barnsley Fern:’ tag.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
The Barnsley Fern: Mathematical Art
This pattern of a fern pictured above was generated by a simple iterative program designed by mathematician Michael Barnsely. I downloaded the Python code from the excellent Tutorialspoint and then modified it slightly to run on repl.it. What we are seeing is the result of 40,000 individual points – each plotted according to a simple algorithm. The algorithm is as follows:
Transformation 1: (0.85 probability of occurrence)
xi+1 = 0.85xi +0.04yi
yi+1= -0.04xi+0.85yi+1.6
Transformation 2: (0.07 probability of occurrence)
xi+1 = 0.2xi -0.26yi
yi+1= 0.23xi+0.22yi+1.6
Transformation 3: (0.07 probability of occurrence)
xi+1 = -0.15xi -0.28yi
yi+1= 0.26xi+0.24yi+0.44
Transformation 4: (0.01 probability of occurrence)
xi+1 = 0
yi+1= 0.16yi
So, I start with (0,0) and then use a random number generator to decide which transformation to use. I can run a generator from 1-100 and assign 1-85 for transformation 1, 86-92 to transformation 2, 93-99 for transformation 3 and 100 for transformation 4. Say I generate the number 36 – therefore I will apply transformation 1.
xi+1 = 0.85(0)+0.04(0)
yi+1= -0.04(0)+0.85(0)+1.6
and my new coordinate is (0,1.6). I mark this on my graph.
I then repeat this process – say this time I generate the number 90. This tells me to do transformation 2. So:
xi+1 = 0.2(0) -0.26(1.6)
yi+1= 0.23(0)+0.22(1.6)+1.6
and my new coordinate is (-0.416, 1.952). I mark this on my graph and carry on again. The graph above was generated with 40,000 iterations – let’s see how it develops over time:
1000 iterations:
10,000 iterations:
100,000 iterations:
500,000 iterations:
If we want to understand what is happening here we can think of each transformation as responsible for a different part of our fern. Transformation 1 is most likely and therefore this fills in the smaller leaflets. Transformations 2 and 3 fill in the bottom left and right leaflet (respectively) and transformation 4 fills in the stem.
It’s quite amazing to think that a simple computer program can create what looks like art – or indeed that is can replicate what we see in nature so well. This fern is an example of a self-similar pattern – i.e one which will look the same at different scales. You could zoom into a detailed picture and see the same patterns repeating. You might want to explore the idea of fractals in delving into this topic in more detail.
Changing the iterations
We can explore what happens when we change the iterations very slightly.
Christmas tree
Crazy spiral
Modern art
You can modify the code to run this here. Have a go!
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
1. Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
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3. Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
4. A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
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All content on this site has been written by Andrew Chambers (MSc. Mathematics, IB Mathematics Examiner).
### New website for International teachers
I’ve just launched a brand new maths site for international schools – over 2000 pdf pages of resources to support IB teachers. If you are an IB teacher this could save you 200+ hours of preparation time.
Explore here!
### Free HL Paper 3 Questions
P3 investigation questions and fully typed mark scheme. Packs for both Applications students and Analysis students.
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You are currently browsing the tag archive for the ‘The Barnsley Fern:’ tag.. If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!. The Barnsley Fern: Mathematical Art. This pattern of a fern pictured above was generated by a simple iterative program designed by mathematician Michael Barnsely. I downloaded the Python code from the excellent Tutorialspoint and then modified it slightly to run on repl.it. What we are seeing is the result of 40,000 individual points – each plotted according to a simple algorithm. The algorithm is as follows:. Transformation 1: (0.85 probability of occurrence). xi+1 = 0.85xi +0.04yi. yi+1= -0.04xi+0.85yi+1.6. Transformation 2: (0.07 probability of occurrence). xi+1 = 0.2xi -0.26yi. yi+1= 0.23xi+0.22yi+1.6. Transformation 3: (0.07 probability of occurrence). xi+1 = -0.15xi -0.28yi. yi+1= 0.26xi+0.24yi+0.44. Transformation 4: (0.01 probability of occurrence). xi+1 = 0. yi+1= 0.16yi. So, I start with (0,0) and then use a random number generator to decide which transformation to use. I can run a generator from 1-100 and assign 1-85 for transformation 1, 86-92 to transformation 2, 93-99 for transformation 3 and 100 for transformation 4. Say I generate the number 36 – therefore I will apply transformation 1.. xi+1 = 0.85(0)+0.04(0). yi+1= -0.04(0)+0.85(0)+1.6. and my new coordinate is (0,1.6). I mark this on my graph.. I then repeat this process – say this time I generate the number 90. This tells me to do transformation 2. So:. xi+1 = 0.2(0) -0.26(1.6). yi+1= 0.23(0)+0.22(1.6)+1.6. and my new coordinate is (-0.416, 1.952). I mark this on my graph and carry on again. The graph above was generated with 40,000 iterations – let’s see how it develops over time:. 1000 iterations:. 10,000 iterations:. 100,000 iterations:. 500,000 iterations:. If we want to understand what is happening here we can think of each transformation as responsible for a different part of our fern.
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Transformation 1 is most likely and therefore this fills in the smaller leaflets. Transformations 2 and 3 fill in the bottom left and right leaflet (respectively) and transformation 4 fills in the stem.. It’s quite amazing to think that a simple computer program can create what looks like art – or indeed that is can replicate what we see in nature so well. This fern is an example of a self-similar pattern – i.e one which will look the same at different scales. You could zoom into a detailed picture and see the same patterns repeating. You might want to explore the idea of fractals in delving into this topic in more detail.. Changing the iterations. We can explore what happens when we change the iterations very slightly.. Christmas tree. Crazy spiral. Modern art. You can modify the code to run this here. Have a go!. Essential Resources for IB Teachers. If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:. 1. Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.. 2. Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.. 3. Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.. 4. A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.. There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!. Essential Resources for both IB teachers and IB students. I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.. ### Website Stats. • 9,263,533 views. All content on this site has been written by Andrew Chambers (MSc. Mathematics, IB Mathematics Examiner).. ### New website for International teachers. I’ve just launched a brand new maths site for international schools – over 2000 pdf pages of resources to support IB teachers. If you are an IB teacher this could save you 200+ hours of preparation time.. Explore here!. ### Free HL Paper 3 Questions. P3 investigation questions and fully typed mark scheme. Packs for both Applications students and Analysis students.
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Question
# Your professor wishes to estimate the proportion of ALL high school students enrolled in college-level courses...
Your professor wishes to estimate the proportion of ALL high school students enrolled in college-level courses each school year. A sample of 1500 students revealed that 18.3% were enrolled in college-level courses. Find the margin of error for a 90% confidence interval for a proportion.
Solution :
Given that,
n = 1500
Point estimate = sample proportion = =18.3%=0.183
1 - = 1-0.183=0.817
At 90% confidence level
= 1 - 90%
= 1 - 0.90 =0.10
/2 = 0.05
Z/2 = Z0.05 = 1.645 ( Using z table )
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.645 (((0.183*0.817) / 1500)
= 0.016
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Question. # Your professor wishes to estimate the proportion of ALL high school students enrolled in college-level courses.... Your professor wishes to estimate the proportion of ALL high school students enrolled in college-level courses each school year. A sample of 1500 students revealed that 18.3% were enrolled in college-level courses. Find the margin of error for a 90% confidence interval for a proportion.. Solution :. Given that,. n = 1500. Point estimate = sample proportion = =18.3%=0.183. 1 - = 1-0.183=0.817. At 90% confidence level. = 1 - 90%. = 1 - 0.90 =0.10. /2 = 0.05. Z/2 = Z0.05 = 1.645 ( Using z table ). Margin of error = E = Z / 2 * (( * (1 - )) / n). = 1.645 (((0.183*0.817) / 1500). = 0.016. #### Earn Coins. Coins can be redeemed for fabulous gifts.. Similar Homework Help Questions. • ### A college admissions director wishes to estimate the mean age of all students currently enrolled. In.... A college admissions director wishes to estimate the mean age of all students currently enrolled. In a random sample of 16 students, the mean age is found to be 21.8 years. From past studies, the ages of enrolled students are normally distributed with a standard deviation of 10.1 years. Construct a 90% confidence interval for the mean age of all students currently enrolled. 1. The critical value: 2. The standard deviation of the sample mean: 3. The margin of error. • ### A college admissions director wishes to estimate the mean age of all students currently enrolled. In.... A college admissions director wishes to estimate the mean age of all students currently enrolled. In a random sample of 16 students, the mean age is found to be 21.8 years. From past studies, the ages of enrolled students are normally distributed with a standard deviation of 10.1 years. Construct a 90% confidence interval for the mean age of all students currently enrolled. 1. The critical value: 2. The standard deviation of the sample mean: 3. The margin of error:.... • ### A college admissions director wishes to estimate the mean age of all students currently enrolled. In.... A college admissions director wishes to estimate the mean age of all students currently enrolled. In a random sample of 24 students, the mean age is found to be 23.1 years. From past studies, the ages of enrolled students are normally distributed with a standard deviation of 10.6 years. Construct a 90% confidence interval for the mean age of all students currently enrolled. 1.
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The critical value: 2. The standard deviation of the sample mean: 3. The margin of error:.... • ### answer B A department of education reported that in 2007, 66% of students enrolled in college.... answer B A department of education reported that in 2007, 66% of students enrolled in college or a trade school within 12 months of graduating from high school. In 2013, a random sample of 320 individuals who graduated from high school 12 months prior was selected. From this sample, 236 students were found to be enrolled in college or a trade school. Complete parts a through c below. a. Construct a 95% confidence interval to estimate the actual proportion of.... • ### A department of education reported that in 2007, 68% of students enrolled in college or a.... A department of education reported that in 2007, 68% of students enrolled in college or a trade school within 12 months of graduating from high school. In 2013, a random sample of 160 individuals who graduated from high school 12 months prior was selected. From this sample, 106 students were found to be enrolled in college or a trade school. Complete parts a through c below. a. Construct a 90% confidence interval to estimate the actual proportion of students enrolled.... • ### A department of education reported that in 2007, 63% of students enrolled in college or a.... A department of education reported that in 2007, 63% of students enrolled in college or a trade school within 12 months of graduating from high school. In 2013, a random sample of 320 individuals who graduated from high school 12 months prior was selected. From this sample, 204 students were found to be enrolled in college or a trade school. Complete parts a through c below. a. Construct a 90% confidence interval to estimate the actual proportion of students enrolled.... • ### A country's education department reported that in2015,66.8% of students enrolled in college or a trade school.... A country's education department reported that in2015,66.8% of students enrolled in college or a trade school within 12months of graduating high school. In 2017, a random sample of171individuals who graduated from high school12months prior was selected. From this sample,108 students were found to be enrolled in college or a trade school. a. Construct a90%confidence interval to estimate the actual proportion of students enrolled in college or a trade school within 12months of graduating from high school in 2017.. • ### A department of education reported that in 2007, 66% of students enrolled in college or a.... A department of education reported that in 2007, 66% of students enrolled in college or a trade school within 12 months of graduating from high school. In 2013, a random sample of 320 individuals who graduated from high school 12 months prior was selected. From this sample, 236 students were found to be enrolled in college or a trade school. Complete parts a through c below. a. Construct a 95% confidence interval to estimate the actual proportion of students enrolled.... • ### A department of education reported that in 2007, 68% of students enrolled in college or a.... A department of education reported that in 2007, 68% of students enrolled in college or a trade school within 12 months of graduating from high school. In 2013, a random sample of 160 individuals who graduated from high school 12 months prior was selected. From this sample, 110 students were found to be enrolled in college or a trade school. Complete parts a through c below. a. Construct a 99% confidence interval to estimate the actual proportion of students enrolled.... • ### What proportion of JMU students like online courses? To estimate this, a random sample of 200.... What proportion of JMU students like online courses? To estimate this, a random sample of 200 students was taken and found that 50 of them like online courses. (1) Are the assumptions for constructing a confidence interval for the population proportion satisfied? Explain.(2) Construct a 99% confidence interval for the population proportion of JMU students who like online courses.(3) What is the margin of error in (2) ?______________(4) Interpret the 99% confidence interval.
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You are on page 1of 7
# Statistics Independent Study
Page 1 of 7
Mixed Practice with Multiple-Choice and Free-Response Questions
Multiple-Choice Section
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You believe that men and women might have differing reactions to the medication. 1. All rights reserved. Answer: C U M R E L F R E Q 1.08 attempts.Statistics Independent Study Page 2 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers Free-Response Section Introduction These questions simulate the Free-Response section of the AP Exam. Answer: Blocks would be desirable here." and should take about 20-30 minutes. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.3 0.9 0. You are also concerned with differences between those 60 and over and those under 60. _____________ © Copyright 2000 Apex Learning Inc.7 0. Question 6 is an "Investigative Task.6 0. . A.2 0. The median is found at the fiftieth percentile. 2. Questions 1-5 should take about 10-15 minutes each.4 0. so the median is 1. so it would be appropriate to block by gender here.8 0. More than half passed on the first trial.1 0.0 1234567 NUMBER OF ATTEMPTS B.0 0. You should do all that you can do in no more than 90 minutes. Answer: µ = (1)(40/75) + (2)(12/75) + (3)(8/75) + (4)(9/75) + 5(5/75) + 6(0/75) + 7(1/75) = 2. A.5 0. so you would also need to divide those under 60 and those 60 and over into different blocks. This material is intended for the exclusive use of registered users only.
345 1. you would want to conduct a two-sample t test for the difference between population means. you would find a P-value of .333. assume there are equal numbers in each block. Answer: H0 : µ d = 0 (The mean difference between the before and after scores is 0. whether you pool your estimates of the population variance. We can use a onesample t procedure. Within each block. The computation you would use for the standard error of the difference would depend on whether you were using the conservative method. randomly assign 1/3 of the subjects to a control group that will receive a placebo. Then break each of these two groups into two additional blocks of those 60 and over and those under 60. and 1/3 to a treatment group that will receive 500 mg tablets. and women 60 and over. women under 60. This gives you four blocks of 45 people each. . 3. or whether you use computer software. 1/3 to a treatment group that will receive 350 mg tablets. If randomization has been done correctly. After a designated period of time.) Since the scores are matched (that is. but not enough to conclude that this sample is not reasonably normal.969 12 df = 11 è . a one-sample t test is appropriate for the differences between the sample values.019. there are approximately 15 people that receive the 350 mg pill and 15 that receive the 500 mg pill. The study should be double-blind: Neither the subjects nor the person administering the tests should know which subjects are receiving which treatment. A boxplot of the data shows no outliers. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.) H a : µ d > 0 (The mean difference is greater than zeroscores were higher after training. All rights reserved. For the sake of discussing the design. Answer: In any one block. C.333 − 0 = 2. Compute an average relief score for each group.) _____________ © Copyright 2000 Apex Learning Inc. n = 12 t = 1. the assumption that the two groups are independent is reasonable. compare pain-relief scores for each of the three groups within each block. If these conditions are satisfied.969. x = 1. there are two different measurements for each individual). There is some skewness to the left. This results in four blocks: men under 60. s = 1. You must also check to see that the distribution of scores for each group does not indicate that the samples might have been drawn from non-normal populations. Answer: Divide your 180 volunteers into two groups based on their gender.01 < p < . This material is intended for the exclusive use of registered users only. men 60 and over.Statistics Independent Study Page 3 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers B.02 (On the TI-83.
6)(.) A one-proportion z-test is appropriate here. P(at least one match in first five drawings) = 1 – P(no matches in the first five drawings) 5 æ 45 ö =1– ç ÷ = . Answer: (6/51)(1 – 6/51)4 = .0418 This value of P is less than the significance level of .01.) Ha : p > . . We are justified in using a z-test here because 35(.071 5. All rights reserved.6 (.05. 4. It is reasonable to conclude that the program was effective in improving mathematics skills in the subjects. so we have good evidence to reject our null hypothesis. since the school believes that their students will do better. you would reach a different conclusion. z = p . ˆ = 26/35 = . The claim that the school's students do better than the national average of . Answer: æ 51ö (6/51)(5/50)(4/49)(3/48)(2/47)(1/46) = 1 /çç ÷÷ = 1/18009460 = .6 is supported by these data.9582 = . è 51 ø C.4) 35 = 1.6 (The proportion of passing grades is . H0 : p = . Note: If you used a significance level of .743. A.465. A.Statistics Independent Study Page 4 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers This P-value is quite low and gives us strong evidence against the null hypothesis. This material is intended for the exclusive use of registered users only.6 (The proportion of passing grades is larger than . Answer: Now. _____________ © Copyright 2000 Apex Learning Inc.0000000555.6. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.4) and 35(.6) are both larger than 10.73 è P-value = 1 – .6.743 − . Answer: It is a one-sided test. the probability of a match remains constant at 6/51. è6 ø B.
9324 = .5. Answer: A histogram of these data (below) shows a pattern that seems to be quite normal.085(500) = 42. 219. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.119 – . we would not have rejected the null hypothesis and would conclude that the evidence was unconvincing that the school is actually doing better than the national average. 6. Answer: The worst thing that could happen in terms of the school's claim is that the missing .5 − 275 = –1.) Expected frequency in the interval (189.034 = .5. All rights reserved. (If you use the TI-83 you will get 42.6 ˆ = 26/36 = . We need to find the area under the normal curve for each interval by using either a calculator or standard z-score techniques.5) = .18 è Upper area = .034 47 Area in the interval (189.722.085.4) 36 è P-value = 1 – .5) = . In this case.Statistics Independent Study Page 5 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers B. Under these conditions.5 could be calculated as follows: z 219.0676. . we would find.084.05. that the area corresponding to the interval 189.82 è Lower area = .722 − .5 = 189. for example. we can use a χ 2 goodness-of-fit test to compare the observed frequencies in each interval with the expected frequencies.19.5 – 219.5 − 275 = –1. This value is greater than the significance level of .119 47 z189.) _____________ © Copyright 2000 Apex Learning Inc.5. z = = 1. This material is intended for the exclusive use of registered users only. (If you use the TI-83 you will get . 219. p .494 student would not have passed.6(.5 = 219. If we assume the data are from a normal distribution. Assuming that the distribution from which the sample was drawn has a mean of 275 and a standard deviation of 47.
44 122.24 31.25 (p = .715 on the TI-83).02 For these data.5 369.5 309. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc. All rights reserved.5 – 219. Thus we have no evidence to argue against the claim that the cholesterol levels of American men are normally distributed.5 – 309. X 2 = 5. _____________ © Copyright 2000 Apex Learning Inc.5 279. df = 9 – 1 = 8 è p > .5 339.5 – 279.22 42.5 – 399.40 9.5 – 369.19 87.Statistics Independent Study Page 6 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers Proceeding similarly for each interval and using the results from the TI-83.5 219.5 – 249.07 2. This material is intended for the exclusive use of registered users only.39.21 115.5 Over 399.5 249.5 – 339. .5 189. we can add a third column to the table: Cholesterol Level Less than 189.20 73.5 Observed Frequency 17 41 80 134 118 71 24 13 2 Expected Frequency 17. A P-value this high indicates that a set of observations such as this is quite likely if the sample were in fact drawn from a normal population.
This material is intended for the exclusive use of registered users only. Overall Score 390 – 560 305 – 389 225 – 304 160 – 224 0 – 159 _____________ AP Approximation 5 4 3 2 1 © Copyright 2000 Apex Learning Inc. You can roughly approximate your score on the AP Exam by using the table below. . No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc. All rights reserved.Statistics Independent Study Page 7 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers Scoring To determine your score. Question 6 is graded on an eight-point scale. Determine your score on this section as follows: Score = question 1 + question 2 + question 3 + question 4 + question 5 + question 6. Free-Response Section Questions 1–5 are graded on a four-point scale. Multiple-Choice Section Determine your score on this section as follows: Score = number of correct multiple-choice questions – 1/4(number of incorrect multiplechoice questions). Overall Score Overall score = (multiple-choice score • 7) + (free-response score • 10). use the following guidelines.
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You are on page 1of 7. # Statistics Independent Study. Page 1 of 7. Mixed Practice with Multiple-Choice and Free-Response Questions. Multiple-Choice Section. _____________. 1.. C. 21.. B. 2.. B. 22.. D. 3.. C. 23.. A. 4.. D. 24.. E. 5.. D. 25.. C. 6.. D. 26.. B. 7.. B. 27.. E. 8.. B. 28.. A. 9.. D. 29.. B. 10.. A. 30.. C. 11.. D. 31.. D. 12.. A. 32.. E. 13.. E. 33.. A. 14.. C. 34.. D. 15.. A. 35.. A. 16.. B. 36.. E. 17.. E. 37.. C. 18.. A. 38.. B. 19.. E. 39.. D. 20.. E. 40.. C. registered users only. No portion of these materials may be reproduced or redistributed in any form without the. express written permission of Apex Learning Inc.. You believe that men and women might have differing reactions to the medication. 1. All rights reserved. Answer: C U M R E L F R E Q 1.08 attempts.Statistics Independent Study Page 2 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers Free-Response Section Introduction These questions simulate the Free-Response section of the AP Exam. Answer: Blocks would be desirable here." and should take about 20-30 minutes. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.3 0.9 0. You are also concerned with differences between those 60 and over and those under 60. _____________ © Copyright 2000 Apex Learning Inc.7 0. Question 6 is an "Investigative Task.6 0. . A.2 0. The median is found at the fiftieth percentile. 2. Questions 1-5 should take about 10-15 minutes each.4 0. so the median is 1. so it would be appropriate to block by gender here.8 0.
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More than half passed on the first trial.1 0.0 1234567 NUMBER OF ATTEMPTS B.0 0. You should do all that you can do in no more than 90 minutes. Answer: µ = (1)(40/75) + (2)(12/75) + (3)(8/75) + (4)(9/75) + 5(5/75) + 6(0/75) + 7(1/75) = 2. A.5 0. so you would also need to divide those under 60 and those 60 and over into different blocks. This material is intended for the exclusive use of registered users only.. 345 1. you would want to conduct a two-sample t test for the difference between population means. you would find a P-value of .333. assume there are equal numbers in each block. Answer: H0 : µ d = 0 (The mean difference between the before and after scores is 0. whether you pool your estimates of the population variance. We can use a onesample t procedure. Within each block. The computation you would use for the standard error of the difference would depend on whether you were using the conservative method. randomly assign 1/3 of the subjects to a control group that will receive a placebo. Then break each of these two groups into two additional blocks of those 60 and over and those under 60. and 1/3 to a treatment group that will receive 500 mg tablets. and women 60 and over. women under 60. This gives you four blocks of 45 people each. . 3. or whether you use computer software. 1/3 to a treatment group that will receive 350 mg tablets. If randomization has been done correctly. After a designated period of time.) Since the scores are matched (that is. but not enough to conclude that this sample is not reasonably normal.969 12 df = 11 è . a one-sample t test is appropriate for the differences between the sample values.019. there are approximately 15 people that receive the 350 mg pill and 15 that receive the 500 mg pill. The study should be double-blind: Neither the subjects nor the person administering the tests should know which subjects are receiving which treatment. A boxplot of the data shows no outliers. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.) H a : µ d > 0 (The mean difference is greater than zeroscores were higher after training. All rights reserved. For the sake of discussing the design. Answer: In any one block. C.333 − 0 = 2. Compute an average relief score for each group.) _____________ © Copyright 2000 Apex Learning Inc. n = 12 t = 1. the assumption that the two groups are independent is reasonable. compare pain-relief scores for each of the three groups within each block. If these conditions are satisfied.969. x = 1. there are two different measurements for each individual). There is some skewness to the left. This results in four blocks: men under 60. s = 1. You must also check to see that the distribution of scores for each group does not indicate that the samples might have been drawn from non-normal populations. Answer: Divide your 180 volunteers into two groups based on their gender.01 < p < . This material is intended for the exclusive use of registered users only. men 60 and over.Statistics Independent Study Page 3 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers B.02 (On the TI-83.. 6)(.) A one-proportion z-test is appropriate here. P(at least one match in first five drawings) = 1 – P(no matches in the first five drawings) 5 æ 45 ö =1– ç ÷ = . Answer: (6/51)(1 – 6/51)4 = .0418 This value of P is less than the significance level of .01.) Ha : p > . . We are justified in using a z-test here because 35(.071 5. All rights reserved.6 (.05. 4. It is reasonable to conclude that the program was effective in improving mathematics skills in the subjects. so we have good evidence to reject our null hypothesis. since the school believes that their students will do better. you would reach a different conclusion. z = p . ˆ = 26/35 = . The claim that the school's students do better than the national average of . Answer: æ 51ö (6/51)(5/50)(4/49)(3/48)(2/47)(1/46) = 1 /çç ÷÷ = 1/18009460 = .6 is supported by these data.9582 = . è 51 ø C.4) 35 = 1.6 (The proportion of passing grades is . H0 : p = . Note: If you used a significance level of .743. A.465. A.Statistics Independent Study Page 4 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers This P-value is quite low and gives us strong evidence against the null hypothesis. This material is intended for the exclusive use of registered users only.6 (The proportion of passing grades is larger than . Answer: Now. _____________ © Copyright 2000 Apex Learning Inc.0000000555.6. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.4) and 35(.6) are both larger than 10.73 è P-value = 1 – .6.743 − . Answer: It is a one-sided test. the probability of a match remains constant at 6/51. è6 ø B.. 9324 = .5. Answer: A histogram of these data (below) shows a pattern that seems to be quite normal.085(500) = 42. 219. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.119 – . we would not have rejected the null hypothesis and would conclude that the evidence was unconvincing that the school is actually doing better than the national average. 6. Answer: The worst thing that could happen in terms of the school's claim is that the missing .5 − 275 = –1.) Expected frequency in the interval (189.034 = .5. All rights reserved. (If you use the TI-83 you will get 42.6 ˆ = 26/36 = . We need to find the area under the normal curve for each interval by using either a calculator or standard z-score techniques.5) = .18 è Upper area = .034 47 Area in the interval (189.722.085.4) 36 è P-value = 1 – .5) = . In this case.Statistics Independent Study Page 5 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers B. Under these conditions.5 could be calculated as follows: z 219.0676. . we would find.084.05. that the area corresponding to the interval 189.82 è Lower area = .722 − .5 = 189. for example. we can use a χ 2 goodness-of-fit test to compare the observed frequencies in each interval with the expected frequencies.19.5 – 219.5 − 275 = –1. This value is greater than the significance level of .119 47 z189.) _____________ © Copyright 2000 Apex Learning Inc.5. z = = 1. This material is intended for the exclusive use of registered users only. (If you use the TI-83 you will get . 219. p .494 student would not have passed.6(.5 = 219. If we assume the data are from a normal distribution. Assuming that the distribution from which the sample was drawn has a mean of 275 and a standard deviation of 47.. 44 122.24 31.25 (p = .715 on the TI-83).02 For these data.5 369.5 309. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc. All rights reserved.5 – 219. Thus we have no evidence to argue against the claim that the cholesterol levels of American men are normally distributed.5 – 309. X 2 = 5. _____________ © Copyright 2000 Apex Learning Inc.5 279. df = 9 – 1 = 8 è p > .5 339.5 – 279.22 42.5 – 399.40 9.5 – 369.19 87.Statistics Independent Study Page 6 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers Proceeding similarly for each interval and using the results from the TI-83.5 219.5 – 249.07 2. This material is intended for the exclusive use of registered users only.39.21 115.5 Over 399.5 249.5 – 339. .5 189. we can add a third column to the table: Cholesterol Level Less than 189.20 73.5 Observed Frequency 17 41 80 134 118 71 24 13 2 Expected Frequency 17. A P-value this high indicates that a set of observations such as this is quite likely if the sample were in fact drawn from a normal population.. This material is intended for the exclusive use of registered users only. Overall Score 390 – 560 305 – 389 225 – 304 160 – 224 0 – 159 _____________ AP Approximation 5 4 3 2 1 © Copyright 2000 Apex Learning Inc. You can roughly approximate your score on the AP Exam by using the table below. . No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc. All rights reserved.Statistics Independent Study Page 7 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers Scoring To determine your score. Question 6 is graded on an eight-point scale. Determine your score on this section as follows: Score = question 1 + question 2 + question 3 + question 4 + question 5 + question 6. Free-Response Section Questions 1–5 are graded on a four-point scale. Multiple-Choice Section Determine your score on this section as follows: Score = number of correct multiple-choice questions – 1/4(number of incorrect multiplechoice questions). Overall Score Overall score = (multiple-choice score • 7) + (free-response score • 10). use the following guidelines.
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## Engage NY Eureka Math 4th Grade Module 3 Lesson 37 Answer Key
### Eureka Math Grade 4 Module 3 Lesson 37 Problem Set Answer Key
Question 1.
Solve 14 × 12 using 4 partial products and 2 partial products. Remember to think in terms of units as you solve.
Write an expression to find the area of each smaller rectangle in the area model.
Explanation:
Solved 14 × 12 = 168 using 4 partial products and 2 partial products.
Wrote an expression to find the area of each smaller rectangle in the area model as shown above.
Question 2.
Solve 32 × 43 using 4 partial products and 2 partial products. Match each partial product to its area on the models. Remember to think in terms of units as you solve.
Explanation:
Solved 32 × 43 = 1,376 using 4 partial products and 2 partial products. Matched each partial product to its area on the models as shown above.
Question 3.
Solve 57 × 15 using 2 partial products. Match each partial product to its rectangle on the area model.
Explanation:
Solved 57 X 15 = 855 using 2 partial products. Matched each partial product to its rectangle on the area model as shown above.
Question 4.
Solve the following using 2 partial products.
a.
Explanation:
Solved the following 46 X 25 = 1,150 using 2 partial products as shown above visualized the area model.
b.
Explanation:
Solved the following 62 X 18 = 1,116 using 2 partial products as shown above visualized the area model.
c.
Explanation:
Solved the following 46 X 39 = 1,794 using 2 partial products as shown above visualized the area model.
d.
Explanation:
Solved the following 23 X 78 = 1,794 using 2 partial products as shown above visualized the area model.
### Eureka Math Grade 4 Module 3 Lesson 37 Exit Ticket Answer Key
Question 1.
Solve 43 × 22 using 4 partial products and 2 partial products.
Remember to think in terms of units as you solve.
Write an expression to find the area of each smaller rectangle in the area model.
Explanation:
Solved 43 × 22 = 946 using 4 partial products and 2 partial products.
Wrote an expression to find the area of each smaller rectangle in the area model as shown above.
Question 2.
Solve the following using 2 partial products.
Explanation:
Solved 15 X 64 = 960 using 2 partial products.
### Eureka Math Grade 4 Module 3 Lesson 37 Homework Answer Key
Question 1.
Solve 26 × 34 using 4 partial products and 2 partial products.
Remember to think in terms of units as you solve.
Write an expression to find the area of each smaller rectangle in the area model.
Explanation:
Solved 26 × 34 = 884 using 4 partial products and 2 partial products.
Wrote an expression to find the area of each smaller rectangle in the area model as shown above.
Question 2.
Solve using 4 partial products and 2 partial products. Remember to think in terms of units as you solve.
Write an expression to find the area of each smaller rectangle in the area model.
Explanation:
Solved 82 × 41 = 3,362 using 4 partial products and 2 partial products.
Wrote an expression to find the area of each smaller rectangle in the area model as shown above.
Question 3.
Solve 52 × 26 using 2 partial products and an area model. Match each partial product to its area on the model.
Explanation:
Solved 52 X 26 = 1,352 using 2 partial products and an area model,
Matched each partial product to its area model as shown above.
Question 4.
Solve the following using 2 partial products.
a.
Explanation:
Solved the following 23 X 68 = 1,564 using 2 partial products as shown above visualized the area model.
b.
Explanation:
Solved the following 33 X 49 = 1,617 using 2 partial products as shown above visualized the area model.
c.
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## Engage NY Eureka Math 4th Grade Module 3 Lesson 37 Answer Key. ### Eureka Math Grade 4 Module 3 Lesson 37 Problem Set Answer Key. Question 1.. Solve 14 × 12 using 4 partial products and 2 partial products. Remember to think in terms of units as you solve.. Write an expression to find the area of each smaller rectangle in the area model.. Explanation:. Solved 14 × 12 = 168 using 4 partial products and 2 partial products.. Wrote an expression to find the area of each smaller rectangle in the area model as shown above.. Question 2.. Solve 32 × 43 using 4 partial products and 2 partial products. Match each partial product to its area on the models. Remember to think in terms of units as you solve.. Explanation:. Solved 32 × 43 = 1,376 using 4 partial products and 2 partial products. Matched each partial product to its area on the models as shown above.. Question 3.. Solve 57 × 15 using 2 partial products. Match each partial product to its rectangle on the area model.. Explanation:. Solved 57 X 15 = 855 using 2 partial products. Matched each partial product to its rectangle on the area model as shown above.. Question 4.. Solve the following using 2 partial products.. a.. Explanation:. Solved the following 46 X 25 = 1,150 using 2 partial products as shown above visualized the area model.. b.. Explanation:. Solved the following 62 X 18 = 1,116 using 2 partial products as shown above visualized the area model.. c.. Explanation:. Solved the following 46 X 39 = 1,794 using 2 partial products as shown above visualized the area model.. d.. Explanation:. Solved the following 23 X 78 = 1,794 using 2 partial products as shown above visualized the area model.. ### Eureka Math Grade 4 Module 3 Lesson 37 Exit Ticket Answer Key. Question 1.. Solve 43 × 22 using 4 partial products and 2 partial products.. Remember to think in terms of units as you solve.
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Write an expression to find the area of each smaller rectangle in the area model.. Explanation:. Solved 43 × 22 = 946 using 4 partial products and 2 partial products.. Wrote an expression to find the area of each smaller rectangle in the area model as shown above.. Question 2.. Solve the following using 2 partial products.. Explanation:. Solved 15 X 64 = 960 using 2 partial products.. ### Eureka Math Grade 4 Module 3 Lesson 37 Homework Answer Key. Question 1.. Solve 26 × 34 using 4 partial products and 2 partial products.. Remember to think in terms of units as you solve.. Write an expression to find the area of each smaller rectangle in the area model.. Explanation:. Solved 26 × 34 = 884 using 4 partial products and 2 partial products.. Wrote an expression to find the area of each smaller rectangle in the area model as shown above.. Question 2.. Solve using 4 partial products and 2 partial products. Remember to think in terms of units as you solve.. Write an expression to find the area of each smaller rectangle in the area model.. Explanation:. Solved 82 × 41 = 3,362 using 4 partial products and 2 partial products.. Wrote an expression to find the area of each smaller rectangle in the area model as shown above.. Question 3.. Solve 52 × 26 using 2 partial products and an area model. Match each partial product to its area on the model.. Explanation:. Solved 52 X 26 = 1,352 using 2 partial products and an area model,. Matched each partial product to its area model as shown above.. Question 4.. Solve the following using 2 partial products.. a.. Explanation:. Solved the following 23 X 68 = 1,564 using 2 partial products as shown above visualized the area model.. b.. Explanation:. Solved the following 33 X 49 = 1,617 using 2 partial products as shown above visualized the area model.. c.
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# The diameter of a sphere is 42 cm.
Question:
The diameter of a sphere is 42 cm. It is melted and drawn into a cylindrical wire of diameter 2.8 cm. Find the length of the wire.
Solution:
Diameter of sphere $=42 \mathrm{~cm}$
Radius of sphere $=21 \mathrm{~cm}$
Volume of sphere $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi \times 21 \times 21 \times 21 \mathrm{~cm}^{3}$
Diameter of wire $=2.8 \mathrm{~cm}$
Radius of wire $=1.4 \mathrm{~cm}$
Let the length of the wire be $/ \mathrm{cm}$.
Volume of the wire $=\pi r^{2} l=\pi \times 1.4 \times 1.4 \times 1$
The volume of the sphere is equal to the volume of the wire.
Therefore,
$\pi \times 1.4 \times 1.4 \times l=\frac{4}{3} \pi \times 21 \times 21 \times 21$
$l=\frac{4 \times 21 \times 21 \times 21}{3 \times 1.4 \times 1.4}=6300 \mathrm{~cm}=63 \mathrm{~m}$
So, the wire is 63 m long.
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# The diameter of a sphere is 42 cm.. Question:. The diameter of a sphere is 42 cm. It is melted and drawn into a cylindrical wire of diameter 2.8 cm. Find the length of the wire.. Solution:. Diameter of sphere $=42 \mathrm{~cm}$. Radius of sphere $=21 \mathrm{~cm}$. Volume of sphere $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi \times 21 \times 21 \times 21 \mathrm{~cm}^{3}$. Diameter of wire $=2.8 \mathrm{~cm}$.
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Radius of wire $=1.4 \mathrm{~cm}$. Let the length of the wire be $/ \mathrm{cm}$.. Volume of the wire $=\pi r^{2} l=\pi \times 1.4 \times 1.4 \times 1$. The volume of the sphere is equal to the volume of the wire.. Therefore,. $\pi \times 1.4 \times 1.4 \times l=\frac{4}{3} \pi \times 21 \times 21 \times 21$. $l=\frac{4 \times 21 \times 21 \times 21}{3 \times 1.4 \times 1.4}=6300 \mathrm{~cm}=63 \mathrm{~m}$. So, the wire is 63 m long.
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# Derivation of Thermodynamic Relations
• Engineering
• lohboys
In summary, the basic thermodynamic relations include the first and second laws of thermodynamics, the Maxwell relations, and the Gibbs-Duhem equation. They describe the behavior of thermodynamic systems and the relationships between different properties. The first law of thermodynamics is derived from the principle of conservation of energy, while the Maxwell relations are derived from the total differential of a property. The Gibbs-Duhem equation is a fundamental relation that describes the relationship between changes in chemical potential, temperature, and pressure. These relations are used in practical applications such as designing heat engines and studying phase transitions.
lohboys
Homework Statement
I'm trying to derive a thermodynamic relation in terms of P,V,T,Cp,Cv and was wondering if my working was correct. I am trying to derive (dG/dV) at constant temperature
Relevant Equations
dG= -SdT + VdP
dG= -SdT + VdP ... now dividing by dV holding temperature constant
(dG/dV)T = -S (dT/dV)T + V (dP/dV)T ... now dT and constant temperature cancel out
(dG/dV)T = V (dP/dV)T
Hi. The mathematics seems all right.
## 1. What is the derivation of thermodynamic relations?
The derivation of thermodynamic relations is the process of mathematically deriving equations that relate different thermodynamic properties, such as temperature, pressure, and volume, to each other. These relations are based on the fundamental laws of thermodynamics and are used to understand and predict the behavior of thermodynamic systems.
## 2. Why is the derivation of thermodynamic relations important?
The derivation of thermodynamic relations is important because it allows us to understand and analyze the behavior of thermodynamic systems using mathematical equations. These equations can be used to make predictions and solve problems related to energy transfer, phase changes, and other thermodynamic processes.
## 3. What are some common thermodynamic relations?
Some common thermodynamic relations include the ideal gas law, the first law of thermodynamics, and the second law of thermodynamics. Other important relations include the Maxwell relations, the Clausius-Clapeyron equation, and the Gibbs-Duhem equation.
## 4. How are thermodynamic relations derived?
Thermodynamic relations are derived using mathematical techniques such as calculus and algebra. These techniques are applied to the fundamental laws of thermodynamics and other thermodynamic equations to derive new relationships between thermodynamic properties.
## 5. What are some real-world applications of thermodynamic relations?
Thermodynamic relations have many real-world applications, including in the design of engines and power plants, the study of atmospheric processes, and the development of new materials and technologies. They are also used in many industrial processes, such as refrigeration, chemical reactions, and energy production.
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# Derivation of Thermodynamic Relations. • Engineering. • lohboys. In summary, the basic thermodynamic relations include the first and second laws of thermodynamics, the Maxwell relations, and the Gibbs-Duhem equation. They describe the behavior of thermodynamic systems and the relationships between different properties. The first law of thermodynamics is derived from the principle of conservation of energy, while the Maxwell relations are derived from the total differential of a property. The Gibbs-Duhem equation is a fundamental relation that describes the relationship between changes in chemical potential, temperature, and pressure. These relations are used in practical applications such as designing heat engines and studying phase transitions.. lohboys. Homework Statement. I'm trying to derive a thermodynamic relation in terms of P,V,T,Cp,Cv and was wondering if my working was correct. I am trying to derive (dG/dV) at constant temperature. Relevant Equations. dG= -SdT + VdP. dG= -SdT + VdP ... now dividing by dV holding temperature constant. (dG/dV)T = -S (dT/dV)T + V (dP/dV)T ... now dT and constant temperature cancel out. (dG/dV)T = V (dP/dV)T. Hi. The mathematics seems all right.. ## 1. What is the derivation of thermodynamic relations?. The derivation of thermodynamic relations is the process of mathematically deriving equations that relate different thermodynamic properties, such as temperature, pressure, and volume, to each other. These relations are based on the fundamental laws of thermodynamics and are used to understand and predict the behavior of thermodynamic systems.. ## 2. Why is the derivation of thermodynamic relations important?. The derivation of thermodynamic relations is important because it allows us to understand and analyze the behavior of thermodynamic systems using mathematical equations. These equations can be used to make predictions and solve problems related to energy transfer, phase changes, and other thermodynamic processes.. ## 3. What are some common thermodynamic relations?. Some common thermodynamic relations include the ideal gas law, the first law of thermodynamics, and the second law of thermodynamics. Other important relations include the Maxwell relations, the Clausius-Clapeyron equation, and the Gibbs-Duhem equation.. ## 4. How are thermodynamic relations derived?. Thermodynamic relations are derived using mathematical techniques such as calculus and algebra. These techniques are applied to the fundamental laws of thermodynamics and other thermodynamic equations to derive new relationships between thermodynamic properties.. ## 5. What are some real-world applications of thermodynamic relations?. Thermodynamic relations have many real-world applications, including in the design of engines and power plants, the study of atmospheric processes, and the development of new materials and technologies. They are also used in many industrial processes, such as refrigeration, chemical reactions, and energy production.. Replies.
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Important Questions: Statistics - Class 10 MCQ
# Important Questions: Statistics - Class 10 MCQ
Test Description
## 10 Questions MCQ Test Mathematics (Maths) Class 10 - Important Questions: Statistics
Important Questions: Statistics for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation. The Important Questions: Statistics questions and answers have been prepared according to the Class 10 exam syllabus.The Important Questions: Statistics MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Important Questions: Statistics below.
Solutions of Important Questions: Statistics questions in English are available as part of our Mathematics (Maths) Class 10 for Class 10 & Important Questions: Statistics solutions in Hindi for Mathematics (Maths) Class 10 course. Download more important topics, notes, lectures and mock test series for Class 10 Exam by signing up for free. Attempt Important Questions: Statistics | 10 questions in 10 minutes | Mock test for Class 10 preparation | Free important questions MCQ to study Mathematics (Maths) Class 10 for Class 10 Exam | Download free PDF with solutions
Important Questions: Statistics - Question 1
### While computing mean of a grouped data, we assume that the frequencies are
Important Questions: Statistics - Question 2
### di is the deviation of xi from assumed mean a.
Detailed Solution for Important Questions: Statistics - Question 2
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Important Questions: Statistics - Question 3
### Construction of a cumulative frequency table is useful in determining the
Detailed Solution for Important Questions: Statistics - Question 3
The correct answer is b
A cumulative frequency table helps to determine the median of the series.
Important Questions: Statistics - Question 4
Mean of 100 items is 49. It was discovered that three items which should have been 60, 70, 80 were wrongly read as 40, 20, 50 respectively. The correct mean is
Detailed Solution for Important Questions: Statistics - Question 4
Sum of 100 observations = 100 x 49 = 4900
Correct sum = 4900 - [40 + 20 + 50] + [60 + 70 + 80] = 5000
∴ Correct mean = 5000/100 = 50
Important Questions: Statistics - Question 5
In the formula, Mode
Detailed Solution for Important Questions: Statistics - Question 5
The formula to find the mode of the grouped data is: Mode = l + [(f1-f0)/(2f1-f0-f2)]×h. Where, l = lower class limit of modal class, h = class size, f1 = frequency of modal class, f0 = frequency of class proceeding to modal class, f2 = frequency of class succeeding to modal class.
Important Questions: Statistics - Question 6
Choose the correct answer from the given four options : In the formula
for finding the mean of grouped data are deviation from a of
Important Questions: Statistics - Question 7
For the following distribution:
The sum of upper limits of the median class and model class is
Important Questions: Statistics - Question 8
While computing mean of grouped data, we assume that the frequencies are
Important Questions: Statistics - Question 9
Consider the following frequency distribution of the height of 60 students of a class:
The sum of the lower limit of the modal class and upper limit of the median class is
Detailed Solution for Important Questions: Statistics - Question 9
Important Questions: Statistics - Question 10
For the following distribution:
the modal class is
## Mathematics (Maths) Class 10
116 videos|420 docs|77 tests
Information about Important Questions: Statistics Page
In this test you can find the Exam questions for Important Questions: Statistics solved & explained in the simplest way possible. Besides giving Questions and answers for Important Questions: Statistics, EduRev gives you an ample number of Online tests for practice
## Mathematics (Maths) Class 10
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Important Questions: Statistics - Class 10 MCQ. # Important Questions: Statistics - Class 10 MCQ. Test Description. ## 10 Questions MCQ Test Mathematics (Maths) Class 10 - Important Questions: Statistics. Important Questions: Statistics for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation. The Important Questions: Statistics questions and answers have been prepared according to the Class 10 exam syllabus.The Important Questions: Statistics MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Important Questions: Statistics below.. Solutions of Important Questions: Statistics questions in English are available as part of our Mathematics (Maths) Class 10 for Class 10 & Important Questions: Statistics solutions in Hindi for Mathematics (Maths) Class 10 course. Download more important topics, notes, lectures and mock test series for Class 10 Exam by signing up for free. Attempt Important Questions: Statistics | 10 questions in 10 minutes | Mock test for Class 10 preparation | Free important questions MCQ to study Mathematics (Maths) Class 10 for Class 10 Exam | Download free PDF with solutions. Important Questions: Statistics - Question 1. ### While computing mean of a grouped data, we assume that the frequencies are. Important Questions: Statistics - Question 2. ### di is the deviation of xi from assumed mean a.. Detailed Solution for Important Questions: Statistics - Question 2. 1 Crore+ students have signed up on EduRev. Have you?. Important Questions: Statistics - Question 3. ### Construction of a cumulative frequency table is useful in determining the. Detailed Solution for Important Questions: Statistics - Question 3. The correct answer is b. A cumulative frequency table helps to determine the median of the series.. Important Questions: Statistics - Question 4. Mean of 100 items is 49. It was discovered that three items which should have been 60, 70, 80 were wrongly read as 40, 20, 50 respectively. The correct mean is. Detailed Solution for Important Questions: Statistics - Question 4. Sum of 100 observations = 100 x 49 = 4900. Correct sum = 4900 - [40 + 20 + 50] + [60 + 70 + 80] = 5000.
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∴ Correct mean = 5000/100 = 50. Important Questions: Statistics - Question 5. In the formula, Mode. Detailed Solution for Important Questions: Statistics - Question 5. The formula to find the mode of the grouped data is: Mode = l + [(f1-f0)/(2f1-f0-f2)]×h. Where, l = lower class limit of modal class, h = class size, f1 = frequency of modal class, f0 = frequency of class proceeding to modal class, f2 = frequency of class succeeding to modal class.. Important Questions: Statistics - Question 6. Choose the correct answer from the given four options : In the formula. for finding the mean of grouped data are deviation from a of. Important Questions: Statistics - Question 7. For the following distribution:. The sum of upper limits of the median class and model class is. Important Questions: Statistics - Question 8. While computing mean of grouped data, we assume that the frequencies are. Important Questions: Statistics - Question 9. Consider the following frequency distribution of the height of 60 students of a class:. The sum of the lower limit of the modal class and upper limit of the median class is. Detailed Solution for Important Questions: Statistics - Question 9. Important Questions: Statistics - Question 10. For the following distribution:. the modal class is. ## Mathematics (Maths) Class 10. 116 videos|420 docs|77 tests. Information about Important Questions: Statistics Page. In this test you can find the Exam questions for Important Questions: Statistics solved & explained in the simplest way possible. Besides giving Questions and answers for Important Questions: Statistics, EduRev gives you an ample number of Online tests for practice. ## Mathematics (Maths) Class 10. 116 videos|420 docs|77 tests.
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http://www.ck12.org/geometry/Proportion-Properties/lesson/Proportion-Properties/r11/
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Proportion Properties
## Two ratios set equal to each other.
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Progress
Practice Proportion Properties
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Proportion Properties
What if you were told that a scale model of a python is in the ratio of 1:24? If the model measures 0.75 feet long, how long is the real python? After completing this Concept, you'll be able to solve problems like this one by using a proportion.
### Watch This
First watch this video.
Now watch this video.
Finally, watch this video.
### Guidance
A proportion is two ratios that are set equal to each other. Usually the ratios in proportions are written in fraction form. An example of a proportion is $\frac{2}{x}={5}{10}$ . To solve a proportion, you need to cross-multiply. The Cross-Multiplication Theorem , which allows us to solve proportions using this method, states that if $a, b, c,$ and $d$ are real numbers, with $b \neq 0$ and $d \neq 0$ and if $\frac{a}{b} = \frac{c}{d}$ , then $ad=bc$ . Cross-multiplying allows us to get rid of the fractions in our equation. The Cross-Multiplication Theorem has several sub-theorems, called corollaries.
Corollary #1: If $a, b, c,$ and $d$ are nonzero and $\frac{a}{b} = \frac{c}{d}$ , then $\frac{a}{c} = \frac{b}{d}$ . Switch $b$ and $c$ .
Corollary #2: If $a, b, c,$ and $d$ are nonzero and $\frac{a}{b} = \frac{c}{d}$ , then $\frac{d}{b} = \frac{c}{a}$ . Switch $a$ and $d$ .
Corollary #3: If $a, b, c,$ and $d$ are nonzero and $\frac{a}{b} = \frac{c}{d}$ , then $\frac{b}{a} = \frac{c}{d}$ . Flip each ratio upside down.
Corollary #4: If $a, b, c,$ and $d$ are nonzero and $\frac{a}{b} = \frac{c}{d}$ , then $\frac{a+b}{b} = \frac{c+d}{d}$ .
Corollary #5: If $a, b, c,$ and $d$ are nonzero and $\frac{a}{b} = \frac{c}{d}$ , then $\frac{a-b}{b} = \frac{c-d}{d}$ .
#### Example A
Solve the proportions.
a) $\frac{4}{5} = \frac{x}{30}$
b) $\frac{y+1}{8} = \frac{5}{20}$
c) $\frac{6}{5} = \frac{2x+4}{x-2}$
Remember, to solve a proportion, you need to cross-multiply.
a)
b)
c)
#### Example B
Your parents have an architect’s drawing of their home. On the paper, the house’s dimensions are 36 in by 30 in. If the shorter length of the house is actually 50 feet, what is the longer length?
To solve, first set up a proportion. If the shorter length is 50 feet, then it lines up with 30 in, the shorter length of the paper dimensions.
$\frac{30}{36} = \frac{50}{x} \longrightarrow \ 30x &=1800\\x &=60 \quad \quad \text{The longer length is 60 feet.}$
#### Example C
Suppose we have the proportion $\frac{2}{5} = \frac{14}{35}$ . Write three true proportions that follow.
First of all, we know this is a true proportion because you would multiply $\frac{2}{5}$ by $\frac{7}{7}$ to get $\frac{14}{35}$ . Using the first three corollaries:
1. $\frac{2}{14} = \frac{5}{35}$
2. $\frac{35}{5} = \frac{14}{2}$
3. $\frac{5}{2} = \frac{35}{14}$
### Guided Practice
1. In the picture, $\frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}$ .
Find the measures of $AC$ and $XY$ .
2. In the picture, $\frac{ED}{AD} = \frac{BC}{AC}$ . Find $y$ .
3. In the picture, $\frac{AB}{BE} = \frac{AC}{CD}$ . Find $BE$ .
1. Plug in the lengths of the sides we know.
2. Substitute in the lengths of the sides we know.
$\frac{6}{y} = \frac{8}{12+8} \ \longrightarrow \ 8y &= 6(20)\\y &= 15$
3. Substitute in the lengths of the sides we know.
$\frac{12}{BE} = \frac{20}{25} \ \longrightarrow \ {20(BE)} & = 12(25)\\BE & = 15$
### Practice
Solve each proportion.
1. $\frac{x}{10} = \frac{42}{35}$
2. $\frac{x}{x-2} = \frac{5}{7}$
3. $\frac{6}{9} = \frac{y}{24}$
4. $\frac{x}{9} = \frac{16}{x}$
5. $\frac{y-3}{8} = \frac{y+6}{5}$
6. $\frac{20}{z+5} = \frac{16}{7}$
7. Shawna drove 245 miles and used 8.2 gallons of gas. At the same rate, if she drove 416 miles, how many gallons of gas will she need? Round to the nearest tenth.
8. The president, vice-president, and financial officer of a company divide the profits is a 4:3:2 ratio. If the company made \$1,800,000 last year, how much did each person receive?
Given the true proportion, $\frac{10}{6}= \frac{15}{d} = \frac{x}{y}$ and $d, x,$ and $y$ are nonzero, determine if the following proportions are also true.
1. $\frac{10}{y} = \frac{x}{6}$
2. $\frac{15}{10} = \frac{d}{6}$
3. $\frac{6+10}{10} = \frac{y+x}{x}$
4. $\frac{15}{x} = \frac{y}{d}$
For questions 13-16, $\frac{AE}{ED} = \frac{BC}{CD}$ and $\frac{ED}{AD} = \frac{CD}{DB} = \frac{EC}{AB}$ .
1. Find $DB$ .
2. Find $EC$ .
3. Find $CB$ .
4. Find $AD$ .
### Vocabulary Language: English Spanish
corollary
corollary
A theorem that follows directly from another theorem.
proportion
proportion
Two ratios that are set equal to each other.
ratio
ratio
A way to compare two numbers. Ratios can be written in three ways: $\frac{a}{b}$, $a:b$, and $a$ to $b$.
Cross Products
Cross Products
To simplify a proportion using cross products, multiply the diagonals of each ratio.
Cross-Multiplication Theorem
Cross-Multiplication Theorem
The Cross-Multiplication theorem states that if $a, b, c$ and $d$ are real numbers, with $b \ne 0$ and $d \ne 0$ and if $\frac{a}{b} = \frac{c}{d}$, then $ad = bc$.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />. You are viewing an older version of this Concept. Go to the latest version.. # Proportion Properties. ## Two ratios set equal to each other.. 0%. Progress. Practice Proportion Properties. Progress. 0%. Proportion Properties. What if you were told that a scale model of a python is in the ratio of 1:24? If the model measures 0.75 feet long, how long is the real python? After completing this Concept, you'll be able to solve problems like this one by using a proportion.. ### Watch This. First watch this video.. Now watch this video.. Finally, watch this video.. ### Guidance. A proportion is two ratios that are set equal to each other. Usually the ratios in proportions are written in fraction form. An example of a proportion is $\frac{2}{x}={5}{10}$ . To solve a proportion, you need to cross-multiply. The Cross-Multiplication Theorem , which allows us to solve proportions using this method, states that if $a, b, c,$ and $d$ are real numbers, with $b \neq 0$ and $d \neq 0$ and if $\frac{a}{b} = \frac{c}{d}$ , then $ad=bc$ . Cross-multiplying allows us to get rid of the fractions in our equation. The Cross-Multiplication Theorem has several sub-theorems, called corollaries.. Corollary #1: If $a, b, c,$ and $d$ are nonzero and $\frac{a}{b} = \frac{c}{d}$ , then $\frac{a}{c} = \frac{b}{d}$ . Switch $b$ and $c$ .. Corollary #2: If $a, b, c,$ and $d$ are nonzero and $\frac{a}{b} = \frac{c}{d}$ , then $\frac{d}{b} = \frac{c}{a}$ . Switch $a$ and $d$ .. Corollary #3: If $a, b, c,$ and $d$ are nonzero and $\frac{a}{b} = \frac{c}{d}$ , then $\frac{b}{a} = \frac{c}{d}$ . Flip each ratio upside down.. Corollary #4: If $a, b, c,$ and $d$ are nonzero and $\frac{a}{b} = \frac{c}{d}$ , then $\frac{a+b}{b} = \frac{c+d}{d}$ .. Corollary #5: If $a, b, c,$ and $d$ are nonzero and $\frac{a}{b} = \frac{c}{d}$ , then $\frac{a-b}{b} = \frac{c-d}{d}$ .. #### Example A. Solve the proportions.. a) $\frac{4}{5} = \frac{x}{30}$. b) $\frac{y+1}{8} = \frac{5}{20}$. c) $\frac{6}{5} = \frac{2x+4}{x-2}$. Remember, to solve a proportion, you need to cross-multiply.. a). b). c). #### Example B. Your parents have an architect’s drawing of their home. On the paper, the house’s dimensions are 36 in by 30 in. If the shorter length of the house is actually 50 feet, what is the longer length?. To solve, first set up a proportion. If the shorter length is 50 feet, then it lines up with 30 in, the shorter length of the paper dimensions.. $\frac{30}{36} = \frac{50}{x} \longrightarrow \ 30x &=1800\\x &=60 \quad \quad \text{The longer length is 60 feet.}$. #### Example C. Suppose we have the proportion $\frac{2}{5} = \frac{14}{35}$ . Write three true proportions that follow.. First of all, we know this is a true proportion because you would multiply $\frac{2}{5}$ by $\frac{7}{7}$ to get $\frac{14}{35}$ . Using the first three corollaries:. 1. $\frac{2}{14} = \frac{5}{35}$. 2. $\frac{35}{5} = \frac{14}{2}$. 3. $\frac{5}{2} = \frac{35}{14}$. ### Guided Practice. 1. In the picture, $\frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}$ .. Find the measures of $AC$ and $XY$ .. 2. In the picture, $\frac{ED}{AD} = \frac{BC}{AC}$ . Find $y$ .. 3. In the picture, $\frac{AB}{BE} = \frac{AC}{CD}$ . Find $BE$ .. 1. Plug in the lengths of the sides we know.
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2. Substitute in the lengths of the sides we know.. $\frac{6}{y} = \frac{8}{12+8} \ \longrightarrow \ 8y &= 6(20)\\y &= 15$. 3. Substitute in the lengths of the sides we know.. $\frac{12}{BE} = \frac{20}{25} \ \longrightarrow \ {20(BE)} & = 12(25)\\BE & = 15$. ### Practice. Solve each proportion.. 1. $\frac{x}{10} = \frac{42}{35}$. 2. $\frac{x}{x-2} = \frac{5}{7}$. 3. $\frac{6}{9} = \frac{y}{24}$. 4. $\frac{x}{9} = \frac{16}{x}$. 5. $\frac{y-3}{8} = \frac{y+6}{5}$. 6. $\frac{20}{z+5} = \frac{16}{7}$. 7. Shawna drove 245 miles and used 8.2 gallons of gas. At the same rate, if she drove 416 miles, how many gallons of gas will she need? Round to the nearest tenth.. 8. The president, vice-president, and financial officer of a company divide the profits is a 4:3:2 ratio. If the company made \$1,800,000 last year, how much did each person receive?. Given the true proportion, $\frac{10}{6}= \frac{15}{d} = \frac{x}{y}$ and $d, x,$ and $y$ are nonzero, determine if the following proportions are also true.. 1. $\frac{10}{y} = \frac{x}{6}$. 2. $\frac{15}{10} = \frac{d}{6}$. 3. $\frac{6+10}{10} = \frac{y+x}{x}$. 4. $\frac{15}{x} = \frac{y}{d}$. For questions 13-16, $\frac{AE}{ED} = \frac{BC}{CD}$ and $\frac{ED}{AD} = \frac{CD}{DB} = \frac{EC}{AB}$ .. 1. Find $DB$ .. 2. Find $EC$ .. 3. Find $CB$ .. 4. Find $AD$ .. ### Vocabulary Language: English Spanish. corollary. corollary. A theorem that follows directly from another theorem.. proportion. proportion. Two ratios that are set equal to each other.. ratio. ratio. A way to compare two numbers. Ratios can be written in three ways: $\frac{a}{b}$, $a:b$, and $a$ to $b$.. Cross Products. Cross Products. To simplify a proportion using cross products, multiply the diagonals of each ratio.. Cross-Multiplication Theorem. Cross-Multiplication Theorem. The Cross-Multiplication theorem states that if $a, b, c$ and $d$ are real numbers, with $b \ne 0$ and $d \ne 0$ and if $\frac{a}{b} = \frac{c}{d}$, then $ad = bc$.
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# Norm of linear functional in $l_1$
I'd like some help to solve the following question:
Let $(a_i)$ be a sequence in $\mathbb{R}$ such that $\sum_{i=1}^\infty a_ix_i< \infty$ for all $(x_i)\in l_1(\mathbb{R})$. Show that $(a_i)\in l_\infty$.
My attempt. For every $n\in \mathbb{N}$, set $f_n:l_1\rightarrow \mathbb{R}$ as $f_n(x)=\sum_{i=1}^na_ix_i$, where $x=(x_i)$. Clearly, $f_n$ is a linear functional in $l_1$. Moreover, since $$|f_n(x)|\leq \sum_{i=1}^n|a_ix_i|\leq \max_{1\leq i\leq n}|a_i| \cdot \|x\|_1,$$ $f_n$ is bounded. I know if I get $\|f_n\|= \max_{1\leq i\leq n}|a_i|$, I can use the uniform boundedness principle to obtain the result. However, I couldn't find an element $x\in l_1$ such that $$\frac{|f_n(x)|}{\|x\|_1}= \max_{1\leq i\leq n}|a_i|.$$ I'll be grateful for any hint. Thanks in advance!
Pick $k \leqslant n$ such that $\lvert a_k\rvert = \max_{1 \leqslant i \leqslant n} \lvert a_i\rvert$. Then
$$\lvert f_n(e_k)\rvert = \lvert a_k\rvert = \max_{1 \leqslant i \leqslant n} \lvert a_i\rvert \cdot \lVert e_k\rVert_1$$
for $e_k$ the $k^{\text{th}}$ standard unit vector, i.e.
$$(e_k)_i = \begin{cases} 1 &\text{if } i = k \\ 0 &\text{if } i \neq k \end{cases}\,.$$
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# Norm of linear functional in $l_1$. I'd like some help to solve the following question:. Let $(a_i)$ be a sequence in $\mathbb{R}$ such that $\sum_{i=1}^\infty a_ix_i< \infty$ for all $(x_i)\in l_1(\mathbb{R})$. Show that $(a_i)\in l_\infty$.. My attempt. For every $n\in \mathbb{N}$, set $f_n:l_1\rightarrow \mathbb{R}$ as $f_n(x)=\sum_{i=1}^na_ix_i$, where $x=(x_i)$. Clearly, $f_n$ is a linear functional in $l_1$. Moreover, since $$|f_n(x)|\leq \sum_{i=1}^n|a_ix_i|\leq \max_{1\leq i\leq n}|a_i| \cdot \|x\|_1,$$ $f_n$ is bounded. I know if I get $\|f_n\|= \max_{1\leq i\leq n}|a_i|$, I can use the uniform boundedness principle to obtain the result. However, I couldn't find an element $x\in l_1$ such that $$\frac{|f_n(x)|}{\|x\|_1}= \max_{1\leq i\leq n}|a_i|.$$ I'll be grateful for any hint.
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Thanks in advance!. Pick $k \leqslant n$ such that $\lvert a_k\rvert = \max_{1 \leqslant i \leqslant n} \lvert a_i\rvert$. Then. $$\lvert f_n(e_k)\rvert = \lvert a_k\rvert = \max_{1 \leqslant i \leqslant n} \lvert a_i\rvert \cdot \lVert e_k\rVert_1$$. for $e_k$ the $k^{\text{th}}$ standard unit vector, i.e.. $$(e_k)_i = \begin{cases} 1 &\text{if } i = k \\ 0 &\text{if } i \neq k \end{cases}\,.$$.
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## Conservation Of Energy
CONSERVATION OF ENERGY THEOREM
Nothing can be destroyed or created in the universe like energy. Suppose that a ball falls from height of 2m, it has only potential energy at the beginning, however, as it falls it gains kinetic energy and its velocity increases. When it hits the ground it has only kinetic energy. Well, where is the potential energy that it has at the beginning? It is totally converted to the kinetic energy, as said in the first sentence nothing can be destroyed or created they just change form. Thus, our potential energy also changes its forms from potential to the kinetic energy. In summary, energy of the system is always constant, they can change their forms but amount of total energy does not change.
Picture shows the energy change of the ball. It has only potential energy 2mgh at the beginning. When it starts to lose height it gains velocity in other word decreasing in the amount of potential energy increases the amount of kinetic energy. At h height it has both potential and kinetic energy and when it hits the ground the potential energy becomes zero and kinetic energy has its maximum value.
Einitial = Efinal
Example: By using the given information in picture given below, find the velocity of the ball at point D.
Ei=Ef (conservation of energy)
mg3h=mg2h+1/2mv²
mgh=1/2mv²
v=√2gh
Example: A block having mass 2kg and velocity 2m/s slide on the inclined plane. If the horizontal surface has friction constant µ=0, 4 find the distance it travels in horizontal before it stops.
We use conservation of energy in solution of this problem.
Einitial=Efinal
Einitial=Ep+Ek=mgh+1/2mv² Efinal=0
Einitial=2kg.10m/s².8m+1/2.2kg. (2m/s) ² Work done by friction=Einitial Einitial=164joule
Wfriction=µ.N.X=0,4.2kg.10m/s².X=Ei
8. X=164joule X=20,5m
Block slides 20,5m in horizontal
Example: Find the final velocity of the box from the given picture.
We again use the conservation of energy theorem.
Einitial must be equal to the Efinal.
Einitial=Ek=1/2mv² Efinal=Ek+Ep=1/2mv’²+mgh
Ei=1/2.2kg.(10m/s) ²=100joule Efinal=1/2.2kg.v’²+2kg.10m/s².4m=80+v’²
100=80+v’² v’=25m/s
Example: Find the amount of compression of the spring if the ball does free fall from 4m and compresses the spring.
From the conservation of energy law we can find the amount of spring’s compression.
Ep=1/2.kx² for spring
X=1/4m
Ball compresses the spring 1/4m.
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## Conservation Of Energy. CONSERVATION OF ENERGY THEOREM. Nothing can be destroyed or created in the universe like energy. Suppose that a ball falls from height of 2m, it has only potential energy at the beginning, however, as it falls it gains kinetic energy and its velocity increases. When it hits the ground it has only kinetic energy. Well, where is the potential energy that it has at the beginning? It is totally converted to the kinetic energy, as said in the first sentence nothing can be destroyed or created they just change form. Thus, our potential energy also changes its forms from potential to the kinetic energy. In summary, energy of the system is always constant, they can change their forms but amount of total energy does not change.. Picture shows the energy change of the ball. It has only potential energy 2mgh at the beginning. When it starts to lose height it gains velocity in other word decreasing in the amount of potential energy increases the amount of kinetic energy. At h height it has both potential and kinetic energy and when it hits the ground the potential energy becomes zero and kinetic energy has its maximum value.. Einitial = Efinal. Example: By using the given information in picture given below, find the velocity of the ball at point D.. Ei=Ef (conservation of energy). mg3h=mg2h+1/2mv². mgh=1/2mv². v=√2gh. Example: A block having mass 2kg and velocity 2m/s slide on the inclined plane. If the horizontal surface has friction constant µ=0, 4 find the distance it travels in horizontal before it stops.. We use conservation of energy in solution of this problem.
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Einitial=Efinal. Einitial=Ep+Ek=mgh+1/2mv² Efinal=0. Einitial=2kg.10m/s².8m+1/2.2kg. (2m/s) ² Work done by friction=Einitial Einitial=164joule. Wfriction=µ.N.X=0,4.2kg.10m/s².X=Ei. 8. X=164joule X=20,5m. Block slides 20,5m in horizontal. Example: Find the final velocity of the box from the given picture.. We again use the conservation of energy theorem.. Einitial must be equal to the Efinal.. Einitial=Ek=1/2mv² Efinal=Ek+Ep=1/2mv’²+mgh. Ei=1/2.2kg.(10m/s) ²=100joule Efinal=1/2.2kg.v’²+2kg.10m/s².4m=80+v’². 100=80+v’² v’=25m/s. Example: Find the amount of compression of the spring if the ball does free fall from 4m and compresses the spring.. From the conservation of energy law we can find the amount of spring’s compression.. Ep=1/2.kx² for spring. X=1/4m. Ball compresses the spring 1/4m.. Author:.
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https://www.kalkulatorku.com/konversi-satuan/berat-massa.php?k1=tons-IMPERIAL&k2=milligrams
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Konversi BERAT-MASSAtons-IMPERIAL ke milligrams
1 Tons IMPERIAL
= 1016050000 Milligrams
Besaran: berat massa
Konversi Satuan: Tons IMPERIAL ke Milligrams
Satuan dasar untuk berat massa adalah kilograms (SI Unit)
Simbol dari [Tons IMPERIAL] adalah: (t [IMPERIAL]), sedangkan simbol untuk [Milligrams] adalah: (mg), keduanya merupakan satuan dari berat massa
Perhitungan cepat konversi Tons IMPERIAL ke Milligrams (t [IMPERIAL] ke mg):
1 t [IMPERIAL] = 1016050000 mg.
1 x 1016050000 mg = 1016050000 Milligrams.
*catatan: kesalahan atau error kecil dalam pembulatan hasil angka desimal bisa terjadi, silakan dicek ulang.
Definisi:
Berdasarkan satuan/unit dari besaran berat massa, yaitu => (kilograms), 1 Tons IMPERIAL (t [IMPERIAL]) sama dengan 1016.05 kilograms, sedangkan 1 Milligrams (mg) = 1.0E-6 kilograms.
oo
Tons IMPERIALto Milligrams (table conversion)
1 t [IMPERIAL] = 1016050000 mg
2 t [IMPERIAL] = 2032100000 mg
3 t [IMPERIAL] = 3048150000 mg
4 t [IMPERIAL] = 4064200000 mg
5 t [IMPERIAL] = 5080250000 mg
6 t [IMPERIAL] = 6096300000 mg
7 t [IMPERIAL] = 7112350000 mg
8 t [IMPERIAL] = 8128400000 mg
9 t [IMPERIAL] = 9144450000 mg
10 t [IMPERIAL] = 10160500000 mg
20 t [IMPERIAL] = 20321000000 mg
30 t [IMPERIAL] = 30481500000 mg
40 t [IMPERIAL] = 40642000000 mg
50 t [IMPERIAL] = 50802500000 mg
60 t [IMPERIAL] = 60963000000 mg
70 t [IMPERIAL] = 71123500000 mg
80 t [IMPERIAL] = 81284000000 mg
90 t [IMPERIAL] = 91444500000 mg
100 t [IMPERIAL] = 101605000000 mg
200 t [IMPERIAL] = 203210000000 mg
300 t [IMPERIAL] = 304815000000 mg
400 t [IMPERIAL] = 406420000000 mg
500 t [IMPERIAL] = 508025000000 mg
600 t [IMPERIAL] = 609630000000 mg
700 t [IMPERIAL] = 711235000000 mg
800 t [IMPERIAL] = 812840000000 mg
900 t [IMPERIAL] = 914445000000 mg
1000 t [IMPERIAL] = 1016050000000 mg
2000 t [IMPERIAL] = 2032100000000 mg
4000 t [IMPERIAL] = 4064200000000 mg
5000 t [IMPERIAL] = 5080250000000 mg
7500 t [IMPERIAL] = 7620375000000 mg
10000 t [IMPERIAL] = 10160500000000 mg
25000 t [IMPERIAL] = 25401250000000 mg
50000 t [IMPERIAL] = 50802500000000 mg
100000 t [IMPERIAL] = 1.01605E+14 mg
1000000 t [IMPERIAL] = 1.01605E+15 mg
1000000000 t [IMPERIAL] = 1.01605E+18 mg
(Tons IMPERIAL) to (Milligrams) conversions
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Konversi BERAT-MASSAtons-IMPERIAL ke milligrams. 1 Tons IMPERIAL. = 1016050000 Milligrams. Besaran: berat massa. Konversi Satuan: Tons IMPERIAL ke Milligrams. Satuan dasar untuk berat massa adalah kilograms (SI Unit). Simbol dari [Tons IMPERIAL] adalah: (t [IMPERIAL]), sedangkan simbol untuk [Milligrams] adalah: (mg), keduanya merupakan satuan dari berat massa. Perhitungan cepat konversi Tons IMPERIAL ke Milligrams (t [IMPERIAL] ke mg):. 1 t [IMPERIAL] = 1016050000 mg.. 1 x 1016050000 mg = 1016050000 Milligrams.. *catatan: kesalahan atau error kecil dalam pembulatan hasil angka desimal bisa terjadi, silakan dicek ulang.. Definisi:. Berdasarkan satuan/unit dari besaran berat massa, yaitu => (kilograms), 1 Tons IMPERIAL (t [IMPERIAL]) sama dengan 1016.05 kilograms, sedangkan 1 Milligrams (mg) = 1.0E-6 kilograms.. oo. Tons IMPERIALto Milligrams (table conversion). 1 t [IMPERIAL] = 1016050000 mg. 2 t [IMPERIAL] = 2032100000 mg. 3 t [IMPERIAL] = 3048150000 mg. 4 t [IMPERIAL] = 4064200000 mg. 5 t [IMPERIAL] = 5080250000 mg. 6 t [IMPERIAL] = 6096300000 mg. 7 t [IMPERIAL] = 7112350000 mg. 8 t [IMPERIAL] = 8128400000 mg. 9 t [IMPERIAL] = 9144450000 mg. 10 t [IMPERIAL] = 10160500000 mg. 20 t [IMPERIAL] = 20321000000 mg. 30 t [IMPERIAL] = 30481500000 mg. 40 t [IMPERIAL] = 40642000000 mg.
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50 t [IMPERIAL] = 50802500000 mg. 60 t [IMPERIAL] = 60963000000 mg. 70 t [IMPERIAL] = 71123500000 mg. 80 t [IMPERIAL] = 81284000000 mg. 90 t [IMPERIAL] = 91444500000 mg. 100 t [IMPERIAL] = 101605000000 mg. 200 t [IMPERIAL] = 203210000000 mg. 300 t [IMPERIAL] = 304815000000 mg. 400 t [IMPERIAL] = 406420000000 mg. 500 t [IMPERIAL] = 508025000000 mg. 600 t [IMPERIAL] = 609630000000 mg. 700 t [IMPERIAL] = 711235000000 mg. 800 t [IMPERIAL] = 812840000000 mg. 900 t [IMPERIAL] = 914445000000 mg. 1000 t [IMPERIAL] = 1016050000000 mg. 2000 t [IMPERIAL] = 2032100000000 mg. 4000 t [IMPERIAL] = 4064200000000 mg. 5000 t [IMPERIAL] = 5080250000000 mg. 7500 t [IMPERIAL] = 7620375000000 mg. 10000 t [IMPERIAL] = 10160500000000 mg. 25000 t [IMPERIAL] = 25401250000000 mg. 50000 t [IMPERIAL] = 50802500000000 mg. 100000 t [IMPERIAL] = 1.01605E+14 mg. 1000000 t [IMPERIAL] = 1.01605E+15 mg. 1000000000 t [IMPERIAL] = 1.01605E+18 mg. (Tons IMPERIAL) to (Milligrams) conversions.
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https://socratic.org/questions/how-to-write-an-equation-in-standard-form-for-a-line-through-5-4-slope-3#142310
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How to write an equation in standard form for a line through (-5,4); slope -3?
1 Answer
May 3, 2015
The most common definition of "standard form for a linear equation is
$A x + B y = C$
For a point $\left({x}_{1} , {y}_{1}\right) = \left(- 5 , 4\right)$ and a slope $m = \left(- 3\right)$
We can start from the point-slope form:
$\frac{y - {y}_{1}}{x - {x}_{1}} = m$
$\frac{y - 4}{x + 5} = - 3$
$y - 4 = - 3 x - 15$
$3 x + y = - 11$
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How to write an equation in standard form for a line through (-5,4); slope -3?. 1 Answer. May 3, 2015. The most common definition of "standard form for a linear equation is. $A x + B y = C$. For a point $\left({x}_{1} , {y}_{1}\right) = \left(- 5 , 4\right)$ and a slope $m = \left(- 3\right)$.
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We can start from the point-slope form:. $\frac{y - {y}_{1}}{x - {x}_{1}} = m$. $\frac{y - 4}{x + 5} = - 3$. $y - 4 = - 3 x - 15$. $3 x + y = - 11$.
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http://ibmaths4u.com/viewtopic.php?f=16&t=108&sid=f2b1c3e33ba9d273719774763e017283
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## MYP Mathematics - Simple Interest
number, algebra, geometry and trigonometry, statistics and probability, discrete mathematics.
### MYP Mathematics - Simple Interest
MYP Mathematics, Financial Mathematics, Simple Interest
How can we calculate the amount borrowed if the interest paid for this loan is \$3000 at a rate of 5% per annum over 24 months?
Thanks
Lucas
Posts: 0
Joined: Mon Jan 28, 2013 8:05 pm
### Re: MYP Mathematics - Simple Interest
The simple interest is given by the following formula
$I=P \cdot r \cdot n$
Where I is the amount of interest, P is the principal, r is the simple interest rate per annum and n is the time in years.
Regarding your question about the calculation of the simple interest we have that
$\3000=P \cdot 0.05 \cdot \frac{24}{12}=>P=\frac{3000}{0.1}=\30000$
Hope these help!
miranda
Posts: 268
Joined: Mon Jan 28, 2013 8:03 pm
### Re: MYP Mathematics - Simple Interest
thanks miranda!!
Lucas
Posts: 0
Joined: Mon Jan 28, 2013 8:05 pm
### Re: MYP Mathematics - Simple Interest
thanks miranda!!
Lucas
Posts: 0
Joined: Mon Jan 28, 2013 8:05 pm
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## MYP Mathematics - Simple Interest. number, algebra, geometry and trigonometry, statistics and probability, discrete mathematics.. ### MYP Mathematics - Simple Interest. MYP Mathematics, Financial Mathematics, Simple Interest. How can we calculate the amount borrowed if the interest paid for this loan is \$3000 at a rate of 5% per annum over 24 months?. Thanks. Lucas. Posts: 0. Joined: Mon Jan 28, 2013 8:05 pm. ### Re: MYP Mathematics - Simple Interest. The simple interest is given by the following formula. $I=P \cdot r \cdot n$. Where I is the amount of interest, P is the principal, r is the simple interest rate per annum and n is the time in years.. Regarding your question about the calculation of the simple interest we have that. $\3000=P \cdot 0.05 \cdot \frac{24}{12}=>P=\frac{3000}{0.1}=\30000$.
|
Hope these help!. miranda. Posts: 268. Joined: Mon Jan 28, 2013 8:03 pm. ### Re: MYP Mathematics - Simple Interest. thanks miranda!!. Lucas. Posts: 0. Joined: Mon Jan 28, 2013 8:05 pm. ### Re: MYP Mathematics - Simple Interest. thanks miranda!!. Lucas. Posts: 0. Joined: Mon Jan 28, 2013 8:05 pm.
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# Is 47 A Prime Or Composite
Is 47 A Prime Or Composite. Is 47 an odd number? Back to is 46 a composite number?
In other words, 47 is only divided by 1 and by itself. Is 47 an odd number? For a number to be classified as a prime number, it should.
### When Finding If A Number Is Prime Or Not, You Have To Divide That Number By All Of Its Factors Up To Its Square Root.
Check 47 prime number or not. A prime number has 2 factors. 47 is not a composite number.
### A Composite Number Has More Than 2 Factors.
Is 47 an odd number? Is 47 composite or prime? What are the factors of 47?
### 1 (One) Is Neither A Prime Nor A Composite Number.
What numbers are composite between 30 and 47? 47 is a prime number. Is 47 a prime number.
### Is 47 Prime Number Or Composite Number ?
Prime numbers are the numbers that have only 2 factors i.e. A composite number is a whole number that is the product of other numbers in addition to 1 and itself. Back to is 46 a composite number?
### Here You Can Find The Answer To Questions Related To:
Take the help of our prime number calculator and determine whether the given 47 number is prime or composite number. Is 67 prime or composite? Let's check 47 number is prime.
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# Is 47 A Prime Or Composite. Is 47 A Prime Or Composite. Is 47 an odd number? Back to is 46 a composite number?. In other words, 47 is only divided by 1 and by itself. Is 47 an odd number? For a number to be classified as a prime number, it should.. ### When Finding If A Number Is Prime Or Not, You Have To Divide That Number By All Of Its Factors Up To Its Square Root.. Check 47 prime number or not. A prime number has 2 factors. 47 is not a composite number.. ### A Composite Number Has More Than 2 Factors.. Is 47 an odd number? Is 47 composite or prime? What are the factors of 47?.
|
### 1 (One) Is Neither A Prime Nor A Composite Number.. What numbers are composite between 30 and 47? 47 is a prime number. Is 47 a prime number.. ### Is 47 Prime Number Or Composite Number ?. Prime numbers are the numbers that have only 2 factors i.e. A composite number is a whole number that is the product of other numbers in addition to 1 and itself. Back to is 46 a composite number?. ### Here You Can Find The Answer To Questions Related To:. Take the help of our prime number calculator and determine whether the given 47 number is prime or composite number. Is 67 prime or composite? Let's check 47 number is prime.
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http://nrich.maths.org/public/leg.php?code=-68&cl=1&cldcmpid=1243
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# Search by Topic
#### Resources tagged with Visualising similar to Pumpkin Patch:
Filter by: Content type:
Stage:
Challenge level:
### Sprouts
##### Stage: 2, 3, 4 and 5 Challenge Level:
A game for 2 people. Take turns joining two dots, until your opponent is unable to move.
### Clocking Off
##### Stage: 2, 3 and 4 Challenge Level:
I found these clocks in the Arts Centre at the University of Warwick intriguing - do they really need four clocks and what times would be ambiguous with only two or three of them?
### Dice, Routes and Pathways
##### Stage: 1, 2 and 3
This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
### Bands and Bridges: Bringing Topology Back
##### Stage: 2 and 3
Lyndon Baker describes how the Mobius strip and Euler's law can introduce pupils to the idea of topology.
### Like a Circle in a Spiral
##### Stage: 2, 3 and 4 Challenge Level:
A cheap and simple toy with lots of mathematics. Can you interpret the images that are produced? Can you predict the pattern that will be produced using different wheels?
### Introducing NRICH TWILGO
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
We're excited about this new program for drawing beautiful mathematical designs. Can you work out how we made our first few pictures and, even better, share your most elegant solutions with us?
### Baravelle
##### Stage: 2, 3 and 4 Challenge Level:
What can you see? What do you notice? What questions can you ask?
### More Pebbles
##### Stage: 2 and 3 Challenge Level:
Have a go at this 3D extension to the Pebbles problem.
### Zooming in on the Squares
##### Stage: 2 and 3
Start with a large square, join the midpoints of its sides, you'll see four right angled triangles. Remove these triangles, a second square is left. Repeat the operation. What happens?
### The Development of Spatial and Geometric Thinking: Co-ordinating Space in Drawings
##### Stage: 1
This second article in the series refers to research about levels of development of spatial thinking and the possible influence of instruction.
### Pattern Power
##### Stage: 1, 2 and 3
Mathematics is the study of patterns. Studying pattern is an opportunity to observe, hypothesise, experiment, discover and create.
### Diagonal Dodge
##### Stage: 2 and 3 Challenge Level:
A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red.
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### 3D Stacks
##### Stage: 2 and 3 Challenge Level:
Can you find a way of representing these arrangements of balls?
### Sea Defences
##### Stage: 2 and 3 Challenge Level:
These are pictures of the sea defences at New Brighton. Can you work out what a basic shape might be in both images of the sea wall and work out a way they might fit together?
### Auditorium Steps
##### Stage: 2 and 3 Challenge Level:
What is the shape of wrapping paper that you would need to completely wrap this model?
### Cubes Within Cubes
##### Stage: 2 and 3 Challenge Level:
We start with one yellow cube and build around it to make a 3x3x3 cube with red cubes. Then we build around that red cube with blue cubes and so on. How many cubes of each colour have we used?
### World of Tan 4 - Monday Morning
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Wai Ping, Wah Ming and Chi Wing?
### World of Tan 21 - Almost There Now
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the lobster, yacht and cyclist?
### Horizontal Vertical
##### Stage: 1 Challenge Level:
Take it in turns to place a domino on the grid. One to be placed horizontally and the other vertically. Can you make it impossible for your opponent to play?
### Change Around
##### Stage: 1 Challenge Level:
Move just three of the circles so that the triangle faces in the opposite direction.
### Seeing Squares
##### Stage: 1 and 2 Challenge Level:
Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.
### World of Tan 27 - Sharing
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Fung at the table?
### Three Squares
##### Stage: 1 and 2 Challenge Level:
What is the greatest number of squares you can make by overlapping three squares?
### Coin Cogs
##### Stage: 2 Challenge Level:
Can you work out what is wrong with the cogs on a UK 2 pound coin?
### World of Tan 3 - Mai Ling
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Mai Ling?
### Building with Cubes
##### Stage: 1 Challenge Level:
Try to picture these buildings of cubes in your head. Can you make them to check whether you had imagined them correctly?
### World of Tan 26 - Old Chestnut
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this brazier for roasting chestnuts?
### World of Tan 24 - Clocks
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of these clocks?
### World of Tan 17 - Weather
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the watering can and man in a boat?
### World of Tan 12 - All in a Fluff
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of these rabbits?
### World of Tan 6 - Junk
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this junk?
### World of Tan 5 - Rocket
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the rocket?
### Cubes Cut Into Four Pieces
##### Stage: 1 Challenge Level:
Eight children each had a cube made from modelling clay. They cut them into four pieces which were all exactly the same shape and size. Whose pieces are the same? Can you decide who made each set?
### World of Tan 13 - A Storm in a Tea Cup
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of these convex shapes?
### Skeleton Shapes
##### Stage: 1 Challenge Level:
How many balls of modelling clay and how many straws does it take to make these skeleton shapes?
### World of Tan 8 - Sports Car
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this sports car?
### The Path of the Dice
##### Stage: 2 Challenge Level:
A game for 1 person. Can you work out how the dice must be rolled from the start position to the finish? Play on line.
### World of Tan 7 - Gat Marn
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this plaque design?
### Makeover
##### Stage: 1 and 2 Challenge Level:
Exchange the positions of the two sets of counters in the least possible number of moves
### World of Tan 28 - Concentrating on Coordinates
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming playing the board game?
### World of Tan 22 - an Appealing Stroll
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the child walking home from school?
### World of Tan 29 - the Telephone
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this telephone?
### L-ateral Thinking
##### Stage: 1 and 2 Challenge Level:
Try this interactive strategy game for 2
### World of Tan 11 - the Past, Present and Future
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the telescope and microscope?
### Counter Roundup
##### Stage: 2 Challenge Level:
A game for 1 or 2 people. Use the interactive version, or play with friends. Try to round up as many counters as possible.
### Put Yourself in a Box
##### Stage: 2 Challenge Level:
A game for 2 players. Given a board of dots in a grid pattern, players take turns drawing a line by connecting 2 adjacent dots. Your goal is to complete more squares than your opponent.
### Tessellating Capitals
##### Stage: 1 Challenge Level:
Have you ever tried tessellating capital letters? Have a look at these examples and then try some for yourself.
### Twice as Big?
##### Stage: 2 Challenge Level:
Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too.
### World of Tan 15 - Millennia
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the workmen?
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# Why is arctan of $-\frac{\sqrt{3}}{3} = -\frac{1}{6}\pi$?
I've been studying the unit circle and inverse trig functions on Khan Academy. One of the questions asked, is, what is the arctan of $-\frac{\sqrt{3}}{3}$.
The solution is $-\frac{1}{6}\pi$, I don't understand why?
If I pull up $-\frac{1}{6}\pi$ on a unit circle tool (in this case on Khan Academy). The $y$ (the sine) value on the same angle $-\frac{1}{6}\pi$ is $-\frac{1}{2}$. I see that the $x$ value (the cosine) on unit circle of $-\frac{1}{6}\pi$ is $\frac{\sqrt{3}}{2}$.
Tan is opposite side / adjacent side, or in the unit circle's case sine/cosine. Which suggests to me that the tan of $-\frac{1}{6}\pi$ is $-(1/2)/(\sqrt{3}/2)$ which equals $-1/\sqrt{3}$ not $-\frac{1}{6}\pi$.
Any insight would be great!
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The arctangent is odd, so $\arctan(-x)=-\arctan(x)$. Now, would you happen to have a drawing of the 30-60-90 triangle lying around? – J. M. Dec 8 '11 at 0:00
It's not. You want $-\pi/6$, not $-1/(6 \pi)$. – David Speyer Dec 8 '11 at 0:09
1. I think you're writing $-1/(6\pi)$ when you mean $-(\pi/6)$. 2. If what you are reporting accurately represents what's on Khan Academy then Khan Academy is wrong. It happens to the best of us. – Gerry Myerson Dec 8 '11 at 0:11
You have shown that $\tan(-\pi/6)=-1/\sqrt{3}=-\sqrt{3}/3$. Since $\arctan x$ is the angle in $(-\pi/2,\pi/2)$ whose $\tan$ is $x$, you have shown that $\arctan(-\sqrt{3}/3)=-\pi/6$. – André Nicolas Dec 8 '11 at 0:23
It is slightly unfortunate that Khan Academy says Note: please answer in terms of radians (ex: "3/4 pi"), partly because whether $\pi$ is part of the denominator or not is unclear, and partly because it would not accept "5/6 pi" in answer to this particular question, instead insisting on "-1/6 pi" as the principal value. – Henry Dec 8 '11 at 0:51
Maybe you're just missing the fact that $1/\sqrt{3}$ is the same as $\sqrt{3}/3$. You get that by rationalizing the denominator:
$$\frac{1}{\sqrt{3}} = \frac{1\cdot\sqrt{3}}{\sqrt{3}\sqrt{3}} = \frac{\sqrt{3}}{3}.$$
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Wow. Yes I completely missed that. That seems to have been my error. Thank you Michael. – drc Dec 8 '11 at 1:02
Maybe you're just messing with the concept of inverse function... Remember:
$\arctan x=y$ if and only if $\tan y =x$ and $y\in ]-\pi/2,\pi/2[$.
Now, you have found $\tan (-\pi/6) =-1/\sqrt{3}=-\sqrt{3}/3$ and you also have $-\pi/6 \in ]-\pi/2, \pi/2[$, hence previous statement applies and it yields $\arctan (-\sqrt{3}/3)=-\pi/6$ as claimed.
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What happens if y is outside of those bounds you specify? – drc Dec 8 '11 at 2:00
Also your use of ][, I'm assuming it indicates an open interval? I'm wondering why you used that instead of (-pi/2 < y < pi/2) – drc Dec 8 '11 at 2:06
@drc: It's not uncommon notation; I've seen it many times. – Arturo Magidin Dec 8 '11 at 2:26
@drc: I usually use the reverse square brakets for the open interval just to avoid confusion with other symbols; in fact, for example, $(a,b)$ is also used to denote the ordered pair with first entry $a$ and second entry $b$. – Pacciu Dec 8 '11 at 12:41
@drc: If $y\notin ]-\pi/2 ,\pi/2[$ then nothing "happens", in the sense that you cannot give any meaning to the formula $\arctan x=y$. In fact, as you should know from your book, the range of $\mathbb{R}\ni x\mapsto \arctan x\in \mathbb{R}$ is the open interval $]-\pi/2, \pi/2[$. – Pacciu Dec 8 '11 at 12:44
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# Why is arctan of $-\frac{\sqrt{3}}{3} = -\frac{1}{6}\pi$?. I've been studying the unit circle and inverse trig functions on Khan Academy. One of the questions asked, is, what is the arctan of $-\frac{\sqrt{3}}{3}$.. The solution is $-\frac{1}{6}\pi$, I don't understand why?. If I pull up $-\frac{1}{6}\pi$ on a unit circle tool (in this case on Khan Academy). The $y$ (the sine) value on the same angle $-\frac{1}{6}\pi$ is $-\frac{1}{2}$. I see that the $x$ value (the cosine) on unit circle of $-\frac{1}{6}\pi$ is $\frac{\sqrt{3}}{2}$.. Tan is opposite side / adjacent side, or in the unit circle's case sine/cosine. Which suggests to me that the tan of $-\frac{1}{6}\pi$ is $-(1/2)/(\sqrt{3}/2)$ which equals $-1/\sqrt{3}$ not $-\frac{1}{6}\pi$.. Any insight would be great!. -. The arctangent is odd, so $\arctan(-x)=-\arctan(x)$. Now, would you happen to have a drawing of the 30-60-90 triangle lying around? – J. M. Dec 8 '11 at 0:00. It's not. You want $-\pi/6$, not $-1/(6 \pi)$. – David Speyer Dec 8 '11 at 0:09. 1. I think you're writing $-1/(6\pi)$ when you mean $-(\pi/6)$. 2. If what you are reporting accurately represents what's on Khan Academy then Khan Academy is wrong. It happens to the best of us. – Gerry Myerson Dec 8 '11 at 0:11. You have shown that $\tan(-\pi/6)=-1/\sqrt{3}=-\sqrt{3}/3$. Since $\arctan x$ is the angle in $(-\pi/2,\pi/2)$ whose $\tan$ is $x$, you have shown that $\arctan(-\sqrt{3}/3)=-\pi/6$. – André Nicolas Dec 8 '11 at 0:23. It is slightly unfortunate that Khan Academy says Note: please answer in terms of radians (ex: "3/4 pi"), partly because whether $\pi$ is part of the denominator or not is unclear, and partly because it would not accept "5/6 pi" in answer to this particular question, instead insisting on "-1/6 pi" as the principal value.
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– Henry Dec 8 '11 at 0:51. Maybe you're just missing the fact that $1/\sqrt{3}$ is the same as $\sqrt{3}/3$. You get that by rationalizing the denominator:. $$\frac{1}{\sqrt{3}} = \frac{1\cdot\sqrt{3}}{\sqrt{3}\sqrt{3}} = \frac{\sqrt{3}}{3}.$$. -. Wow. Yes I completely missed that. That seems to have been my error. Thank you Michael. – drc Dec 8 '11 at 1:02. Maybe you're just messing with the concept of inverse function... Remember:. $\arctan x=y$ if and only if $\tan y =x$ and $y\in ]-\pi/2,\pi/2[$.. Now, you have found $\tan (-\pi/6) =-1/\sqrt{3}=-\sqrt{3}/3$ and you also have $-\pi/6 \in ]-\pi/2, \pi/2[$, hence previous statement applies and it yields $\arctan (-\sqrt{3}/3)=-\pi/6$ as claimed.. -. What happens if y is outside of those bounds you specify? – drc Dec 8 '11 at 2:00. Also your use of ][, I'm assuming it indicates an open interval? I'm wondering why you used that instead of (-pi/2 < y < pi/2) – drc Dec 8 '11 at 2:06. @drc: It's not uncommon notation; I've seen it many times. – Arturo Magidin Dec 8 '11 at 2:26. @drc: I usually use the reverse square brakets for the open interval just to avoid confusion with other symbols; in fact, for example, $(a,b)$ is also used to denote the ordered pair with first entry $a$ and second entry $b$. – Pacciu Dec 8 '11 at 12:41. @drc: If $y\notin ]-\pi/2 ,\pi/2[$ then nothing "happens", in the sense that you cannot give any meaning to the formula $\arctan x=y$. In fact, as you should know from your book, the range of $\mathbb{R}\ni x\mapsto \arctan x\in \mathbb{R}$ is the open interval $]-\pi/2, \pi/2[$. – Pacciu Dec 8 '11 at 12:44.
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# Maths no problem. Maths no problem can help students to understand the material and improve their grades. We can solving math problem.. ## The Best Maths no problem. Maths no problem is a mathematical instrument that assists to solve math equations. At the same time, because the book is often turned, the material has some missing corners, which is a great challenge to the scanner's three-dimensional flattening technology. The stability of the scanner's software processing determines the imaging effect. On the whole, the picture that looks strong is still slightly overexposed, and the clarity is OK, but the handwriting is shallow and still a little fuzzy; As for the use of computer drawing, the author of the paper points out that when it comes to the illustrations of the 11th set of primary school mathematics textbooks of the people's education society, it is to first draw the outline by hand, and then scan the outline to the computer with a scanner for later coloring and combining, so as to complete the creation of the whole illustration. The figure outline in the illustration is hand drawn, which is very casual, easy and complete in one go. The items and props are drawn by computer.. Our proposed formula allows general constraints to be handled in location-based settings. The location-based explicit solver is easy to understand and implement. Many differential equations can be solved by integrating directly, but some differential equations are not. In other words, it is difficult to find suitable differential homeomorphisms directly for these differential equations to rectify the original equations. For this reason, Newton thought of using Taylor expansion to solve it.. However, in the real situation, we can't meet so well-designed data, and the resulting equation may not be reducible on mathbq, so we can't even do the first step This forces us to seek more general methods to obtain such good things as the root finding formula of quadratic equations, so that we can solve arbitrary polynomial equations.
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This is what people have pursued in history The core of the infinite self similarity method is to establish a shape such as R_ infty=R_ {infty + 1}. In this method, because series and parallel exist at the same time, it is often necessary to solve the quadratic equation, which usually has two roots. For pure resistance networks, these two roots can clearly judge the positive and negative, so that one of them can be easily excluded. However, when there are inductance, capacitance and other components in the circuit, this judgment method needs to be improved.. The form compensation method is to supplement a figure into a triangle, which is the most common one of our contact form compensation methods. Its form is also relatively variable. The main complementable behaviors are general triangles, isosceles triangles, right triangles and equilateral triangles. How should we complement them? We should make changes according to the existing form in the conditions.. The building area of the relocation house exceeds 15 square meters per capita? The excess part shall be purchased at the price of local affordable housing. For households with housing difficulties whose per capita area is less than 5 square meters. Resettlement shall be carried out according to the standard of per capita construction area of 5 square meters... ## Math solver you can trust. Amazing I found that when I used the app for the first time it showed me how to use the app and I'd recommend to anyone struggling on a problem in math So amazing and so beneficial. This app as helped me a lot. However, the questions are, it will solve it for u.
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# Example: An insulating solid sphere of radius R has a uniform positive volume charge density and total charge Q. a)Find the electric potential at a point.
## Presentation on theme: "Example: An insulating solid sphere of radius R has a uniform positive volume charge density and total charge Q. a)Find the electric potential at a point."— Presentation transcript:
Example: An insulating solid sphere of radius R has a uniform positive volume charge density and total charge Q. a)Find the electric potential at a point outside the sphere (r > R). Take V = 0 at r = b)Find the electric potential at a point inside the sphere (r < R). a) r > R: Outside the sphere we can determine the electric field using Gauss’s Law and we get the Electric field of a point charge 0 0 Electric potential of a point charge
b) r < R: The electric field inside the sphere can be determined using Gauss’s Law Electric potential at the surface of the sphere Radial component of E To simplify expression
Electric potential due to a charged conductor We are now going to examine the electric potential of a conductor. In order to do so we must recall some important information about conductors. 1) What is the strength of the electric field inside a conductor? 2) What is the strength and direction of the electric field outside of a conductor relative to the surface of the conductor? 3) Where is all the excess charge located for a conductor? Zero – the electric field inside a conductor is always zero. The electric field outside a conductor has a strength of 0 and is directed perpendicular to the surface (parallel to the area vector). All excess charge is located on the surface of a conductor.
Let us now look at the electric potential of a conductor 1) If we have two points on the surface of a conductor, what is the potential difference between these two points? 2) What is the potential difference between a point inside the conductor and a point on the surface of the conductor? E and ds are perpendicular as you move along the surface 0 All points on the surface are electrically connected and therefore at the same electric potential. 0 All points on the surface are electrically connected to all point inside the conductor and therefore at the same electric potential. The electric field is always zero inside a conductor This means that all points anywhere on or in the conductor are all at the same potential !!
3) How much work must be done to move a charge from one point on a conductor to another? 0 All points are at the same electric potential. Charges move freely in a conductor with no addition or loss of energy. What about a hollow conductor, what is the potential difference between two points on opposite sides of the cavity? What is the electric field inside the cavity of a conductor? The electric potential difference is still zero since all parts of the shell are still electrically connected. The net electric field is still zero inside the cavity of the conductor.
A solid spherical conductor is given a net nonzero charge. The electrostatic potential of the conductor is 1. largest at the center. 2. largest on the surface. 3. largest somewhere between center and surface. 4. constant throughout the volume.
Consider two isolated spherical conductors each having net charge Q. The spheres have radii a and b, where b > a. Which sphere has the higher potential? 1. the sphere of radius a 2. the sphere of radius b 3. They have the same potential. r = a or b
A conducting shell is primarily used as shielding. Anything that is inside the conductor is completely protected from any electric fields. A conducting shell used for this purpose is often called a Faraday Cage. Faraday cages are widely used to protect sensitive electronics from external influences. Faraday cages can be made of solid conductors or a conducting mesh.
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Presentation is loading. Please wait.. # Example: An insulating solid sphere of radius R has a uniform positive volume charge density and total charge Q. a)Find the electric potential at a point.. ## Presentation on theme: "Example: An insulating solid sphere of radius R has a uniform positive volume charge density and total charge Q. a)Find the electric potential at a point."— Presentation transcript:. Example: An insulating solid sphere of radius R has a uniform positive volume charge density and total charge Q. a)Find the electric potential at a point outside the sphere (r > R). Take V = 0 at r = b)Find the electric potential at a point inside the sphere (r < R). a) r > R: Outside the sphere we can determine the electric field using Gauss’s Law and we get the Electric field of a point charge 0 0 Electric potential of a point charge. b) r < R: The electric field inside the sphere can be determined using Gauss’s Law Electric potential at the surface of the sphere Radial component of E To simplify expression. Electric potential due to a charged conductor We are now going to examine the electric potential of a conductor. In order to do so we must recall some important information about conductors. 1) What is the strength of the electric field inside a conductor? 2) What is the strength and direction of the electric field outside of a conductor relative to the surface of the conductor? 3) Where is all the excess charge located for a conductor? Zero – the electric field inside a conductor is always zero. The electric field outside a conductor has a strength of 0 and is directed perpendicular to the surface (parallel to the area vector). All excess charge is located on the surface of a conductor.. Let us now look at the electric potential of a conductor 1) If we have two points on the surface of a conductor, what is the potential difference between these two points? 2) What is the potential difference between a point inside the conductor and a point on the surface of the conductor? E and ds are perpendicular as you move along the surface 0 All points on the surface are electrically connected and therefore at the same electric potential. 0 All points on the surface are electrically connected to all point inside the conductor and therefore at the same electric potential. The electric field is always zero inside a conductor This means that all points anywhere on or in the conductor are all at the same potential !!. 3) How much work must be done to move a charge from one point on a conductor to another? 0 All points are at the same electric potential. Charges move freely in a conductor with no addition or loss of energy. What about a hollow conductor, what is the potential difference between two points on opposite sides of the cavity? What is the electric field inside the cavity of a conductor? The electric potential difference is still zero since all parts of the shell are still electrically connected. The net electric field is still zero inside the cavity of the conductor.. A solid spherical conductor is given a net nonzero charge. The electrostatic potential of the conductor is 1. largest at the center.
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2. largest on the surface. 3. largest somewhere between center and surface. 4. constant throughout the volume.. Consider two isolated spherical conductors each having net charge Q. The spheres have radii a and b, where b > a. Which sphere has the higher potential? 1. the sphere of radius a 2. the sphere of radius b 3. They have the same potential. r = a or b. A conducting shell is primarily used as shielding. Anything that is inside the conductor is completely protected from any electric fields. A conducting shell used for this purpose is often called a Faraday Cage. Faraday cages are widely used to protect sensitive electronics from external influences. Faraday cages can be made of solid conductors or a conducting mesh.. Download ppt "Example: An insulating solid sphere of radius R has a uniform positive volume charge density and total charge Q. a)Find the electric potential at a point.". Similar presentations. Ads by Google.
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http://gmatclub.com/forum/all-trainees-in-a-certain-aviator-training-program-must-take-47829.html?sort_by_oldest=true
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# All trainees in a certain aviator training program must take
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All trainees in a certain aviator training program must take [#permalink] 26 Jun 2007, 10:39
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All trainees in a certain aviator training program must take both a written test and a flight test. If 70 percent of the trainees passed the written test, and 80 percent of the trainees passed the flight test, what percent of the trainees passed both tests?
(1) 10 percent of the trainees did not pass either test.
(2) 20 percent of the trainees passed only the flight test.
Director
Joined: 06 Sep 2006
Posts: 746
Followers: 1
Kudos [?]: 15 [0], given: 0
Set Theory
Total = Set1 + Set2 + Neither - Both (Common)
1)
100 = 70 + 80 + 10 - Both
Both = 60
2)
80 percent of the trainees passed the flight test.
20 percent of the trainees passed only the flight test.
Both = 80 - 20 = 60
D.
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All trainees in a certain aviator training program must take 4 14 Jan 2010, 10:43
All trainees in a certain aviator training program must take 1 24 Jun 2009, 02:11
All trainees in a certain aviator training program must take 1 20 Sep 2008, 01:19
All trainees in a certain aviator training program must take 3 26 Apr 2005, 09:34
If all of the telephone extensions in a certain company must 16 22 Jan 2005, 13:03
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Find all School-related info fast with the new School-Specific MBA Forum. It is currently 14 Sep 2014, 18:09. ### GMAT Club Daily Prep. #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.. Customized. for You. we will pick new questions that match your level based on your Timer History. Track. every week, we’ll send you an estimated GMAT score based on your performance. Practice. Pays. we will pick new questions that match your level based on your Timer History. # Events & Promotions. ###### Events & Promotions in June. Open Detailed Calendar. # All trainees in a certain aviator training program must take. Author Message. TAGS:. Director. Joined: 09 Aug 2006. Posts: 529. Followers: 2. Kudos [?]: 23 [0], given: 0. All trainees in a certain aviator training program must take [#permalink] 26 Jun 2007, 10:39. 00:00. Difficulty:. (N/A). Question Stats:. 0% (00:00) correct 100% (00:02) wrong based on 1 sessions.
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All trainees in a certain aviator training program must take both a written test and a flight test. If 70 percent of the trainees passed the written test, and 80 percent of the trainees passed the flight test, what percent of the trainees passed both tests?. (1) 10 percent of the trainees did not pass either test.. (2) 20 percent of the trainees passed only the flight test.. Director. Joined: 06 Sep 2006. Posts: 746. Followers: 1. Kudos [?]: 15 [0], given: 0. Set Theory. Total = Set1 + Set2 + Neither - Both (Common). 1). 100 = 70 + 80 + 10 - Both. Both = 60. 2). 80 percent of the trainees passed the flight test.. 20 percent of the trainees passed only the flight test.. Both = 80 - 20 = 60. D.. Similar topics Replies Last post. Similar. Topics:. All trainees in a certain aviator training program must take 4 14 Jan 2010, 10:43. All trainees in a certain aviator training program must take 1 24 Jun 2009, 02:11. All trainees in a certain aviator training program must take 1 20 Sep 2008, 01:19. All trainees in a certain aviator training program must take 3 26 Apr 2005, 09:34. If all of the telephone extensions in a certain company must 16 22 Jan 2005, 13:03. Display posts from previous: Sort by.
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Instructions
What approximate value will come in place of, the question mark (?) in the following questions ? (You are not expected to calculate the exact value).
Question 37
# $$159\div39.99+\frac{4}{5}\times180-120.05$$
Solution
The given equation can be written as $$\frac{160}{40} + \frac{4}{5}* 180 - 120$$
$$4+144-120= 28$$
Option A is the right answer.
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Instructions. What approximate value will come in place of, the question mark (?) in the following questions ? (You are not expected to calculate the exact value).. Question 37. # $$159\div39.99+\frac{4}{5}\times180-120.05$$. Solution.
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The given equation can be written as $$\frac{160}{40} + \frac{4}{5}* 180 - 120$$. $$4+144-120= 28$$. Option A is the right answer.
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https://www.codevscolor.com/javascript-find-sphere-volume
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# JavaScript program to find the volume of a sphere
## JavaScript program to find the volume of a sphere:
In this post, we will learn how to find the volume of a sphere in JavaScript. With this program, you will learn how to do basic mathematical calculations in JavaScript.
To find the volume of a sphere, we need the radius. We can use just a simple formula to find the volume.
Below formula is used to find the volume of a sphere:
``Volume = 4/3 * πr^3``
Where, r is the radius of the sphere. π or PI is a mathematical constant. It is defined in the Math class in JavaScript. We can create one variable initialized with the value of π or we can simply use the Math class.
Now, to calculate the volume, we need just the radius of the sphere. We can then use this formula to find the volume.
In this post, we will learn different ways to solve this problem. Let’s check these one by one.
### Method 1: Finding the sphere volume with a given radius value:
Let’s try to find the volume of a sphere with a given radius value. The value of the radius is given and this program will simply print the volume of the sphere.
``````let radius = 5;
let volume = (4/3)* Math.PI * Math.pow(radius, 3);
console.log('Volume of Sphere: '+volume.toFixed(2));``````
Here,
• radius is the given radius of the sphere.
• The volume is holding the volume of the sphere. It uses the same formula we discussed above to calculate the volume.
• Finally, the last line is printing this value. It is formatting it to two digit after the decimal.
If you run this program, it will print the below output:
``Volume of Sphere: 523.60``
### Method 2: Finding the sphere volume with a user given radius value:
We can also take the radius as input from the user. The other part of the program will run in a same way. Just read the radius and calculate the volume.
``````let radius = parseInt(prompt("Enter the radius", "0"), 0);
let volume = (4/3)* Math.PI * Math.pow(radius, 3);
console.log('Volume of Sphere: '+volume.toFixed(2));``````
You need to run it in a browser. Go to the console tab in the developer menu and paste the code. It will ask the user to enter the radius value with a prompt. Enter it and hit ok, it will show the volume in the console.
You will get similar result as the above example.
### Method 3: Using with HTML:
We can also use the JavaScript code with a HTML page. Create one file called index.html with the below code:
``````<input type="number" id="input_radius" placeholder="Enter the radius">
<input type="button" value="Find" id="input_button" >
<script>
function calculateVolume(){
let volume = (4/3)* Math.PI * Math.pow(radius, 3);
alert("The volume of a sphere: "+volume.toFixed(2));
}
let button=document.getElementById("input_button");
button.onclick=calculateVolume;
</script>``````
This is a HTML file with the javascript code included to find the volume. If you open this file in a browser, it will show one input box with a button. This input box is to get the radius and the button is to calculate that value.
If you enter any value in that input box and click on the button, it will show one alert dialog that will show the volume of the sphere as like below:
You can add css to style the input box and the button as you like.
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# JavaScript program to find the volume of a sphere. ## JavaScript program to find the volume of a sphere:. In this post, we will learn how to find the volume of a sphere in JavaScript. With this program, you will learn how to do basic mathematical calculations in JavaScript.. To find the volume of a sphere, we need the radius. We can use just a simple formula to find the volume.. Below formula is used to find the volume of a sphere:. ``Volume = 4/3 * πr^3``. Where, r is the radius of the sphere. π or PI is a mathematical constant. It is defined in the Math class in JavaScript. We can create one variable initialized with the value of π or we can simply use the Math class.. Now, to calculate the volume, we need just the radius of the sphere. We can then use this formula to find the volume.. In this post, we will learn different ways to solve this problem. Let’s check these one by one.. ### Method 1: Finding the sphere volume with a given radius value:. Let’s try to find the volume of a sphere with a given radius value. The value of the radius is given and this program will simply print the volume of the sphere.. ``````let radius = 5;. let volume = (4/3)* Math.PI * Math.pow(radius, 3);. console.log('Volume of Sphere: '+volume.toFixed(2));``````. Here,. • radius is the given radius of the sphere.. • The volume is holding the volume of the sphere. It uses the same formula we discussed above to calculate the volume.. • Finally, the last line is printing this value. It is formatting it to two digit after the decimal.. If you run this program, it will print the below output:. ``Volume of Sphere: 523.60``. ### Method 2: Finding the sphere volume with a user given radius value:.
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We can also take the radius as input from the user. The other part of the program will run in a same way. Just read the radius and calculate the volume.. ``````let radius = parseInt(prompt("Enter the radius", "0"), 0);. let volume = (4/3)* Math.PI * Math.pow(radius, 3);. console.log('Volume of Sphere: '+volume.toFixed(2));``````. You need to run it in a browser. Go to the console tab in the developer menu and paste the code. It will ask the user to enter the radius value with a prompt. Enter it and hit ok, it will show the volume in the console.. You will get similar result as the above example.. ### Method 3: Using with HTML:. We can also use the JavaScript code with a HTML page. Create one file called index.html with the below code:. ``````<input type="number" id="input_radius" placeholder="Enter the radius">. <input type="button" value="Find" id="input_button" >. <script>. function calculateVolume(){. let volume = (4/3)* Math.PI * Math.pow(radius, 3);. alert("The volume of a sphere: "+volume.toFixed(2));. }. let button=document.getElementById("input_button");. button.onclick=calculateVolume;. </script>``````. This is a HTML file with the javascript code included to find the volume. If you open this file in a browser, it will show one input box with a button. This input box is to get the radius and the button is to calculate that value.. If you enter any value in that input box and click on the button, it will show one alert dialog that will show the volume of the sphere as like below:. You can add css to style the input box and the button as you like.
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https://www.vedantu.com/question-answer/a-tv-marked-price-rs-16000-is-available-on-15-class-8-maths-cbse-5ed97ce8251062488817e2e0
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# A TV marked price Rs 16000 is available on 15 percent off. What is its selling price?(A) Rs 13600(B) Rs 11000(C) Rs 15760(D) Rs 15000
Last updated date: 19th Jul 2024
Total views: 453.6k
Views today: 10.53k
Given, the marked price of TV is ${\text{M}}{\text{.P}} = {\text{Rs 16000}}$
As we know that selling price of any item is given by ${\text{S}}{\text{.P}} = {\text{M}}{\text{.P}} - \left[ {\left( {\dfrac{{{\text{Percent off}}}}{{100}}} \right) \times \left( {{\text{M}}{\text{.P}}} \right)} \right]$
The selling price of the TV is ${\text{S}}{\text{.P}} = 16000 - \left[ {\left( {\dfrac{{{\text{15}}}}{{100}}} \right) \times \left( {{\text{16000}}} \right)} \right] = 16000 - 2400 = {\text{Rs }}13600$.
Note- In these type of problems, the amount of discount given on the marked price is $\left[ {\left( {\dfrac{{{\text{Percent off}}}}{{100}}} \right) \times \left( {{\text{M}}{\text{.P}}} \right)} \right]$ and finally this discount amount is subtracted from the marked price in order to get the selling price.
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Courses. Courses for Kids. Free study material. Offline Centres. More. Store. # A TV marked price Rs 16000 is available on 15 percent off. What is its selling price?(A) Rs 13600(B) Rs 11000(C) Rs 15760(D) Rs 15000.
|
Last updated date: 19th Jul 2024. Total views: 453.6k. Views today: 10.53k. Given, the marked price of TV is ${\text{M}}{\text{.P}} = {\text{Rs 16000}}$. As we know that selling price of any item is given by ${\text{S}}{\text{.P}} = {\text{M}}{\text{.P}} - \left[ {\left( {\dfrac{{{\text{Percent off}}}}{{100}}} \right) \times \left( {{\text{M}}{\text{.P}}} \right)} \right]$. The selling price of the TV is ${\text{S}}{\text{.P}} = 16000 - \left[ {\left( {\dfrac{{{\text{15}}}}{{100}}} \right) \times \left( {{\text{16000}}} \right)} \right] = 16000 - 2400 = {\text{Rs }}13600$.. Note- In these type of problems, the amount of discount given on the marked price is $\left[ {\left( {\dfrac{{{\text{Percent off}}}}{{100}}} \right) \times \left( {{\text{M}}{\text{.P}}} \right)} \right]$ and finally this discount amount is subtracted from the marked price in order to get the selling price.
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Bodhee Prep-CAT Online Preparation
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Number of Factors Concepts and Formulas for CAT exam
Number of factors based questions have regularly appeared in various competitive exams including CAT. Primarily, these questions are based on the prime factorization of a number. Generally, factors based questions are of following types:
For certain kind of questions, we already have direct formula and shortcut techniques which we apply to get the answer. However, if an aspirant wants to master this topic, must also understand the conceptual approach to solve these types of questions.
In this article, we will focus more on the concepts and approach to tackle question based on factors.
Prime factorisation
We all know that every composite number can be written as a product of some prime numbers. For example, we can write 90 as $2 \times {3^2} \times 5$. This process is called prime factorisation, and it is the very first step to solve any questions related to factors.
Number of factors of a number
Factors are those numbers that divide the given number completely. For example, below is the list of all factors of number 72.
Factors: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 48 and 72.
Observe that the number 1 is always a factor of every number, and the number itself will be the factor of the number. Here we can see that there are 12 factors of 72.
Now let us understand the formula for finding the number of factors of any composite number. We will take the above example of 72.
If the prime factorization 72 we get, $72 = {2^3} \times {3^2}$
Clearly, the number 72 is divisible by each of ${2^0},{2^1},{2^2},{2^3}$ but not by any higher power
of 2 like ${2^4},{2^5},…$
Similarly, the number 72 is divisible by each of ${3^0},{3^1},{3^2}$ but not by ${3^3},{3^4},….$
We should also observe that the number can be divisible by any ‘combination’ of one
of ${2^0},{2^1},{2^2},{2^3}$ and one of ${3^0},{3^1},{3^2}$ i.e. by numbers of the type ${2^1} \times {3^2}$ or ${2^2} \times {3^1}$ or ${2^1} \times {3^1}$ and so on.
Thus, with ${2^0}$, we could have a total of 3 ‘combinations’ i.e.
Each of these 3 numbers would divide ${2^3} \times {3^2}$ and thus would be a factor.
Similarly, with ${2^1}$, we could have 3 more combinations
And each of these 3 factors would be distinct from the earlier 3 factors.
Similarly with EACH of ${2^2}$ and ${2^3}$, we would get 3 more distinct factors and thus the
the total number of factors would be 4 × 3 = 12.
If observed, each of the factors is are distinct numbers because the power of 2 or 3 differs in each of the combinations.
Since the exponent of 2 could assume 4 different values (from 0 to 3), the exponent of 3 could assume 3 distinct values (from 0 to 2), the total number of combinations is 4 × 3 = 12.
From the above example, it is clear that all the factors of 72 would be in the form ${2^a} \times {3^b}$ where a could assume any value from 0 to 3 i.e., 4 different values, and b could consider any value from 0 to 2 i.e., 3 different values. Therefore, the number of different combinations and the number of factors would be 4 × 3 i.e. 12.
The above concepts is lucidly explained in the below video tutorials on number of factors.
Important Formula
Number of factors
If we have to find the number of factors of any number say N, then we should follow below steps:
Step 1: Prime factorize $N = {p^a} \times {q^b} \times {r^c} \times …$
Step 2: The number of factors of N= (a+1)(b+1)(c+1)…
Sum of factors
To find the sum of all the factors of a number (say N), we follow below two steps:
Step 1: Prime factorize $N = {p^a} \times {q^b} \times {r^c} \times …$
Step 2: Sum of factors =$\left( {\frac{{{p^{a + 1}} – 1}}{{p – 1}}} \right)\left( {\frac{{{q^{b + 1}} – 1}}{{q – 1}}} \right)\left( {\frac{{{r^{c + 1}} – 1}}{{r – 1}}} \right) \ldots$
Product of Factors
To find the product of all the factors of a number (say N), we follow below three steps:
Step 1: Prime factorize $N = {p^a} \times {q^b} \times {r^c} \times …$
Step 2: Let the number of factors of N be x. therefore, x= (a+1)(b+1)(c+1)…
Step 3: Product of factors =${N^{\frac{x}{2}}}$
Let us take an example to understand the working of the above formulas.
Question: Find the number of factors, the sum of factors and product of factors of 1800.
Solution:
Prime factorization of $1800 = {2^3} \times {3^2} \times {5^2}$
Number of factors 1800= (3+1)(2+1)(2+1) = 36
Sum of factors of 1800
$\begin{array}{l} = \left( {\frac{{{2^{3 + 1}} – 1}}{{2 – 1}}} \right)\left( {\frac{{{3^{2 + 1}} – 1}}{{3 – 1}}} \right)\left( {\frac{{{5^{2 + 1}} – 1}}{{5 – 1}}} \right)\\ = 15 \times 4 \times 31 = 1860\end{array}$
Product of Factors of 1800=${1800^{\frac{{36}}{2}}} = {1800^{18}}$
Number of factors of specific types
The questions which appear in competitive exams from factors are not restricted to the above three types only. Variety of questions can be created based on finding the factors. The best way to approach such questions is to go by the conceptual way rather than remembering the formula for each kind.
Let us take some varieties of questions which appear frequently.
Number of odd and even factors
Question: Find the number of Even and Odd factors of 1200.
Solution:
Prime factorization of $1200 = {2^4} \times {3^1} \times {5^2}$
We know the fact that all the factors of 1200 will be in the form of ${2^a} \times {3^b} \times {5^c}$.
Where a will range from 0 to 4, b from 0 to 1 and c from 0 to 2. However, it is quite evident that any factor which has 2 as one of the prime factors is always even. similarly, and your factors will not have 2 as its prime factor.
Therefore, for ODD factors, the exponent of 2 i.e., a has to be 0 always. Or, the number of ways using 2 for making the combination is 1, i.e., ${2^0}$. Also, the number of values which the exponent of 3 and 5, i.e., b, and c can take are 2 and 3 respectively.
Hence the number of odd factors of 1200 = 1x2x3=6.
Extending the logic, we can say that for a better to be Even, it must contain 2 at least once.
So, the number of values a, b, and c can take are 4, 2, and 3 respectively.
Therefore, the total number of even factors of 1200 = 4x2x3=24.
Number of factors which are perfect squares
Question: Find the number of factors which are perfect squares of 10800.
Solution:
Prime factorization of $10800 = {2^4} \times {3^3} \times {5^2}$
If we prime factorise any number which is a perfect square, we would observe that in all cases the exponent of all the prime factors of the number to be even only.
For example, 36 is perfect square $36 = {2^2} \times {3^2}$. here we can see that the exponent of both 2 and 3 are even.
Again, any factor 10800 will be in the form of ${2^a}{ \times ^b} \times {5^c}$. For the factors to be perfect squares, all the values a, b, and c has to be even only.
Or, the possible values which a can take = 0, 2, 4, i.e. 3 values only. Similarly, b can take 0, 2 i.e. 2 values and c can take 0, 2 i.e. 2 values.
Therefore, the different combinations we can have = 3x2x2 = 12.
Hence, 10800 has 12 factors which are perfect squares.
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Bodhee Prep-CAT Online Preparation. | Best Online CAT PreparationFor Enquiry CALL @ +91-95189-40261. Number of Factors Concepts and Formulas for CAT exam. Number of factors based questions have regularly appeared in various competitive exams including CAT. Primarily, these questions are based on the prime factorization of a number. Generally, factors based questions are of following types:. For certain kind of questions, we already have direct formula and shortcut techniques which we apply to get the answer. However, if an aspirant wants to master this topic, must also understand the conceptual approach to solve these types of questions.. In this article, we will focus more on the concepts and approach to tackle question based on factors.. Prime factorisation. We all know that every composite number can be written as a product of some prime numbers. For example, we can write 90 as $2 \times {3^2} \times 5$. This process is called prime factorisation, and it is the very first step to solve any questions related to factors.. Number of factors of a number. Factors are those numbers that divide the given number completely. For example, below is the list of all factors of number 72.. Factors: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 48 and 72.. Observe that the number 1 is always a factor of every number, and the number itself will be the factor of the number. Here we can see that there are 12 factors of 72.. Now let us understand the formula for finding the number of factors of any composite number. We will take the above example of 72.. If the prime factorization 72 we get, $72 = {2^3} \times {3^2}$. Clearly, the number 72 is divisible by each of ${2^0},{2^1},{2^2},{2^3}$ but not by any higher power. of 2 like ${2^4},{2^5},…$. Similarly, the number 72 is divisible by each of ${3^0},{3^1},{3^2}$ but not by ${3^3},{3^4},….$. We should also observe that the number can be divisible by any ‘combination’ of one. of ${2^0},{2^1},{2^2},{2^3}$ and one of ${3^0},{3^1},{3^2}$ i.e. by numbers of the type ${2^1} \times {3^2}$ or ${2^2} \times {3^1}$ or ${2^1} \times {3^1}$ and so on.. Thus, with ${2^0}$, we could have a total of 3 ‘combinations’ i.e.. Each of these 3 numbers would divide ${2^3} \times {3^2}$ and thus would be a factor.. Similarly, with ${2^1}$, we could have 3 more combinations. And each of these 3 factors would be distinct from the earlier 3 factors.. Similarly with EACH of ${2^2}$ and ${2^3}$, we would get 3 more distinct factors and thus the. the total number of factors would be 4 × 3 = 12.. If observed, each of the factors is are distinct numbers because the power of 2 or 3 differs in each of the combinations.. Since the exponent of 2 could assume 4 different values (from 0 to 3), the exponent of 3 could assume 3 distinct values (from 0 to 2), the total number of combinations is 4 × 3 = 12.. From the above example, it is clear that all the factors of 72 would be in the form ${2^a} \times {3^b}$ where a could assume any value from 0 to 3 i.e., 4 different values, and b could consider any value from 0 to 2 i.e., 3 different values. Therefore, the number of different combinations and the number of factors would be 4 × 3 i.e. 12.. The above concepts is lucidly explained in the below video tutorials on number of factors.. Important Formula. Number of factors. If we have to find the number of factors of any number say N, then we should follow below steps:. Step 1: Prime factorize $N = {p^a} \times {q^b} \times {r^c} \times …$. Step 2: The number of factors of N= (a+1)(b+1)(c+1)…. Sum of factors. To find the sum of all the factors of a number (say N), we follow below two steps:. Step 1: Prime factorize $N = {p^a} \times {q^b} \times {r^c} \times …$. Step 2: Sum of factors =$\left( {\frac{{{p^{a + 1}} – 1}}{{p – 1}}} \right)\left( {\frac{{{q^{b + 1}} – 1}}{{q – 1}}} \right)\left( {\frac{{{r^{c + 1}} – 1}}{{r – 1}}} \right) \ldots$. Product of Factors. To find the product of all the factors of a number (say N), we follow below three steps:. Step 1: Prime factorize $N = {p^a} \times {q^b} \times {r^c} \times …$. Step 2: Let the number of factors of N be x. therefore, x= (a+1)(b+1)(c+1)…. Step 3: Product of factors =${N^{\frac{x}{2}}}$. Let us take an example to understand the working of the above formulas.. Question: Find the number of factors, the sum of factors and product of factors of 1800.. Solution:. Prime factorization of $1800 = {2^3} \times {3^2} \times {5^2}$. Number of factors 1800= (3+1)(2+1)(2+1) = 36.
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Sum of factors of 1800. $\begin{array}{l} = \left( {\frac{{{2^{3 + 1}} – 1}}{{2 – 1}}} \right)\left( {\frac{{{3^{2 + 1}} – 1}}{{3 – 1}}} \right)\left( {\frac{{{5^{2 + 1}} – 1}}{{5 – 1}}} \right)\\ = 15 \times 4 \times 31 = 1860\end{array}$. Product of Factors of 1800=${1800^{\frac{{36}}{2}}} = {1800^{18}}$. Number of factors of specific types. The questions which appear in competitive exams from factors are not restricted to the above three types only. Variety of questions can be created based on finding the factors. The best way to approach such questions is to go by the conceptual way rather than remembering the formula for each kind.. Let us take some varieties of questions which appear frequently.. Number of odd and even factors. Question: Find the number of Even and Odd factors of 1200.. Solution:. Prime factorization of $1200 = {2^4} \times {3^1} \times {5^2}$. We know the fact that all the factors of 1200 will be in the form of ${2^a} \times {3^b} \times {5^c}$.. Where a will range from 0 to 4, b from 0 to 1 and c from 0 to 2. However, it is quite evident that any factor which has 2 as one of the prime factors is always even. similarly, and your factors will not have 2 as its prime factor.. Therefore, for ODD factors, the exponent of 2 i.e., a has to be 0 always. Or, the number of ways using 2 for making the combination is 1, i.e., ${2^0}$. Also, the number of values which the exponent of 3 and 5, i.e., b, and c can take are 2 and 3 respectively.. Hence the number of odd factors of 1200 = 1x2x3=6.. Extending the logic, we can say that for a better to be Even, it must contain 2 at least once.. So, the number of values a, b, and c can take are 4, 2, and 3 respectively.. Therefore, the total number of even factors of 1200 = 4x2x3=24.. Number of factors which are perfect squares. Question: Find the number of factors which are perfect squares of 10800.. Solution:. Prime factorization of $10800 = {2^4} \times {3^3} \times {5^2}$. If we prime factorise any number which is a perfect square, we would observe that in all cases the exponent of all the prime factors of the number to be even only.. For example, 36 is perfect square $36 = {2^2} \times {3^2}$. here we can see that the exponent of both 2 and 3 are even.. Again, any factor 10800 will be in the form of ${2^a}{ \times ^b} \times {5^c}$. For the factors to be perfect squares, all the values a, b, and c has to be even only.. Or, the possible values which a can take = 0, 2, 4, i.e. 3 values only. Similarly, b can take 0, 2 i.e. 2 values and c can take 0, 2 i.e. 2 values.. Therefore, the different combinations we can have = 3x2x2 = 12.. Hence, 10800 has 12 factors which are perfect squares.. Other Articles of Number system. [PDF] CAT 2021 Question Paper (slot 1, 2 & 3) with Solutions. CAT 2021 question paper PDF is available on this page. The page has the CAT 2021 question paper PDFs of all the three slots. There. All About CAT Mock Test Series. Table of Content for CAT Mock Tests Ideal number of CAT Mock Test Series How many CAT mocks should one write What is the right. CAT success stories of our 2021 and 2020 batch. The stories that we are sharing here are some of those students whom we mentored right from the start of their preparation. Having mentored them,. [PDF] CAT 2020 Question Paper (slot 1,2 &3) with Solution. CAT 2020 question paper threw a number of surprises. Not only was there a change in exam pattern but also the difficulty level of almost. CAT 2020 Analysis : Slot (1 2 and 3) – cutoffs. Much of CAT 2020 turned out to be as expected, both in terms of pattern and difficulty. Following the announcement of the change in pattern,. CAT Online Coaching Course for Preparation of CAT 2022. If you are new to CAT preparation, and are looking for a full-fledged CAT online coaching, then this is the page that you must completely. CAT 2023Classroom Course. We are starting classroom course for CAT 2023 in Gurugram from the month of December.
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Unformatted text preview: Kapoor (mk9499) – oldhomework 16 – Turner – (60230) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points A 156 Ω lamp, a 27 Ω electric heater, and a 41 Ω fan are connected in parallel across a 119 V line. What total current is supplied to the cir- cuit? Correct answer: 8 . 07267 A. Explanation: Let : R 1 = 156 Ω , R 2 = 27 Ω , R 3 = 41 Ω , and V = 119 V . For the parallel circuit, 1 R = 1 R 1 + 1 R 2 + 1 R 3 = R 1 R 2 + R 2 R 3 + R 3 R 1 R 1 R 2 R 3 R = R 1 R 2 R 3 R 1 R 2 + R 2 R 3 + R 3 R 1 R 1 R 2 + R 2 R 3 + R 3 R 1 = (156 Ω) (27 Ω) + (27 Ω) (41 Ω) + (41 Ω) (156 Ω) = 11715 Ω 2 , so R = (156 Ω) (27 Ω) (41 Ω) 11715 Ω 2 = 14 . 7411 Ω and I = V R = 119 V 14 . 7411 Ω = 8 . 07267 A . 002 (part 2 of 4) 10.0 points What is the voltage across the fan? Correct answer: 119 V. Explanation: Because the circuit elements are connected in parallel, the voltage across the fan is the same as that across the whole circuit. 003 (part 3 of 4) 10.0 points What is the current in the lamp? Correct answer: 0 . 762821 A. Explanation: The voltage is the same in all 3 appliances, so the current through the lamp is I 1 = V R 1 = 119 V 156 Ω = . 762821 A . 004 (part 4 of 4) 10.0 points What power is expended in the heater? Correct answer: 524 . 481 W. Explanation: The power consumed by the heater is P 2 = V 2 R 2 = (119 V) 2 27 Ω = 524 . 481 W . 005 (part 1 of 4) 10.0 points A particle of mass m and charge q starts from rest at the origin (point A in the figure below). E B A C G X Y B There is a uniform electric field vector E in the positive y-direction and a uniform magnetic field vector B directed towards the reader. It can be shown that the path is a cycloid whose radius of curvature at the top point is twice the y-coordinate at that level. What is the relation between kinetic energy of the charge at points A and B ?...
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{[ promptMessage ]}. Bookmark it. {[ promptMessage ]}. 16 - Kapoor(mk9499 – oldhomework 16 – Turner –(60230.... This preview shows pages 1–2. Sign up to view the full content.. This preview has intentionally blurred sections. Sign up to view the full version.. View Full Document. This is the end of the preview. Sign up to access the rest of the document.. Unformatted text preview: Kapoor (mk9499) – oldhomework 16 – Turner – (60230) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points A 156 Ω lamp, a 27 Ω electric heater, and a 41 Ω fan are connected in parallel across a 119 V line. What total current is supplied to the cir- cuit? Correct answer: 8 . 07267 A. Explanation: Let : R 1 = 156 Ω , R 2 = 27 Ω , R 3 = 41 Ω , and V = 119 V . For the parallel circuit, 1 R = 1 R 1 + 1 R 2 + 1 R 3 = R 1 R 2 + R 2 R 3 + R 3 R 1 R 1 R 2 R 3 R = R 1 R 2 R 3 R 1 R 2 + R 2 R 3 + R 3 R 1 R 1 R 2 + R 2 R 3 + R 3 R 1 = (156 Ω) (27 Ω) + (27 Ω) (41 Ω) + (41 Ω) (156 Ω) = 11715 Ω 2 , so R = (156 Ω) (27 Ω) (41 Ω) 11715 Ω 2 = 14 . 7411 Ω and I = V R = 119 V 14 . 7411 Ω = 8 . 07267 A . 002 (part 2 of 4) 10.0 points What is the voltage across the fan? Correct answer: 119 V.
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Explanation: Because the circuit elements are connected in parallel, the voltage across the fan is the same as that across the whole circuit. 003 (part 3 of 4) 10.0 points What is the current in the lamp? Correct answer: 0 . 762821 A. Explanation: The voltage is the same in all 3 appliances, so the current through the lamp is I 1 = V R 1 = 119 V 156 Ω = . 762821 A . 004 (part 4 of 4) 10.0 points What power is expended in the heater? Correct answer: 524 . 481 W. Explanation: The power consumed by the heater is P 2 = V 2 R 2 = (119 V) 2 27 Ω = 524 . 481 W . 005 (part 1 of 4) 10.0 points A particle of mass m and charge q starts from rest at the origin (point A in the figure below). E B A C G X Y B There is a uniform electric field vector E in the positive y-direction and a uniform magnetic field vector B directed towards the reader. It can be shown that the path is a cycloid whose radius of curvature at the top point is twice the y-coordinate at that level. What is the relation between kinetic energy of the charge at points A and B ?.... View Full Document. {[ snackBarMessage ]}. Page1 / 5. 16 - Kapoor(mk9499 – oldhomework 16 – Turner –(60230.... This preview shows document pages 1 - 2. Sign up to view the full document.. View Full Document. Ask a homework question - tutors are online.
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# DIFFICULT PROBLEMS ON GEOMETRIC SERIES
Difficult Problems on Geometric Series :
Here we are going to see some difficult problems in geometric series.
## Difficult Problems on Geometric Series - Questions
Question 1 :
Kumar writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with the instruction that they continue the process similarly. Assuming that the process is unaltered and it costs 2 to mail one letter, find the amount spent on postage when 8th set of letters is mailed.
Solution :
By writing the number of letters as series, we get
4 + 16 + 64 + ...................
It forms a geometric series. Now we have to find the sum of the series upto 8 terms.
S = a(rn - 1)/(r - 1)
n = 8, a = 4 and r = 16/4 = 4
S = 4(48 - 1)/(4 - 1)
= 4(65535)/3
= 4(21845)
= 87380
So far, we get the number of letters posted. Amount spend for one post is 2.
Required cost = 2 (87380)
= 174760
Question 2 :
Find the rational form of the number
Solution :
x = 0.123 123 123............. ------(1)
Multiply each side by 1000, we get
1000x = 123.123 123.............. ------(2)
(2) - (1)
1000x - x = 123.123 123.............. - 0.123 123 123..............
999x = 123
x = 123/999
x = 41/333
Hence the rational form of the given number is 41/333.
Question 3 :
If Sn = (x + y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3) + ...... n terms then prove that
(x - y) S= {[x2(xn - 1)/(x - 1)] - [y2(yn - 1)/(y - 1)]}
Solution :
= (x + y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3) + ...... n
Multiply and divide it by (x - y)
[x2 + x3 + x4 + ................n terms] ----(1)
Sn = [x2 (xn - 1)/(x - 1)]
[y2 + y3 + y4 + ................n terms] ----(1)
Sn = [y2 (yn - 1)/(y - 1)]
= 1/(x -y){[x2(xn-1)/(x-1)] - [y2(yn - 1)/(y - 1)]}
Hence proved.
After having gone through the stuff given above, we hope that the students would have understood, "Difficult Problems on Geometric Series".
Apart from the stuff given in this section "Difficult Problems on Geometric Series"if you need any other stuff in math, please use our google custom search here.
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# DIFFICULT PROBLEMS ON GEOMETRIC SERIES. Difficult Problems on Geometric Series :. Here we are going to see some difficult problems in geometric series.. ## Difficult Problems on Geometric Series - Questions. Question 1 :. Kumar writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with the instruction that they continue the process similarly. Assuming that the process is unaltered and it costs 2 to mail one letter, find the amount spent on postage when 8th set of letters is mailed.. Solution :. By writing the number of letters as series, we get. 4 + 16 + 64 + .................... It forms a geometric series. Now we have to find the sum of the series upto 8 terms.. S = a(rn - 1)/(r - 1). n = 8, a = 4 and r = 16/4 = 4. S = 4(48 - 1)/(4 - 1). = 4(65535)/3. = 4(21845). = 87380. So far, we get the number of letters posted. Amount spend for one post is 2.. Required cost = 2 (87380). = 174760. Question 2 :. Find the rational form of the number.
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Solution :. x = 0.123 123 123............. ------(1). Multiply each side by 1000, we get. 1000x = 123.123 123.............. ------(2). (2) - (1). 1000x - x = 123.123 123.............. - 0.123 123 123............... 999x = 123. x = 123/999. x = 41/333. Hence the rational form of the given number is 41/333.. Question 3 :. If Sn = (x + y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3) + ...... n terms then prove that. (x - y) S= {[x2(xn - 1)/(x - 1)] - [y2(yn - 1)/(y - 1)]}. Solution :. = (x + y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3) + ...... n. Multiply and divide it by (x - y). [x2 + x3 + x4 + ................n terms] ----(1). Sn = [x2 (xn - 1)/(x - 1)]. [y2 + y3 + y4 + ................n terms] ----(1). Sn = [y2 (yn - 1)/(y - 1)]. = 1/(x -y){[x2(xn-1)/(x-1)] - [y2(yn - 1)/(y - 1)]}. Hence proved.. After having gone through the stuff given above, we hope that the students would have understood, "Difficult Problems on Geometric Series".. Apart from the stuff given in this section "Difficult Problems on Geometric Series"if you need any other stuff in math, please use our google custom search here.
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https://web2.0calc.com/questions/finnd-the-total-area-of-a-cube-with-an-edge-of-9
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+0
Finnd the total area of a cube with an edge of 9.
0
315
1
+252
Finnd the total area of a cube with an edge of 9.
A) 324 square units
B) 486 square units
C) 162 square units
D) 398 square units
Nov 2, 2017
Each edge = 9, so each side has an area of 81 units2. Since a cube has 6 sides, the overall area is (side area) times 6. $$81 \times 6 = 486$$ units2, so the answer is b.
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+0. Finnd the total area of a cube with an edge of 9.. 0. 315. 1. +252. Finnd the total area of a cube with an edge of 9.. A) 324 square units.
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B) 486 square units. C) 162 square units. D) 398 square units. Nov 2, 2017. Each edge = 9, so each side has an area of 81 units2. Since a cube has 6 sides, the overall area is (side area) times 6. $$81 \times 6 = 486$$ units2, so the answer is b.
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1. Two masses, m and M are involved in a glacing collision as seen below where θ and ø= pi/2.
If M = nm what must n be such that the collision is elastic?
Remember if θ+ø=pi/2 then cos(θ)=sin(ø) and cos(ø)=sin(θ)
http://ift.tt/1liMFfp
2. I am suppose to find an number for n.
3. ∑KEo=∑KEf
1/2m$_{1}$v$^{2}_{0}$+0=1/2m$_{1}$v$^{2}_{f1}$+1/2m$_{2}$v$^{2}_{f2}$
substitute m$_{1}$n for m$_{2}$ and cancel the 1/2m$_{1}$
v$^{2}_{o1}$=v$^{2}_{f1}$+nv$^{2}_{f2}$
n=$\frac{v^{2}_{o1}-v^{2}_{f1}}{v^{2}_{f2}}$
Not sure what to do from here please help. I know I’m probably suppose to use the θ and ø, but I’m not sure how to incorporate it.
1. The problem statement, all variables and given/known data
Known: M=nm, θ+ø=pi/2
Unknown: n
2. Relevant equations
W$_{NC}$=ΔKE+ΔPE
KE=$\frac{1}{2}$mv$^{2}$
Momentum=∑p=mv$_{f}$-mv$_{o}$
3. The attempt at a solution
PS sorry I kinda messed this up it is my first post.
http://ift.tt/1tk60Cl
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1. Two masses, m and M are involved in a glacing collision as seen below where θ and ø= pi/2.. If M = nm what must n be such that the collision is elastic?. Remember if θ+ø=pi/2 then cos(θ)=sin(ø) and cos(ø)=sin(θ). http://ift.tt/1liMFfp. 2. I am suppose to find an number for n.. 3. ∑KEo=∑KEf. 1/2m$_{1}$v$^{2}_{0}$+0=1/2m$_{1}$v$^{2}_{f1}$+1/2m$_{2}$v$^{2}_{f2}$. substitute m$_{1}$n for m$_{2}$ and cancel the 1/2m$_{1}$. v$^{2}_{o1}$=v$^{2}_{f1}$+nv$^{2}_{f2}$. n=$\frac{v^{2}_{o1}-v^{2}_{f1}}{v^{2}_{f2}}$. Not sure what to do from here please help. I know I’m probably suppose to use the θ and ø, but I’m not sure how to incorporate it.
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1. The problem statement, all variables and given/known data. Known: M=nm, θ+ø=pi/2. Unknown: n. 2. Relevant equations. W$_{NC}$=ΔKE+ΔPE. KE=$\frac{1}{2}$mv$^{2}$. Momentum=∑p=mv$_{f}$-mv$_{o}$. 3. The attempt at a solution. PS sorry I kinda messed this up it is my first post.. http://ift.tt/1tk60Cl.
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can you solve this math problems??? - XP Math - Forums
XP Math - Forums can you solve this math problems???
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07-24-2007 #1 KeNaruto Guest Posts: n/a can you solve this math problems??? 1. Two hours after a truck leaves Phoenix traveling at 45 miles per hour, a car leaves to overtake the truck. If it takes the car ten hours to catch the truck, what was the speed of the car?2. A 20 gallon salt solution is 30% salt. What is the greatest number of gallons of water that can be evaporated so the solution is still less than 45% salt?pls. answer this.....
07-24-2007 #2 cassandras_evil_sister 1 Guest Posts: n/a the car had to go 54 mph to catch the truck .2hrs x 45 mph =90+ 10 hrs x 45 mph for the truck and car to meet=54 mph or another way to say it, truck traveled 540 miles in 12 hrs time, 10x 45+90=540, the car went 54 mph to catch it, if the truck never stopped and both left the same point 2 hrs apart.about the solution q, i dont know
07-24-2007 #3 jeetu N Guest Posts: n/a ans for 2:water(gallons) salt 20 30x 44x=600/44=13.63gallonsgallons of evaporated water =20-13.63 = 6.37gallons
07-24-2007 #4 Atman Guest Posts: n/a 1.the car catches the bus 10 hrs after it (the car) leaves i.e. 12 hrs after the bus leaves phoenix. in 12 hrs, the bus travels a distance 540 miles (45miles/hr*12hrs). that means the car travels 540 miles in 10 hrs. therefore, its speed is 540/10=54miles/hr.2.the salt is 30% in 20 gallons. i.e. it is [(30/100)*20] gallons.it comes out to be 6 gallons. for it to be 45%, [(45/100)*x]=6 gallons. where x is the total amount of solution left after evaporating. on calculating, x comes out as 13.33. hence, amount of water that can be evaporated so as to keep the solution less than 45% is less than 6.66 gallons
07-24-2007 #5 zadatak Guest Posts: n/a 30%*20= 6 gallons saltsalt-----= %salttotal6/x=45% (.45)x=6/.4513.3333333so 13.33333 gallons of water will be 45% saltso you could evaporate 6.666666 gallons of water
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can you solve this math problems??? - XP Math - Forums. XP Math - Forums can you solve this math problems???. User Name Remember Me? Password. Register Arcade Members List Mark Forums Read. Thread Tools Display Modes. 07-24-2007 #1 KeNaruto Guest Posts: n/a can you solve this math problems??? 1. Two hours after a truck leaves Phoenix traveling at 45 miles per hour, a car leaves to overtake the truck. If it takes the car ten hours to catch the truck, what was the speed of the car?2. A 20 gallon salt solution is 30% salt. What is the greatest number of gallons of water that can be evaporated so the solution is still less than 45% salt?pls. answer this...... 07-24-2007 #2 cassandras_evil_sister 1 Guest Posts: n/a the car had to go 54 mph to catch the truck .2hrs x 45 mph =90+ 10 hrs x 45 mph for the truck and car to meet=54 mph or another way to say it, truck traveled 540 miles in 12 hrs time, 10x 45+90=540, the car went 54 mph to catch it, if the truck never stopped and both left the same point 2 hrs apart.about the solution q, i dont know. 07-24-2007 #3 jeetu N Guest Posts: n/a ans for 2:water(gallons) salt 20 30x 44x=600/44=13.63gallonsgallons of evaporated water =20-13.63 = 6.37gallons. 07-24-2007 #4 Atman Guest Posts: n/a 1.the car catches the bus 10 hrs after it (the car) leaves i.e. 12 hrs after the bus leaves phoenix. in 12 hrs, the bus travels a distance 540 miles (45miles/hr*12hrs). that means the car travels 540 miles in 10 hrs.
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therefore, its speed is 540/10=54miles/hr.2.the salt is 30% in 20 gallons. i.e. it is [(30/100)*20] gallons.it comes out to be 6 gallons. for it to be 45%, [(45/100)*x]=6 gallons. where x is the total amount of solution left after evaporating. on calculating, x comes out as 13.33. hence, amount of water that can be evaporated so as to keep the solution less than 45% is less than 6.66 gallons. 07-24-2007 #5 zadatak Guest Posts: n/a 30%*20= 6 gallons saltsalt-----= %salttotal6/x=45% (.45)x=6/.4513.3333333so 13.33333 gallons of water will be 45% saltso you could evaporate 6.666666 gallons of water. Thread Tools Display Modes Linear Mode. Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules. Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Welcome XP Math News Off-Topic Discussion Mathematics XP Math Games Worksheets Homework Help Problems Library Math Challenges. All times are GMT -4. The time now is 08:10 AM.. Contact Us - XP Math - Forums - Archive - Privacy Statement - Top.
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https://www.physicsforums.com/threads/angular-diameter.250596/
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Angular diameter
1. Aug 18, 2008
yiuscott
1. The problem statement, all variables and given/known data
This is the questions:
1) The angular diameter of the sun measured from the earth is 0.52 degrees. Calculate:
i) the angular diameter in radians
ii) the sun's radius in meters
iii) The surface area of the sun, assuming that it is a sphere
2) Using L=aAT4 where L is the luminosity, a is the stefan-Boltzmann constant: 5.67X10^-8, A is the surface area, T is the temperature.
Calculate the total power and the power per m^2 radiated by the earth at a temperature of 228K. You can assume that the earth is a sphere of radius 6400km.
3. The attempt at a solution
Question 1:
Question i) is quite easy and found out that it is 0.00908
Question ii) is the major problem. I tried to assume that it is a triangle and used the sine rule (since i know the distance of the sun to the earth is 149X10^11m while the other angles in the triangle should be (180-0.52)/2. However, this gave me an answer of 1.4X10^9. If i divide by 2 (to find the radius), i ger 7X10^8m.
The answer at the back of the book is 1.4X10^9m. which is double my answer. I suspect that the book may be wrong but just wish to double check
Question iii) if i knew the answer to question ii), i would be able to do this.
Question 2: I thought that this question should be easy where i just put in the numbers into the equation... but i couldn't get the right answer...
L=aAT^4
L = (5.67X10^-8)*(4*pi*6400000^2)*(228^4)
L = 7.89X10^16
The book's answer is 201X10^15 W
Thank you very much
Last edited: Aug 18, 2008
2. Aug 18, 2008
tiny-tim
Hi yiuscott!
I think the book is wrong, and you are right … width = radius * radians.
erm … 228K? … I don't think so!
Try 288K!
3. Aug 18, 2008
yiuscott
Thanks for the reply.
lol thanks. I can't imagine a textbook getting 2 questions wrong in a row. I should have spotted the problem with a 228K earth though...
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Angular diameter. 1. Aug 18, 2008. yiuscott. 1. The problem statement, all variables and given/known data. This is the questions:. 1) The angular diameter of the sun measured from the earth is 0.52 degrees. Calculate:. i) the angular diameter in radians. ii) the sun's radius in meters. iii) The surface area of the sun, assuming that it is a sphere. 2) Using L=aAT4 where L is the luminosity, a is the stefan-Boltzmann constant: 5.67X10^-8, A is the surface area, T is the temperature.. Calculate the total power and the power per m^2 radiated by the earth at a temperature of 228K. You can assume that the earth is a sphere of radius 6400km.. 3. The attempt at a solution. Question 1:. Question i) is quite easy and found out that it is 0.00908. Question ii) is the major problem. I tried to assume that it is a triangle and used the sine rule (since i know the distance of the sun to the earth is 149X10^11m while the other angles in the triangle should be (180-0.52)/2. However, this gave me an answer of 1.4X10^9. If i divide by 2 (to find the radius), i ger 7X10^8m.. The answer at the back of the book is 1.4X10^9m. which is double my answer.
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I suspect that the book may be wrong but just wish to double check. Question iii) if i knew the answer to question ii), i would be able to do this.. Question 2: I thought that this question should be easy where i just put in the numbers into the equation... but i couldn't get the right answer.... L=aAT^4. L = (5.67X10^-8)*(4*pi*6400000^2)*(228^4). L = 7.89X10^16. The book's answer is 201X10^15 W. Thank you very much. Last edited: Aug 18, 2008. 2. Aug 18, 2008. tiny-tim. Hi yiuscott!. I think the book is wrong, and you are right … width = radius * radians.. erm … 228K? … I don't think so!. Try 288K!. 3. Aug 18, 2008. yiuscott. Thanks for the reply.. lol thanks. I can't imagine a textbook getting 2 questions wrong in a row. I should have spotted the problem with a 228K earth though...
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## Real Mathematics – Life vs. Maths #2
Matches
For thousands of years people tried find a precise value for the number π (3,1415192…). At first this special number was thought to be seen only when there is a circle around. Within time π started to appear in places where scientist didn’t expect it to be. One of them was an 18th century scientist Georges Buffon.
Buffon came up with a probability problem named “Buffon’s needle problem” in 1777 when he came across with the number π. As I didn’t possess that many needles, I modified the problem as “Serkan’s matches problem”.
Buffon’s Needle Problem: Take a piece of paper and draw perpendicular lines on it with specific amount of space between them. Buffon wondered if one can calculate the probability of a needle that will land on one of the lines.
To start Serkan’s matches problem you need at least 100 matches, a piece of empty paper, a ruler, pen/pencil and a calculator.
First of all, draw perpendicular lines with 2 matches-length spaces between them.
Then just throw the matches on the paper randomly.
Start collecting the matches which land on a line. At last you should use your calculator to divide the total number of matches to the number of matches landed on a line.
In my experiment out of 100 matches, 32 of them landed on a line. That gave me 3,125 which is close to the magical number π.
In fact, 100 matches are not enough for this experiment. In my second try 34 matches landed on one of the lines which gave 100/34=2,9411… Obviously this is not close to π. More matches we use, closer we will get to π.
In an experiment back in 1980 2000 needles were used to analyze Buffon’s needle problem. Result was 3,1430… which is seriously close to the number π.
##### You could go to https://mste.illinois.edu/activity/buffon/ and use this simulator which uses 1000 needle. In my first try I got 3,1496… You should try and see the result yourself.
In the future I will be talking about why a needle (or a match) is connected to the number π.
One wonders…
Try to do your own experiment and repeat Buffon’s needle problem for five times. Take the arithmetic average of your solutions and see how close you are to π?
M. Serkan Kalaycıoğlu
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## Real Mathematics – Life vs. Maths #2. Matches. For thousands of years people tried find a precise value for the number π (3,1415192…). At first this special number was thought to be seen only when there is a circle around. Within time π started to appear in places where scientist didn’t expect it to be. One of them was an 18th century scientist Georges Buffon.. Buffon came up with a probability problem named “Buffon’s needle problem” in 1777 when he came across with the number π. As I didn’t possess that many needles, I modified the problem as “Serkan’s matches problem”.. Buffon’s Needle Problem: Take a piece of paper and draw perpendicular lines on it with specific amount of space between them. Buffon wondered if one can calculate the probability of a needle that will land on one of the lines.. To start Serkan’s matches problem you need at least 100 matches, a piece of empty paper, a ruler, pen/pencil and a calculator.. First of all, draw perpendicular lines with 2 matches-length spaces between them.. Then just throw the matches on the paper randomly.. Start collecting the matches which land on a line. At last you should use your calculator to divide the total number of matches to the number of matches landed on a line.. In my experiment out of 100 matches, 32 of them landed on a line.
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That gave me 3,125 which is close to the magical number π.. In fact, 100 matches are not enough for this experiment. In my second try 34 matches landed on one of the lines which gave 100/34=2,9411… Obviously this is not close to π. More matches we use, closer we will get to π.. In an experiment back in 1980 2000 needles were used to analyze Buffon’s needle problem. Result was 3,1430… which is seriously close to the number π.. ##### You could go to https://mste.illinois.edu/activity/buffon/ and use this simulator which uses 1000 needle. In my first try I got 3,1496… You should try and see the result yourself.. In the future I will be talking about why a needle (or a match) is connected to the number π.. One wonders…. Try to do your own experiment and repeat Buffon’s needle problem for five times. Take the arithmetic average of your solutions and see how close you are to π?. M. Serkan Kalaycıoğlu.
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Question
# QUESTION 27 A uniform disk of radius 0.40 m and mass 31.0 kg rolls on a...
QUESTION 27
1. A uniform disk of radius 0.40 m and mass 31.0 kg rolls on a plane without slipping with angular speed 3.0 rad/s. The rotational kinetic energy of the disk is __________. The moment of inertia of the disk is given by 0.5MR2.
Given mass=31kg
Translational velocityv =wr=1.2m/s
Moment of inertia I =0.5MR²=2.48kgm²
Total kinetic energy of the disc will be equal to the sum of kinetic energy due to tranlational as well as due to its rotational motion
Kinetic energy due to translational motion=(1/2)mv²=(1/2)*2.48*(1.2*1.2)=22.32j
Kinetic energy due to roational motion=(1/2)Iw²
=(1/2)(1/2)(31)(0.4*0.4)(3*3)=11.16j
So total kinetic energy is 33.48joule
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Question. # QUESTION 27 A uniform disk of radius 0.40 m and mass 31.0 kg rolls on a.... QUESTION 27. 1. A uniform disk of radius 0.40 m and mass 31.0 kg rolls on a plane without slipping with angular speed 3.0 rad/s. The rotational kinetic energy of the disk is __________. The moment of inertia of the disk is given by 0.5MR2.. Given mass=31kg.
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Translational velocityv =wr=1.2m/s. Moment of inertia I =0.5MR²=2.48kgm². Total kinetic energy of the disc will be equal to the sum of kinetic energy due to tranlational as well as due to its rotational motion. Kinetic energy due to translational motion=(1/2)mv²=(1/2)*2.48*(1.2*1.2)=22.32j. Kinetic energy due to roational motion=(1/2)Iw². =(1/2)(1/2)(31)(0.4*0.4)(3*3)=11.16j. So total kinetic energy is 33.48joule.
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Problems With Expected Utility
# Problems With Expected Utility
## Problems With Expected Utility
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##### Presentation Transcript
1. Problems With Expected Utility or if the axioms fit, use them but...
2. Completeness (Consistency) • Which do you prefer? A: a 1 out of 100 chance of losing \$1000 B: buy insurance for \$10 to protect you from this loss • Which do you prefer? C: a 1 out of 100 chance of losing \$1000 D: lose \$10 for sure
3. Completeness - Response Modes • Suppose you own a lottery ticket that gives you a 8/9 chance to receive \$4. What is the lowest price you would take to sell this ticket? • Suppose you own a lottery ticket that gives you a 1/9 chance to receive \$40. What is the lowest price you would take to sell this ticket? • Suppose you can choose only one of the above lotteries. Which do you prefer?
4. Intransitivity • Consider the following five gambles: A) 7/24 chance of receiving \$5.00 B) 8/24 chance of receiving \$4.75 C) 9/24 chance of receiving \$4.50 D) 10/24 chance of receiving \$4.25 E) 11/24 chance of receiving \$4.00 • Do you prefer A or B? B or C? C or D? D or E? A or E?
5. The Money Pump • Your grocery store has three brands of canned peas, each 12 oz.: National brand name \$2.49 Store brand \$2.09 Generic \$1.69 If you prefer N>S>G, but G>N, then I can trade you S for G and \$.41, N for S and \$.41, G and \$.79 for N, etc.
6. Allais Paradox (Savage’s Independence Principle) • Which would you prefer: A. receive \$1 Million for sure B. a gamble in which you have an 89% chance of getting \$1 Million, 10% chance of getting \$5 Million, and 1% chance of receiving nothing • Which would you prefer: C. a gamble in which you have an 11% chance of getting \$1 Million, otherwise nothing D. a gamble in which you have a 10% chance of getting \$5 Million, otherwise nothing
7. Lottery Ticket Formulation Lottery Tickets
8. Ellsberg’s Paradox Two jars contain Red and White balls. You may choose Red or White and pick one ball from one jar. If you pick your chosen color you will be paid \$100. • Jar 1 contains 50% Red and 50% White • Jar 2 contains a randomly chosen unknown % of Red and White Which jar do you choose to pick from?
9. Solvability • Which would you prefer, A. win \$30 for sure B. play a gamble in which you have an 80% chance of winning \$45, otherwise \$0 • Which would you prefer, C. play a gamble with a 75% chance of winning \$30, otherwise \$0 D. play a gamble with a 60% chance of winning \$45, otherwise \$0
10. Solvability, continued • Suppose you must choose the second part of a two-stage gamble now. In the first stage, there is a 25% chance that you will get \$0, and a 75% chance that you will go on to the second stage. In the second stage, you either A. win \$30 for sure B. play a gamble in which you have an 80% chance of winning \$45, otherwise \$0 • You must choose A or B before you know the first-stage result. Which do you choose?
11. Errors and Biases • The above examples illustrate persistent problems with the EU model • People behave in ways that are inconsistent with the axioms, predictably, not randomly • Some theorists have developed models that retain the form of the EU rule with additional modifications to probability and value functions, while others catalog decision rules that bear no direct resemblance to EU
12. Exercise #2: Discussion Sales (Events) 1000 5000 9000 \$4000 -2000 6000 14000 machine \$10000 -6400 8000 22400 machine p .3 .5 .2 Outcome = price*sales - fixed cost - var. cost*sales e.g., \$4*5000 - \$4000 - \$2*5000 = \$6000
13. Choosing by Expected Value • For the \$4000 machine, EV = .3(-2000) +.5(6000) +.2(14000) = \$5200 • For the \$10000 machine, EV = .3(-6400)+.5(8000)+.2(22400) = \$6560 • Thus, on EV grounds, go with the more expensive machine!
14. Choosing by Expected Utility Risk averse utility: • U(\$0) = 0 • U(\$2000) = 1000 • U(-6400) = -4000 • U(-2000) = -1500 • U(6000) = 2500 • U(8000) = 3200 • U(14000) = 4800 • U(22400) = 6500 4 0 -4 -6.4 0 \$22.4
15. Choosing by Expected Utility • For the \$4000 machine EU = .3(-1500) +.5(2500) +.2(4800) =1760 • For the \$10000 machine EU = .3(-4000) +.5(3200) +.2(6500) =1700 Therefore, choose the \$4000 machine!
16. Additional Complexities • Within the model: continuous sales estimates ambiguous probabilities other costs (your time?) other benefits (polish your resume?) • Outside the model: self-esteem, anxiety change of self-concept/preferences reputation/respect/status
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Problems With Expected Utility. # Problems With Expected Utility. ## Problems With Expected Utility. - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -. ##### Presentation Transcript. 1. Problems With Expected Utility or if the axioms fit, use them but.... 2. Completeness (Consistency) • Which do you prefer? A: a 1 out of 100 chance of losing \$1000 B: buy insurance for \$10 to protect you from this loss • Which do you prefer? C: a 1 out of 100 chance of losing \$1000 D: lose \$10 for sure. 3. Completeness - Response Modes • Suppose you own a lottery ticket that gives you a 8/9 chance to receive \$4. What is the lowest price you would take to sell this ticket? • Suppose you own a lottery ticket that gives you a 1/9 chance to receive \$40. What is the lowest price you would take to sell this ticket? • Suppose you can choose only one of the above lotteries. Which do you prefer?. 4. Intransitivity • Consider the following five gambles: A) 7/24 chance of receiving \$5.00 B) 8/24 chance of receiving \$4.75 C) 9/24 chance of receiving \$4.50 D) 10/24 chance of receiving \$4.25 E) 11/24 chance of receiving \$4.00 • Do you prefer A or B? B or C? C or D? D or E? A or E?. 5. The Money Pump • Your grocery store has three brands of canned peas, each 12 oz.: National brand name \$2.49 Store brand \$2.09 Generic \$1.69 If you prefer N>S>G, but G>N, then I can trade you S for G and \$.41, N for S and \$.41, G and \$.79 for N, etc.. 6. Allais Paradox (Savage’s Independence Principle) • Which would you prefer: A. receive \$1 Million for sure B. a gamble in which you have an 89% chance of getting \$1 Million, 10% chance of getting \$5 Million, and 1% chance of receiving nothing • Which would you prefer: C. a gamble in which you have an 11% chance of getting \$1 Million, otherwise nothing D. a gamble in which you have a 10% chance of getting \$5 Million, otherwise nothing. 7. Lottery Ticket Formulation Lottery Tickets. 8. Ellsberg’s Paradox Two jars contain Red and White balls. You may choose Red or White and pick one ball from one jar.
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If you pick your chosen color you will be paid \$100. • Jar 1 contains 50% Red and 50% White • Jar 2 contains a randomly chosen unknown % of Red and White Which jar do you choose to pick from?. 9. Solvability • Which would you prefer, A. win \$30 for sure B. play a gamble in which you have an 80% chance of winning \$45, otherwise \$0 • Which would you prefer, C. play a gamble with a 75% chance of winning \$30, otherwise \$0 D. play a gamble with a 60% chance of winning \$45, otherwise \$0. 10. Solvability, continued • Suppose you must choose the second part of a two-stage gamble now. In the first stage, there is a 25% chance that you will get \$0, and a 75% chance that you will go on to the second stage. In the second stage, you either A. win \$30 for sure B. play a gamble in which you have an 80% chance of winning \$45, otherwise \$0 • You must choose A or B before you know the first-stage result. Which do you choose?. 11. Errors and Biases • The above examples illustrate persistent problems with the EU model • People behave in ways that are inconsistent with the axioms, predictably, not randomly • Some theorists have developed models that retain the form of the EU rule with additional modifications to probability and value functions, while others catalog decision rules that bear no direct resemblance to EU. 12. Exercise #2: Discussion Sales (Events) 1000 5000 9000 \$4000 -2000 6000 14000 machine \$10000 -6400 8000 22400 machine p .3 .5 .2 Outcome = price*sales - fixed cost - var. cost*sales e.g., \$4*5000 - \$4000 - \$2*5000 = \$6000. 13. Choosing by Expected Value • For the \$4000 machine, EV = .3(-2000) +.5(6000) +.2(14000) = \$5200 • For the \$10000 machine, EV = .3(-6400)+.5(8000)+.2(22400) = \$6560 • Thus, on EV grounds, go with the more expensive machine!. 14. Choosing by Expected Utility Risk averse utility: • U(\$0) = 0 • U(\$2000) = 1000 • U(-6400) = -4000 • U(-2000) = -1500 • U(6000) = 2500 • U(8000) = 3200 • U(14000) = 4800 • U(22400) = 6500 4 0 -4 -6.4 0 \$22.4. 15. Choosing by Expected Utility • For the \$4000 machine EU = .3(-1500) +.5(2500) +.2(4800) =1760 • For the \$10000 machine EU = .3(-4000) +.5(3200) +.2(6500) =1700 Therefore, choose the \$4000 machine!. 16. Additional Complexities • Within the model: continuous sales estimates ambiguous probabilities other costs (your time?) other benefits (polish your resume?) • Outside the model: self-esteem, anxiety change of self-concept/preferences reputation/respect/status.
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https://math.stackexchange.com/questions/3203908/extended-stars-and-bars-where-the-upper-bound-of-a-variable-is-one-of-the-other
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# Extended Stars-and-Bars where the upper bound of a variable is one of the other variables
How many non-negative integer solutions are there for the equation x + y + z + w = 20, where x>y.
I started by substituting x for a new variable x'=x-y:
x > y
x-y > 0
x'>0
However, this led me to a point from which I didn't know how to make progress:
x'+ 2y + z + w = 20
How should I deal with the 2y?
• I'm not sure one can apply Stars-and-Bars with such modifications. In general introducing coefficients not equal to one makes the problem (counting non-negative integer solutions) more difficult. Would you entertain an alternative approach? – hardmath Apr 26 '19 at 23:53
By symmetry, the number of solutions with $$x>y$$ is the same as the number with $$x so the answer is "half the number of solutions where $$x\neq y$$." I would go about this by subtracting the solutions with $$x=y$$ from the total number of solutions.
Stars and bars gives the total solutions, so now we need to know the number of solutions to $$2x+w+z=20$$ in nonnegative integers. For a given $$x$$ this is just $$21-2x$$ so it's easy to add them up.
• Your logic makes perfect sense. Sadly, I feel it comes back to the problem I encountered in my attempt: not knowing how to make progress with Stars-and-Bars when 2 of the variables are the same. – pug Apr 27 '19 at 0:13
• @pug - saulspatz is saying this: iterate through $x=0,1,...,10$ and you have $w+z = 20-2x$ which is itself a stars-and-bars problem. however, this problem is so small ($2$ variables) that the no. of solutions is obviously ${20-2x + 2 -1 \choose 2 - 1} = 21 - 2x$ (he has a typo). so now you sum through all $x=0,1,...10$ etc – antkam Apr 27 '19 at 3:02
• @antkam Thanks for pointing out the typo. It's fixed. – saulspatz Apr 27 '19 at 11:14
$$x+y+z+w=20\tag{1}$$
Firstly it says $$x \gt y$$ so $$x=y+x'+1$$ where $$x'\ge 0$$.
This gives
$$x'+2y+z+w=19\, .$$
So since $$2y$$ is always even, $$x'+z+w$$ must be odd which means either all three variables are odd or just one is.
1. If all three are odd then say $$x'=2x''+1$$, $$z=2z''+1$$, $$w=2w''+1$$, giving
\begin{align}&& 2(x''+y+z''+w'')&=16\\[1ex] &\implies& x''+y+z''+w''&=8\, ,\end{align}
which has $$\binom{8+3}{3}$$ non-negative integer solutions by bars and stars.
1. If just one is odd, say $$x'=2x''+1$$, then the other two must be even: $$z=2z''$$, $$w=2w''$$, this gives
\begin{align}&& 2(x''+y+z''+w'')&=18\\[1ex] &\implies & x''+y+z''+w''&=9\, ,\end{align}
which has $$\binom{9+3}{3}$$ non-negative integer solutions by bars and stars. However, there are 3 choices for our odd variable so that's $$3\binom{9+3}{3}$$ total solutions for this case.
Adding both cases together we have a grand total of
$$\binom{8+3}{3}+3\binom{9+3}{3}=825\tag{Answer}$$
non-negative integer solutions to $$(1)$$ with $$x\gt y$$.
• A nice approach. However, the equation $x'' + y + z'' + w'' = 8$ has only $\binom{\color{red}{8} + 3}{3}$ solutions in the nonnegative integers. – N. F. Taussig Apr 27 '19 at 9:09
• Whoops! Thank you @N.F.Taussig. Where was my head when I wrote that? I don't know. I'll edit. – N. Shales Apr 27 '19 at 9:35
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# Extended Stars-and-Bars where the upper bound of a variable is one of the other variables. How many non-negative integer solutions are there for the equation x + y + z + w = 20, where x>y.. I started by substituting x for a new variable x'=x-y:. x > y. x-y > 0. x'>0. However, this led me to a point from which I didn't know how to make progress:. x'+ 2y + z + w = 20. How should I deal with the 2y?. • I'm not sure one can apply Stars-and-Bars with such modifications. In general introducing coefficients not equal to one makes the problem (counting non-negative integer solutions) more difficult. Would you entertain an alternative approach? – hardmath Apr 26 '19 at 23:53. By symmetry, the number of solutions with $$x>y$$ is the same as the number with $$x so the answer is "half the number of solutions where $$x\neq y$$." I would go about this by subtracting the solutions with $$x=y$$ from the total number of solutions.. Stars and bars gives the total solutions, so now we need to know the number of solutions to $$2x+w+z=20$$ in nonnegative integers. For a given $$x$$ this is just $$21-2x$$ so it's easy to add them up.. • Your logic makes perfect sense. Sadly, I feel it comes back to the problem I encountered in my attempt: not knowing how to make progress with Stars-and-Bars when 2 of the variables are the same. – pug Apr 27 '19 at 0:13. • @pug - saulspatz is saying this: iterate through $x=0,1,...,10$ and you have $w+z = 20-2x$ which is itself a stars-and-bars problem. however, this problem is so small ($2$ variables) that the no. of solutions is obviously ${20-2x + 2 -1 \choose 2 - 1} = 21 - 2x$ (he has a typo). so now you sum through all $x=0,1,...10$ etc – antkam Apr 27 '19 at 3:02. • @antkam Thanks for pointing out the typo. It's fixed. – saulspatz Apr 27 '19 at 11:14. $$x+y+z+w=20\tag{1}$$. Firstly it says $$x \gt y$$ so $$x=y+x'+1$$ where $$x'\ge 0$$.
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This gives. $$x'+2y+z+w=19\, .$$. So since $$2y$$ is always even, $$x'+z+w$$ must be odd which means either all three variables are odd or just one is.. 1. If all three are odd then say $$x'=2x''+1$$, $$z=2z''+1$$, $$w=2w''+1$$, giving. \begin{align}&& 2(x''+y+z''+w'')&=16\\[1ex] &\implies& x''+y+z''+w''&=8\, ,\end{align}. which has $$\binom{8+3}{3}$$ non-negative integer solutions by bars and stars.. 1. If just one is odd, say $$x'=2x''+1$$, then the other two must be even: $$z=2z''$$, $$w=2w''$$, this gives. \begin{align}&& 2(x''+y+z''+w'')&=18\\[1ex] &\implies & x''+y+z''+w''&=9\, ,\end{align}. which has $$\binom{9+3}{3}$$ non-negative integer solutions by bars and stars. However, there are 3 choices for our odd variable so that's $$3\binom{9+3}{3}$$ total solutions for this case.. Adding both cases together we have a grand total of. $$\binom{8+3}{3}+3\binom{9+3}{3}=825\tag{Answer}$$. non-negative integer solutions to $$(1)$$ with $$x\gt y$$.. • A nice approach. However, the equation $x'' + y + z'' + w'' = 8$ has only $\binom{\color{red}{8} + 3}{3}$ solutions in the nonnegative integers. – N. F. Taussig Apr 27 '19 at 9:09. • Whoops! Thank you @N.F.Taussig. Where was my head when I wrote that? I don't know. I'll edit. – N. Shales Apr 27 '19 at 9:35.
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https://www.gamedev.net/forums/topic/106455-how-do-you-simulate-gravity/
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#### Archived
This topic is now archived and is closed to further replies.
# How do you simulate gravity?
## Recommended Posts
[removed] [Edited by - DevLiquidKnight on February 14, 2007 9:21:39 AM]
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Hey,
You just add a velocity vector to you models and increase that with the gravity (9.8m/sec) or your own values
for example:
struct strMyObject {
Vector Pos, Vec;
}
strMyObject Player;
LOOP ->:
Player.Vec.y -= GRAVITY/Framespersec (or just a constant value)
Player.Pos += Vec;
So each frames you add the velocity vector to the players possision to make him accelerate down if nothing is stopping him
(like a floor =
<- LOOP
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To elaborate on the AP's response, gravity is a force acting between all objects that have mass. I'll assume for the time being that you want to simulate gravity as it would be felt on the surface of a planet. Let's stick with Earth, since it's nice and easy to relate to.
First, take Newton's First Law:
F = M*a (1)
Force = Mass * Acceleration
The Gravitational force between two objects is proportional to their masses and inversely proprtional to the square of the separation between them, so that
G M1 M2F = ------- (2) R2
G is the gravitational constant and is equal to 6.672*10-11N m2 kg-2
If you hold one mass (M1) constant, say that of the Earth, then the acceleration that the other mass will undergo, relative to the first mass, is given by combining (1) and (2) to give
G M1a = ------- (2) R2
Substituting in the radius of the Earth (around 6,400 km on average) and the Mass of the Earth (about 5.98*1024 kg) gives an acceleration around 9.8 m s-2.
This is the same acceleration that all bodies undergo at the surface of the Earth. The Force that each body undergoes though is different and is given by F = 9.8 * mass of the body.
Since you know the acceleration of the body, you can work out it's vertical velocity (which is independant of the horizontal velocity) by integrating the acceleration over a fixed time interval, since acceleration is the first derivative of velocity with respect to time. You need to know the starting velocity to do this, since
v(t+dt) = v(t) + integraltt+dt a dt
Furthermore, if you know the position of the object at time t then you can work out its position at time t+dt by integrating the velocity. The result is that for any object near the surface of the Earth, you get the following equations for the height of the object and it's velocity:
s(t+dt) = s(t) + 1/2a*dt2
v(t+dt) = v(t) + a*dt
So if you know a (which you set yourself), the initial height, s(t), and the initial velocity, v(t) m/s (down... negative means up) then you can work out where the object is some time later!
I hope this explains the solution for you.
Good luck,
Timkin
[edited by - Timkin on July 29, 2002 11:36:28 PM]
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i have an extremley simple way of simulating gravity, just make an mz variable for your object. every frame, add the gravity variable to mz. and every frame, add mz to z, so that way, it accelerates down until it hits something, then make mz = 0 and only add gravity if bOnGround is false or something
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quote:
Original post by Timkin
G M1a = ------- (2) R2
Hey,
I''m just having a problem with this formula: the problem of it is that the acceleration isn''t constant here, and if I calculate physics in a space game every frame, I have to assume that the acceleration is constant during the 1/85th of a second this frame takes, and while this may seem only a small problem, the results of it are dramatic and the physics get more than 100% inaccurate. Do you have an idea how to calculate exactly what happened during one frame, knowing that not only the magnitude but also the direction of the gravity is changing during the time of one frame?
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coming across one problem i simulated gravity and now when i walk on the platforms i made i get stuck so i tried to make it subtract 1 from the y variable when he is on a platform so he could move side to side and it causes the guy to bounce up and down really fast
[edited by - DevLiquidKnight on July 30, 2002 8:35:10 PM]
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quote:
Original post by Lode
I''m just having a problem with this formula: the problem of it is that the acceleration isn''t constant here, and if I calculate physics in a space game every frame, I have to assume that the acceleration is constant during the 1/85th of a second this frame takes, and while this may seem only a small problem, the results of it are dramatic and the physics get more than 100% inaccurate. Do you have an idea how to calculate exactly what happened during one frame, knowing that not only the magnitude but also the direction of the gravity is changing during the time of one frame?
Unless things are travelling quite fast, then 1/85th of a second is a very small time to integrate over. Are you using an Euler integration (which I suspect you are)?. You may want to consider using a predictor-corrector integration scheme.
Good luck,
Timkin
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The 1/85th of a second becomes to large to keep it accurate in two cases:
-if the planet has such a big mass that it generates very strong accelerations if you get close to it, but then for example my ship gets an acceleration of 10000000000m/s² at a certain point and in the 1/85th of a second it''s of course suddenly very far away of the planet... so far that it doesn''t get the negative -10000000000m/s² that it should get when it just crossed the planet anymore.
-if I speed up the time of the simulation: if I speed up the simulation with a factor of 1000, the 1/85th of a second becomes like 11.7seconds and that is too long to keep it accurate.
So indeed I should somehow predict where the ship will be after that time... but what kind of predictor-corrector integration scheme should I be looking for?
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Okay, I didn''t realise you were talking about such large accelerations. Since you are, you really cannot treat this problem with classical (Newtonian) physics. You need to deal with the simulation from the perspective of the gravitational potential energy and the conversion from this potential energy into kinetic energy (strictly speaking you should be using General Relativity to get things correct, but that''s more trouble than it''s worth!).
Taking the ''energy approach'' will enable you to bring in relativistic effects that limit the speed of the craft to the speed of light. It will also mean that at high velocities, a huge gravitational potential will only impart a small additional velocity.
If you''ve got an introductory college text on physics everything you need to know is in there. If you don''t, then try google... there''s certainly lots of physics stuff around. Failing that, holler here and I''ll write something up for you (I''m a bit flat out at work today otherwise I''d just do it now for you).
Good luck,
Timkin
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quote:
Original post by DevLiquidKnight
coming across one problem i simulated gravity and now when i walk on the platforms i made i get stuck so i tried to make it subtract 1 from the y variable when he is on a platform so he could move side to side and it causes the guy to bounce up and down really fast
I think you''re looking for something simpler than what''s being discussed here!
As the player is falling down towards the platform, you need to calculate whether the players current position + the players current (downward) velocity results in collision with the platform.
If this is the case, you need to ''snap'' the player somehow to the platform so that he appears to be standing on it (how easily you accomplish this depends on how you have stored the platform tiles).
Once the player has been re-positioned on the platform, set the y-acceleration (gravity) to zero and then set a flag saying not to start the downward acceleration again until a platform can no longer be detected underneath the player.
(this is only one way to avoid the ''bobbing'' you talk of, but you could use the ''snapping'' part of your code, but it dpeends on how you structure it).
hope this is of some help. I can elaborate if you like.
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quote:
Original post by Carrot
I think you''re looking for something simpler than what''s being discussed here!
As the player is falling down towards the platform, you need to calculate whether the players current position + the players current (downward) velocity results in collision with the platform.
Read the poster''s last post. His application is considerably more complex than Super Mario Brothers.
Don''t listen to me. I''ve had too much coffee.
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I was just re-reading the posts to make sure I wasn''t misinterpreting your problem... I''m curious as to why the direction of the acceleration it changing! What frame of reference are you performing the calculations in?
You should compute them in the frame of reference of the planet as this will make life a lot easier for you (unless you have several large planets all close by... but then you have the problem of asking yourself HOW do these planets stay close by without crashing into each other!
Cheers,
Timkin
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quote:
Original post by DevLiquidKnight
coming across one problem i simulated gravity and now when i walk on the platforms i made i get stuck so i tried to make it subtract 1 from the y variable when he is on a platform so he could move side to side and it causes the guy to bounce up and down really fast
This sounds like Mario to me.
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Okay, the problem is that there are actually two questions in this thread. The first asked by DevLiquidKnight, which I answered in my first post... and the second asked by Lode.
If it isn''t obvious which question you are responding to, please point it out. Perhaps quoting the question in question!
Thanks,
Timkin
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Woah - I''m not sure he was after the full-on universal gravitation maths.
Gravity is very simple to do:
If your character is at x,y,z
and every frame you add its velocity vx,vy,vz
then simply accelerate it towards the ground by adding a constant to vz every frame. That constant is the acceleration due to gravity. Note that ALL objects accelerate towards the ground at the same rate. It doesn''t make any difference how heavy it is.
Carrot did a good job of describing how to snap the character to the platform.
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quote:
Original post by Anonymous Poster
Woah - I''m not sure he was after the full-on universal gravitation maths.
Again, people really should consider which question and answer they''re making comments about. Yep, DevLiquidKnight only needs the basic Newtonian results I gave at the end of my first post. Lode on the other hand is dealing with a different problem, where a simple Euler integration (as you suggest) is insufficient due to the nonlinearities in the problem.
It''s particularly frustrating to hear people disparage other people''s answers because they haven''t taken note of all of the posts in the thread.
Timkin
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quote:
Original post by Timkin
Your right, but surely when someone posts a reply it relates to the original poster''s question by default.
I wasn''t intentionally disparaging anyone, I just thought the original poster was looking for a simple discussion on the implementation of gravity in a 2D platformer.
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It seems to me that if you simply add the velocity to the position, at higher velocities, like say 100, the player is going to be skipping 100 pixels at each iteration, possibly missing a whole lot of collisions in the process (platforms, projectiles, etc.) Even at lower velocities it seems easy to miss collisions with thin or small objects.
However, this seems a lot easier than moving the player one pixel per iteration and calculating velocity in terms of the amount of time per iteration. I''m just doubtful about how well it works in practice - maybe the velocities never get high enough for it to be a problem?
Thanks in advance if you can clear this up for me.
Derek
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quote:
Original post by derekyu
It seems to me that if you simply add the velocity to the position, at higher velocities, like say 100, the player is going to be skipping 100 pixels at each iteration, possibly missing a whole lot of collisions in the process (platforms, projectiles, etc.)
If each platform or NPC has a collision area described by a bounding box (just for simplicity), and the player''s velocity is described by a two-dimensional vector then a simple ''vector/box intersection'' test will catch any collisions.
So basically you need a function that will test the players velocity-vector (located at player''s current position) with the bounding box (which is just four vector/vector tests).
Hope this helps.
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quote:
Original post by Carrot
Your right, but surely when someone posts a reply it relates to the original poster''s question by default.
What is it they say about assumption???
quote:
Original post by Carrot
I wasn''t intentionally disparaging anyone
That comment wasn''t aimed specifically at you (and not even generally, although it does apply to everyone)!
Cheers,
Timkin
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quote:
Original post by Carrot
If each platform or NPC has a collision area described by a bounding box (just for simplicity), and the player''s velocity is described by a two-dimensional vector then a simple ''vector/box intersection'' test will catch any collisions.
There was an interesting discussion about a month ago about how to deal with collision detection when collision timescales were expected to be less than integration timescales. If anyone is particularly interested there''s probably a link through my profile... or you can do a search.
Cheers,
Timkin
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#### Archived. This topic is now archived and is closed to further replies.. # How do you simulate gravity?. ## Recommended Posts. [removed] [Edited by - DevLiquidKnight on February 14, 2007 9:21:39 AM]. ##### Share on other sites. Hey,. You just add a velocity vector to you models and increase that with the gravity (9.8m/sec) or your own values. for example:. struct strMyObject {. Vector Pos, Vec;. }. strMyObject Player;. LOOP ->:. Player.Vec.y -= GRAVITY/Framespersec (or just a constant value). Player.Pos += Vec;. So each frames you add the velocity vector to the players possision to make him accelerate down if nothing is stopping him. (like a floor =. <- LOOP. ##### Share on other sites. To elaborate on the AP's response, gravity is a force acting between all objects that have mass. I'll assume for the time being that you want to simulate gravity as it would be felt on the surface of a planet. Let's stick with Earth, since it's nice and easy to relate to.. First, take Newton's First Law:. F = M*a (1). Force = Mass * Acceleration. The Gravitational force between two objects is proportional to their masses and inversely proprtional to the square of the separation between them, so that. G M1 M2F = ------- (2) R2. G is the gravitational constant and is equal to 6.672*10-11N m2 kg-2. If you hold one mass (M1) constant, say that of the Earth, then the acceleration that the other mass will undergo, relative to the first mass, is given by combining (1) and (2) to give. G M1a = ------- (2) R2. Substituting in the radius of the Earth (around 6,400 km on average) and the Mass of the Earth (about 5.98*1024 kg) gives an acceleration around 9.8 m s-2.. This is the same acceleration that all bodies undergo at the surface of the Earth. The Force that each body undergoes though is different and is given by F = 9.8 * mass of the body.. Since you know the acceleration of the body, you can work out it's vertical velocity (which is independant of the horizontal velocity) by integrating the acceleration over a fixed time interval, since acceleration is the first derivative of velocity with respect to time. You need to know the starting velocity to do this, since. v(t+dt) = v(t) + integraltt+dt a dt. Furthermore, if you know the position of the object at time t then you can work out its position at time t+dt by integrating the velocity. The result is that for any object near the surface of the Earth, you get the following equations for the height of the object and it's velocity:. s(t+dt) = s(t) + 1/2a*dt2. v(t+dt) = v(t) + a*dt. So if you know a (which you set yourself), the initial height, s(t), and the initial velocity, v(t) m/s (down... negative means up) then you can work out where the object is some time later!. I hope this explains the solution for you.. Good luck,. Timkin. [edited by - Timkin on July 29, 2002 11:36:28 PM]. ##### Share on other sites. i have an extremley simple way of simulating gravity, just make an mz variable for your object. every frame, add the gravity variable to mz. and every frame, add mz to z, so that way, it accelerates down until it hits something, then make mz = 0 and only add gravity if bOnGround is false or something. ##### Share on other sites. quote:. Original post by Timkin. G M1a = ------- (2) R2. Hey,. I''m just having a problem with this formula: the problem of it is that the acceleration isn''t constant here, and if I calculate physics in a space game every frame, I have to assume that the acceleration is constant during the 1/85th of a second this frame takes, and while this may seem only a small problem, the results of it are dramatic and the physics get more than 100% inaccurate. Do you have an idea how to calculate exactly what happened during one frame, knowing that not only the magnitude but also the direction of the gravity is changing during the time of one frame?. ##### Share on other sites. coming across one problem i simulated gravity and now when i walk on the platforms i made i get stuck so i tried to make it subtract 1 from the y variable when he is on a platform so he could move side to side and it causes the guy to bounce up and down really fast. [edited by - DevLiquidKnight on July 30, 2002 8:35:10 PM]. ##### Share on other sites. quote:. Original post by Lode. I''m just having a problem with this formula: the problem of it is that the acceleration isn''t constant here, and if I calculate physics in a space game every frame, I have to assume that the acceleration is constant during the 1/85th of a second this frame takes, and while this may seem only a small problem, the results of it are dramatic and the physics get more than 100% inaccurate. Do you have an idea how to calculate exactly what happened during one frame, knowing that not only the magnitude but also the direction of the gravity is changing during the time of one frame?. Unless things are travelling quite fast, then 1/85th of a second is a very small time to integrate over. Are you using an Euler integration (which I suspect you are)?. You may want to consider using a predictor-corrector integration scheme.. Good luck,. Timkin. ##### Share on other sites. The 1/85th of a second becomes to large to keep it accurate in two cases:. -if the planet has such a big mass that it generates very strong accelerations if you get close to it, but then for example my ship gets an acceleration of 10000000000m/s² at a certain point and in the 1/85th of a second it''s of course suddenly very far away of the planet... so far that it doesn''t get the negative -10000000000m/s² that it should get when it just crossed the planet anymore.. -if I speed up the time of the simulation: if I speed up the simulation with a factor of 1000, the 1/85th of a second becomes like 11.7seconds and that is too long to keep it accurate.. So indeed I should somehow predict where the ship will be after that time... but what kind of predictor-corrector integration scheme should I be looking for?. ##### Share on other sites. Okay, I didn''t realise you were talking about such large accelerations. Since you are, you really cannot treat this problem with classical (Newtonian) physics. You need to deal with the simulation from the perspective of the gravitational potential energy and the conversion from this potential energy into kinetic energy (strictly speaking you should be using General Relativity to get things correct, but that''s more trouble than it''s worth!).. Taking the ''energy approach'' will enable you to bring in relativistic effects that limit the speed of the craft to the speed of light. It will also mean that at high velocities, a huge gravitational potential will only impart a small additional velocity.. If you''ve got an introductory college text on physics everything you need to know is in there. If you don''t, then try google... there''s certainly lots of physics stuff around. Failing that, holler here and I''ll write something up for you (I''m a bit flat out at work today otherwise I''d just do it now for you).. Good luck,. Timkin. ##### Share on other sites. quote:. Original post by DevLiquidKnight. coming across one problem i simulated gravity and now when i walk on the platforms i made i get stuck so i tried to make it subtract 1 from the y variable when he is on a platform so he could move side to side and it causes the guy to bounce up and down really fast. I think you''re looking for something simpler than what''s being discussed here!. As the player is falling down towards the platform, you need to calculate whether the players current position + the players current (downward) velocity results in collision with the platform.. If this is the case, you need to ''snap'' the player somehow to the platform so that he appears to be standing on it (how easily you accomplish this depends on how you have stored the platform tiles).. Once the player has been re-positioned on the platform, set the y-acceleration (gravity) to zero and then set a flag saying not to start the downward acceleration again until a platform can no longer be detected underneath the player.. (this is only one way to avoid the ''bobbing'' you talk of, but you could use the ''snapping'' part of your code, but it dpeends on how you structure it).. hope this is of some help.
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I can elaborate if you like.. ##### Share on other sites. quote:. Original post by Carrot. I think you''re looking for something simpler than what''s being discussed here!. As the player is falling down towards the platform, you need to calculate whether the players current position + the players current (downward) velocity results in collision with the platform.. Read the poster''s last post. His application is considerably more complex than Super Mario Brothers.. Don''t listen to me. I''ve had too much coffee.. ##### Share on other sites. I was just re-reading the posts to make sure I wasn''t misinterpreting your problem... I''m curious as to why the direction of the acceleration it changing! What frame of reference are you performing the calculations in?. You should compute them in the frame of reference of the planet as this will make life a lot easier for you (unless you have several large planets all close by... but then you have the problem of asking yourself HOW do these planets stay close by without crashing into each other!. Cheers,. Timkin. ##### Share on other sites. quote:. Original post by DevLiquidKnight. coming across one problem i simulated gravity and now when i walk on the platforms i made i get stuck so i tried to make it subtract 1 from the y variable when he is on a platform so he could move side to side and it causes the guy to bounce up and down really fast. This sounds like Mario to me.. ##### Share on other sites. Okay, the problem is that there are actually two questions in this thread. The first asked by DevLiquidKnight, which I answered in my first post... and the second asked by Lode.. If it isn''t obvious which question you are responding to, please point it out. Perhaps quoting the question in question!. Thanks,. Timkin. ##### Share on other sites. Woah - I''m not sure he was after the full-on universal gravitation maths.. Gravity is very simple to do:. If your character is at x,y,z. and every frame you add its velocity vx,vy,vz. then simply accelerate it towards the ground by adding a constant to vz every frame. That constant is the acceleration due to gravity. Note that ALL objects accelerate towards the ground at the same rate. It doesn''t make any difference how heavy it is.. Carrot did a good job of describing how to snap the character to the platform.. ##### Share on other sites. quote:. Original post by Anonymous Poster. Woah - I''m not sure he was after the full-on universal gravitation maths.. Again, people really should consider which question and answer they''re making comments about. Yep, DevLiquidKnight only needs the basic Newtonian results I gave at the end of my first post. Lode on the other hand is dealing with a different problem, where a simple Euler integration (as you suggest) is insufficient due to the nonlinearities in the problem.. It''s particularly frustrating to hear people disparage other people''s answers because they haven''t taken note of all of the posts in the thread.. Timkin. ##### Share on other sites. quote:. Original post by Timkin. Your right, but surely when someone posts a reply it relates to the original poster''s question by default.. I wasn''t intentionally disparaging anyone, I just thought the original poster was looking for a simple discussion on the implementation of gravity in a 2D platformer.. ##### Share on other sites. It seems to me that if you simply add the velocity to the position, at higher velocities, like say 100, the player is going to be skipping 100 pixels at each iteration, possibly missing a whole lot of collisions in the process (platforms, projectiles, etc.) Even at lower velocities it seems easy to miss collisions with thin or small objects.. However, this seems a lot easier than moving the player one pixel per iteration and calculating velocity in terms of the amount of time per iteration. I''m just doubtful about how well it works in practice - maybe the velocities never get high enough for it to be a problem?. Thanks in advance if you can clear this up for me.. Derek. ##### Share on other sites. quote:. Original post by derekyu. It seems to me that if you simply add the velocity to the position, at higher velocities, like say 100, the player is going to be skipping 100 pixels at each iteration, possibly missing a whole lot of collisions in the process (platforms, projectiles, etc.). If each platform or NPC has a collision area described by a bounding box (just for simplicity), and the player''s velocity is described by a two-dimensional vector then a simple ''vector/box intersection'' test will catch any collisions.. So basically you need a function that will test the players velocity-vector (located at player''s current position) with the bounding box (which is just four vector/vector tests).. Hope this helps.. ##### Share on other sites. quote:. Original post by Carrot. Your right, but surely when someone posts a reply it relates to the original poster''s question by default.. What is it they say about assumption???. quote:. Original post by Carrot. I wasn''t intentionally disparaging anyone. That comment wasn''t aimed specifically at you (and not even generally, although it does apply to everyone)!. Cheers,. Timkin. ##### Share on other sites. quote:. Original post by Carrot. If each platform or NPC has a collision area described by a bounding box (just for simplicity), and the player''s velocity is described by a two-dimensional vector then a simple ''vector/box intersection'' test will catch any collisions.. There was an interesting discussion about a month ago about how to deal with collision detection when collision timescales were expected to be less than integration timescales. If anyone is particularly interested there''s probably a link through my profile... or you can do a search.. Cheers,. Timkin. • ### Forum Statistics. • Total Topics. 628353. • Total Posts. 2982228. • 10. • 9. • 11. • 24. • 11.
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Is 86 Divisible By 5?
In this quick and easy guide, we'll work out whether 86 is divisible by 5. There are some simple rules we can follow to decide whether one number is divisible by another without ever needing to even do the division!
First up, let's clarify what we mean by "86 is divisible by 5". What we want to check is whether 86 can be divided by 5 without any remainder (i.e the answer is a whole number).
Checking whether 86 is divisible by 5 is one of the easiest divisibility checks you can make. Basically, if the number ends with either a 5 or a 0, it is divisible by 5. In this case, it ends with 6.
We can see that 86 DOES NOT end with a 5 or a 0, which means that 86 IS NOT divisible by 5.
Another way you can figure out if 86 is divisible by 5 is by actually doing the calculation and dividing 86 by 5:
86 / 5 = 17.2
As you can see, when we do this division we have a decimal of 0.2. Since the division does not result in a whole number, this shows us that 86 is not divisible by 5.
Hopefully now you know exactly how to work out whether one number is divisible by another. Could I have just told you to divide 86 by 5 and check if it is a whole number? Yes, but aren't you glad you learned the process?
Give this a go for yourself and try to calculate a couple of these without using our calculator. Grab a pencil and a piece of paper and pick a couple of numbers to try it with.
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Is 86 Divisible By 5?. In this quick and easy guide, we'll work out whether 86 is divisible by 5. There are some simple rules we can follow to decide whether one number is divisible by another without ever needing to even do the division!. First up, let's clarify what we mean by "86 is divisible by 5". What we want to check is whether 86 can be divided by 5 without any remainder (i.e the answer is a whole number).. Checking whether 86 is divisible by 5 is one of the easiest divisibility checks you can make. Basically, if the number ends with either a 5 or a 0, it is divisible by 5. In this case, it ends with 6.. We can see that 86 DOES NOT end with a 5 or a 0, which means that 86 IS NOT divisible by 5.. Another way you can figure out if 86 is divisible by 5 is by actually doing the calculation and dividing 86 by 5:.
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86 / 5 = 17.2. As you can see, when we do this division we have a decimal of 0.2. Since the division does not result in a whole number, this shows us that 86 is not divisible by 5.. Hopefully now you know exactly how to work out whether one number is divisible by another. Could I have just told you to divide 86 by 5 and check if it is a whole number? Yes, but aren't you glad you learned the process?. Give this a go for yourself and try to calculate a couple of these without using our calculator. Grab a pencil and a piece of paper and pick a couple of numbers to try it with.
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math equations
My four equations are
1. 6x+2= 4x+3 = (2x+1=1/2)
2. 3x+6= 2x+4 = (-1x=+2)
3. -2x+5= 3x+1 = 5x= +4=4/5
4. -2x+5 -5x-4 = 1/6
what i have learned about grade 9 fractions
• if its addition and subtraction you have to make the denominator the same and you have to make it equavelent too. $\frac {-6}/{-4}$
• tug of war $\frac {-5}+{9}$
• just do it- its when it a multiplication question
• operations with integers, multiplication and division
• speed multiplication, for speed multiplication you have to divide the denominator and the numerator to make it the smallest fraction.
• we learned how to put fractions on a numberline and we also learned how to compare fractions. for a numberline you make the same denominator, for comparing fractions you also have to make it the same denominator and you can tell which ones bigger.
• we learned how to divide and multiply fractions together, for divsion u make it into a reciprocal fraction and you multiply. for multiplication you can just multiply straight across. Divion= $\frac {-5}{8}\div\frac{5}{3}$
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math equations. My four equations are. 1. 6x+2= 4x+3 = (2x+1=1/2). 2. 3x+6= 2x+4 = (-1x=+2). 3. -2x+5= 3x+1 = 5x= +4=4/5. 4. -2x+5 -5x-4 = 1/6. what i have learned about grade 9 fractions.
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• if its addition and subtraction you have to make the denominator the same and you have to make it equavelent too. $\frac {-6}/{-4}$. • tug of war $\frac {-5}+{9}$. • just do it- its when it a multiplication question. • operations with integers, multiplication and division. • speed multiplication, for speed multiplication you have to divide the denominator and the numerator to make it the smallest fraction.. • we learned how to put fractions on a numberline and we also learned how to compare fractions. for a numberline you make the same denominator, for comparing fractions you also have to make it the same denominator and you can tell which ones bigger.. • we learned how to divide and multiply fractions together, for divsion u make it into a reciprocal fraction and you multiply. for multiplication you can just multiply straight across. Divion= $\frac {-5}{8}\div\frac{5}{3}$.
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https://math.answers.com/basic-math/Least_Common_Multiples_of_35_and_75
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0
# Least Common Multiples of 35 and 75?
Updated: 4/28/2022
Wiki User
13y ago
The LCM is: 525
Wiki User
13y ago
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Q: Least Common Multiples of 35 and 75?
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Related questions
### What are the least common multiples of 75 and 105?
525 is the least common multiple.
### What are the least common multiples of 30 36 and 75?
The Least Common Multiple (LCM) for 30 36 75 is 900.
### Does 75 and 110 have multiples in common?
Yes. The least common multiple is: 1,650
### Least common multiple of 75 and 250?
The multiples of 75: 75, 150, 225, 300, 375, 450, 525, 600, 675, 750, 825, ... The multiples of 250: 250, 500, 750, 1000, ... The least common multiple is the number in both lists that is the least. For 75 and 250, the least common multiple is 750.
600
### Common multiples of 15 and 25?
Common multiples of 15 and 25 are all multiples of 75. That is because 75 is the least common multiple of 15 and 25. So they are 75, 150, 225, etc. Just keep multiplying 75 by 4,5,6...etc.
### What are the common multiples of 25 and 75?
Since 75 is a multiple of 25, all of its multiples are common.
### What is the least common multiples of 75 and 15?
Since 75 is a multiple of 15, the least common multiple is 75.75 x 1 = 7515 x 5 = 75751*75 = 755*15 = 75
### What is the least common multiple of 25 50 and 75?
The LCM of 25, 50, and 75 is 150. The multiples of 25 are 25, 50, 75, 100, 125, 150, and so on. The multiples of 50 are 50, 100, 150, and so on. The multiples of 75 are 75, 150, and so on. So the LCM is 150.
### What are the common multiples of 15 and 25?
The common multiples of 15 and 25 are all multiples of their LCM, which is 75. So, the common multiples are 75, 150, 225, 300, 375, 450, 525, 600, and so on.
### What is the least common multiple of 35 and 75?
The answer is 525. The prime factorization for 35 is 7*5, and the one for 75 is 3*5*5.Take the 5s from 75, and the 3, and take 7 from 35. Multiply it to get 525!
75 and 150
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0. # Least Common Multiples of 35 and 75?. Updated: 4/28/2022. Wiki User. 13y ago. The LCM is: 525. Wiki User. 13y ago. Earn +20 pts. Q: Least Common Multiples of 35 and 75?. Submit. Still have questions?. Related questions. ### What are the least common multiples of 75 and 105?. 525 is the least common multiple.. ### What are the least common multiples of 30 36 and 75?. The Least Common Multiple (LCM) for 30 36 75 is 900.. ### Does 75 and 110 have multiples in common?. Yes. The least common multiple is: 1,650. ### Least common multiple of 75 and 250?. The multiples of 75: 75, 150, 225, 300, 375, 450, 525, 600, 675, 750, 825, ... The multiples of 250: 250, 500, 750, 1000, ... The least common multiple is the number in both lists that is the least. For 75 and 250, the least common multiple is 750.. 600.
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### Common multiples of 15 and 25?. Common multiples of 15 and 25 are all multiples of 75. That is because 75 is the least common multiple of 15 and 25. So they are 75, 150, 225, etc. Just keep multiplying 75 by 4,5,6...etc.. ### What are the common multiples of 25 and 75?. Since 75 is a multiple of 25, all of its multiples are common.. ### What is the least common multiples of 75 and 15?. Since 75 is a multiple of 15, the least common multiple is 75.75 x 1 = 7515 x 5 = 75751*75 = 755*15 = 75. ### What is the least common multiple of 25 50 and 75?. The LCM of 25, 50, and 75 is 150. The multiples of 25 are 25, 50, 75, 100, 125, 150, and so on. The multiples of 50 are 50, 100, 150, and so on. The multiples of 75 are 75, 150, and so on. So the LCM is 150.. ### What are the common multiples of 15 and 25?. The common multiples of 15 and 25 are all multiples of their LCM, which is 75. So, the common multiples are 75, 150, 225, 300, 375, 450, 525, 600, and so on.. ### What is the least common multiple of 35 and 75?. The answer is 525. The prime factorization for 35 is 7*5, and the one for 75 is 3*5*5.Take the 5s from 75, and the 3, and take 7 from 35. Multiply it to get 525!. 75 and 150.
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# How much grass is there on the planet?
## How much grass is there on the planet?
The surface of earth is roughly 70.8% water (Ibid), which means that 29.2% is land. Of these 29.2% approximately 20% is grass (http://www.answerbag.com/q_view/27304).
How many grass are there?
How many types of grass are there? There are over 12,000 species of grass around the world and dozens of them can be found in different regions of the United States.
### How much of the US is grass?
Turf grasses occupy 2% of the land in the continental US. That’s more acreage than the amount of land used to grow irrigated corn.
How do you count grass?
Walk the length of your lawn, figuring that one pace equals about 3 feet. Do the same with the width of the lawn. Then multiply the length by the width to arrive at the total. Make sure you subtract the square footage of your home and driveway when calculating the total square footage of your lawn.
## How many blades of grass are in a inch?
You multiply 60,912 and 6,400 because there are 6,400 square yards in a football field and there are 60,912 blades of grass in a square yard so you multiply. Our class found the average of blades of grass in a square inch: 47 blades of grass.
How heavy is the earth?
5.972 × 10^24 kg
Earth/Mass
### How many blades of grass are in a yard?
This website cites the Oklahoma Museum of Natural history as stating that, “The average square foot of grass has 3,000 blades.” This article claims average lawn size is about an acre. That’s 43560 square feet, or about 130 million blades of grass.
Is bamboo a grass?
Bamboo is in fact a type of grass – a very fast growing and giant grass. Bamboo grows in a short but strong growth spurt during summer and then remains near dormant over winter. During the ‘growth spurt’, a bamboo will start with new shoots from the ground which will grow to full height in two to three months.
## Why the American lawn is bad?
Their maintenance produces more greenhouse gases than they absorb, and they are biodiversity deserts that have contributed to vanishing insect populations. Residential lawns cover 2% of US land and require more irrigation than any agricultural crop grown in the country.
Is a lawn worth it?
A healthy lawn out-competes weeds, helps control erosion on your property and limits runoff into overburdened storm drains. Lawn care products should also be safe for people, as well as for pets. It takes time, effort and an investment for a lawn to thrive, but the consensus is obviously that it’s worth the effort.
### How many blades of grass does a lawn have?
The amount of blades of grass in a square foot of lawn depends on the type of grass and the density. The average square foot of grass has 3,000 blades, according to the Oklahoma Museum of Natural History.
How big is a piece of turf?
The average turf is roughly 1.3ft or 0.41m wide and 8.16ft or 2.49m long. Carbutts turves are sold in standard sizes.
## How many blades of grass are there per square meter?
Let’s assume there are 10 blades of grass per square centimeter. This means there are 100.000 blades of grass per square meter and 100.000.000.000 blades of grass per square kilometer. The total surface area of earth is 5.1 * 10 8 km2 (http://chartsbin.com/view/wwu).
How to calculate the amount of grass seed needed per hectare?
If you need to calculate amount of grass seed needed per hectare, then you will need to calculate it out via a longer method: 1.81 kg ÷ 92.90 square meters = 0.0195 kg/square meters; since 1 hectare = 10,000 square meters, 0.0195 x 10,000 = 195 kg of seed is needed per hectare.
### How tall does grass grow in the ground?
Grasses vary in size from 2.1 m (7 ft) tall with roots extending down into the soil 1.8 m (6 ft), to the short grasses growing to a height of only 20 to 25 cm (8 to 10 in) tall. These short grasses can have roots that extend 1 m (about 3 ft) deep. The height of grass correlates with the amount of rainfall it receives.
How many fescue seeds per pound of grass?
Think of it this way: There are 225,000 tall fescue seeds per pound, whereas there are around 2,180,000 kentucky bluegrass seeds per pound. So you need more of the larger seed to cover the same area as you would with less of the smaller seed.
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# How much grass is there on the planet?. ## How much grass is there on the planet?. The surface of earth is roughly 70.8% water (Ibid), which means that 29.2% is land. Of these 29.2% approximately 20% is grass (http://www.answerbag.com/q_view/27304).. How many grass are there?. How many types of grass are there? There are over 12,000 species of grass around the world and dozens of them can be found in different regions of the United States.. ### How much of the US is grass?. Turf grasses occupy 2% of the land in the continental US. That’s more acreage than the amount of land used to grow irrigated corn.. How do you count grass?. Walk the length of your lawn, figuring that one pace equals about 3 feet. Do the same with the width of the lawn. Then multiply the length by the width to arrive at the total. Make sure you subtract the square footage of your home and driveway when calculating the total square footage of your lawn.. ## How many blades of grass are in a inch?. You multiply 60,912 and 6,400 because there are 6,400 square yards in a football field and there are 60,912 blades of grass in a square yard so you multiply. Our class found the average of blades of grass in a square inch: 47 blades of grass.. How heavy is the earth?. 5.972 × 10^24 kg. Earth/Mass. ### How many blades of grass are in a yard?. This website cites the Oklahoma Museum of Natural history as stating that, “The average square foot of grass has 3,000 blades.” This article claims average lawn size is about an acre. That’s 43560 square feet, or about 130 million blades of grass.. Is bamboo a grass?. Bamboo is in fact a type of grass – a very fast growing and giant grass. Bamboo grows in a short but strong growth spurt during summer and then remains near dormant over winter. During the ‘growth spurt’, a bamboo will start with new shoots from the ground which will grow to full height in two to three months.
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## Why the American lawn is bad?. Their maintenance produces more greenhouse gases than they absorb, and they are biodiversity deserts that have contributed to vanishing insect populations. Residential lawns cover 2% of US land and require more irrigation than any agricultural crop grown in the country.. Is a lawn worth it?. A healthy lawn out-competes weeds, helps control erosion on your property and limits runoff into overburdened storm drains. Lawn care products should also be safe for people, as well as for pets. It takes time, effort and an investment for a lawn to thrive, but the consensus is obviously that it’s worth the effort.. ### How many blades of grass does a lawn have?. The amount of blades of grass in a square foot of lawn depends on the type of grass and the density. The average square foot of grass has 3,000 blades, according to the Oklahoma Museum of Natural History.. How big is a piece of turf?. The average turf is roughly 1.3ft or 0.41m wide and 8.16ft or 2.49m long. Carbutts turves are sold in standard sizes.. ## How many blades of grass are there per square meter?. Let’s assume there are 10 blades of grass per square centimeter. This means there are 100.000 blades of grass per square meter and 100.000.000.000 blades of grass per square kilometer. The total surface area of earth is 5.1 * 10 8 km2 (http://chartsbin.com/view/wwu).. How to calculate the amount of grass seed needed per hectare?. If you need to calculate amount of grass seed needed per hectare, then you will need to calculate it out via a longer method: 1.81 kg ÷ 92.90 square meters = 0.0195 kg/square meters; since 1 hectare = 10,000 square meters, 0.0195 x 10,000 = 195 kg of seed is needed per hectare.. ### How tall does grass grow in the ground?. Grasses vary in size from 2.1 m (7 ft) tall with roots extending down into the soil 1.8 m (6 ft), to the short grasses growing to a height of only 20 to 25 cm (8 to 10 in) tall. These short grasses can have roots that extend 1 m (about 3 ft) deep. The height of grass correlates with the amount of rainfall it receives.. How many fescue seeds per pound of grass?. Think of it this way: There are 225,000 tall fescue seeds per pound, whereas there are around 2,180,000 kentucky bluegrass seeds per pound. So you need more of the larger seed to cover the same area as you would with less of the smaller seed.
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### CFA Practice Question
There are 923 practice questions for this topic.
### CFA Practice Question
Currency A is selling for \$0.61 and the buying rate for Currency B is \$0.19. The currency A : currency B cross rate is ______.
A. A:B = 0.31
B. A:B = 0.81
C. Neither of these answers
Correct Answer: C
A:B = 0.61/0.19 = 3.21
### User Contributed Comments6
User Comment
VenkatB from the dealer point of view..
A:\$ = unknown bid - 0.61\$/A
B:\$ = 0.19\$/B - unknown ask
The question asks us to find A:B = B/A = number f units of B needed to buy one unit of A
From the investor standpoint : We start with 1 unit of B, and sell it to dealer and get 0.19\$,
we then take that 0.19\$ and buy A (one unit of A sells for 0.61\$, so 1\$ will get us 1/.61 of A and 0.19\$ will get us .19/.61 of A = .311475 of A)
so with one unit of B we can buy .311475 units of A
To buy 1 unit of A, we need 1/.311475 units of B = 3.210526
____________________________________________
or
A:\$ = unknown bid - 0.61\$/A
B:\$ = 0.19\$/B - unknown ask
A:B = B/A = (\$/A) / (\$/B) = 0.61/0.19
BryonBUI I think this question doesn't involve a spread.
Simply: A:B=(A:\$)"divide"(B:\$)=0.61/0.19
Notation used as in CFAI text.
quanttrader it's a direct quote -- ie quote amt of fc in terms of dc
ankurwa10 I second ByronBUI's approach.
And Venkat, I believe the earlier reading mentioned that when S= A/B, implies the number of units of B needed to buy A. So it won't make sense, unless it is a direct quote. Which is the number of units of B needed to buy A would be expressed as B/A instead of A/B.
mtsimone .61/x * x/.19 = .61/.19 = 3.211. Use a form where the unknown cancels out.
skarthi146 given:
USD/A = unknown bid / 0.61 offer
USD/B = 0.19 bid / unknown offer
question is: B/A = (USD/A) / (USD/B) = 0.61 / 0.19 = 3.21
You need to log in first to add your comment.
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### CFA Practice Question. There are 923 practice questions for this topic.. ### CFA Practice Question. Currency A is selling for \$0.61 and the buying rate for Currency B is \$0.19. The currency A : currency B cross rate is ______.. A. A:B = 0.31. B. A:B = 0.81. C. Neither of these answers. Correct Answer: C. A:B = 0.61/0.19 = 3.21. ### User Contributed Comments6. User Comment. VenkatB from the dealer point of view... A:\$ = unknown bid - 0.61\$/A. B:\$ = 0.19\$/B - unknown ask. The question asks us to find A:B = B/A = number f units of B needed to buy one unit of A. From the investor standpoint : We start with 1 unit of B, and sell it to dealer and get 0.19\$,. we then take that 0.19\$ and buy A (one unit of A sells for 0.61\$, so 1\$ will get us 1/.61 of A and 0.19\$ will get us .19/.61 of A = .311475 of A). so with one unit of B we can buy .311475 units of A.
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To buy 1 unit of A, we need 1/.311475 units of B = 3.210526. ____________________________________________. or. A:\$ = unknown bid - 0.61\$/A. B:\$ = 0.19\$/B - unknown ask. A:B = B/A = (\$/A) / (\$/B) = 0.61/0.19. BryonBUI I think this question doesn't involve a spread.. Simply: A:B=(A:\$)"divide"(B:\$)=0.61/0.19. Notation used as in CFAI text.. quanttrader it's a direct quote -- ie quote amt of fc in terms of dc. ankurwa10 I second ByronBUI's approach.. And Venkat, I believe the earlier reading mentioned that when S= A/B, implies the number of units of B needed to buy A. So it won't make sense, unless it is a direct quote. Which is the number of units of B needed to buy A would be expressed as B/A instead of A/B.. mtsimone .61/x * x/.19 = .61/.19 = 3.211. Use a form where the unknown cancels out.. skarthi146 given:. USD/A = unknown bid / 0.61 offer. USD/B = 0.19 bid / unknown offer. question is: B/A = (USD/A) / (USD/B) = 0.61 / 0.19 = 3.21. You need to log in first to add your comment.
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https://xmphysics.com/2019/02/09/sign-convention/
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# Sign Convention
• Displacement, velocity and acceleration are all vector quantities. They thus have both magnitude and direction.
• Rectilinear motion refers to motion which occurs strictly along a straight line. Since only two directions are possible, a positive sign indicates one direction, and the negative indicates the opposite direction.
• When velocity is negative
1. “decreasing” velocity actually implies increasing speed.
2. positive acceleration actually implies decreasing speed and vice versa.
• For example, let’s adopt the “rightward is positive” convention for the motions below.
• Since the sheep is moving rightward and speeding up, its velocity is represented by more and more positive numbers.
• Since the velocity is increasing, the acceleration is positive.
• Since the sheep is moving leftward and speeding up, its velocity is represented by more and more negative numbers.
• Since the velocity is “decreasing”, the acceleration is negative.
• Since the sheep is moving rightward but slowing down, its velocity is represented by less and less positive numbers.
• Since the velocity is decreasing, the acceleration is negative.
• Since the sheep is moving leftward but slowing down, its velocity is represented by less and less negative numbers.
• Since the velocity is “increasing”, the acceleration is positive.
• In a nutshell, the speed increases if the velocity and acceleration have the same sign, and decreases if the velocity and acceleration have opposite signs.
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# Sign Convention. • Displacement, velocity and acceleration are all vector quantities. They thus have both magnitude and direction.. • Rectilinear motion refers to motion which occurs strictly along a straight line. Since only two directions are possible, a positive sign indicates one direction, and the negative indicates the opposite direction.. • When velocity is negative. 1. “decreasing” velocity actually implies increasing speed.. 2. positive acceleration actually implies decreasing speed and vice versa.. • For example, let’s adopt the “rightward is positive” convention for the motions below.
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• Since the sheep is moving rightward and speeding up, its velocity is represented by more and more positive numbers.. • Since the velocity is increasing, the acceleration is positive.. • Since the sheep is moving leftward and speeding up, its velocity is represented by more and more negative numbers.. • Since the velocity is “decreasing”, the acceleration is negative.. • Since the sheep is moving rightward but slowing down, its velocity is represented by less and less positive numbers.. • Since the velocity is decreasing, the acceleration is negative.. • Since the sheep is moving leftward but slowing down, its velocity is represented by less and less negative numbers.. • Since the velocity is “increasing”, the acceleration is positive.. • In a nutshell, the speed increases if the velocity and acceleration have the same sign, and decreases if the velocity and acceleration have opposite signs.
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MCAT Content / Work / Work Done By A Constant Force W Fd Cos0
### Work done by a constant force: W = Fd cosθ
Topic: Work
The work done on a system by a constant force is the product of the component of the force in the direction of motion times the distance through which the force acts.
W is work, F is the magnitude of the force on the system, d is the magnitude of the displacement of the system, and θ is the angle between the force vector F and the displacement vector d.
Examples of work:
1. The work done by the force F on this lawnmower is Fdcosθ.
2. A person holding a briefcase does no work on it because there is no motion. No energy is transferred to or from the briefcase. The person moving the briefcase horizontally at a constant speed does no work on it and transfers no energy to it.
3. Work is done on the briefcase by carrying it upstairs at a constant speed because there is necessarily a component of force F in the direction of the motion. Energy is transferred to the briefcase and could, in turn, be used to do work.
The work can be calculated as the area under the force-displacement graph. This is true for work done by a constant force (like the examples shown here), as well as work done by a variable force. Work done by an elastic force, for example, follows the equation W = ksx2/2, where ks is the spring constant of the force and x is the displacement. Work can also be calculated in these systems from force-displacement graphs.
Practice Questions
Coronary heart disease and blood pressure
MCAT Official Prep (AAMC)
Physics Online Flashcards Question 10
Physics Question Pack Passage 7 Question 45
Physics Question Pack Question 57
Physics Question Pack Question 74
Physics Question Pack Passage 13 Question 79
Physics Question Pack Passage 13 Question 82
Official Guide C/P Section Passage 2 Question 7
Practice Exam 3 B/B Section Passage 1 Question 1
Practice Exam 3 B/B Section Passage 1 Question 4
Key Points
• Work is the transfer of energy by a force acting on an object as it is displaced.
• The work done by a force is zero if the displacement is either zero or perpendicular to the force.
• The work done is positive if the force and displacement have the same direction, and the work done is negative if they have opposite direction.
Key Terms
Energy: A quantity that denotes the ability to do work and is measured in a unit dimensioned in mass × distance²/time² (ML²/T²) or the equivalent
Force-displacement graph: A graph of the force applied on the y-axis, and distance moved on the x-axis. The area under the curve between two points is the work done.
Billing Information
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MCAT Content / Work / Work Done By A Constant Force W Fd Cos0. ### Work done by a constant force: W = Fd cosθ. Topic: Work. The work done on a system by a constant force is the product of the component of the force in the direction of motion times the distance through which the force acts.. W is work, F is the magnitude of the force on the system, d is the magnitude of the displacement of the system, and θ is the angle between the force vector F and the displacement vector d.. Examples of work:. 1. The work done by the force F on this lawnmower is Fdcosθ.. 2. A person holding a briefcase does no work on it because there is no motion. No energy is transferred to or from the briefcase. The person moving the briefcase horizontally at a constant speed does no work on it and transfers no energy to it.. 3. Work is done on the briefcase by carrying it upstairs at a constant speed because there is necessarily a component of force F in the direction of the motion. Energy is transferred to the briefcase and could, in turn, be used to do work.. The work can be calculated as the area under the force-displacement graph. This is true for work done by a constant force (like the examples shown here), as well as work done by a variable force. Work done by an elastic force, for example, follows the equation W = ksx2/2, where ks is the spring constant of the force and x is the displacement. Work can also be calculated in these systems from force-displacement graphs.. Practice Questions. Coronary heart disease and blood pressure.
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MCAT Official Prep (AAMC). Physics Online Flashcards Question 10. Physics Question Pack Passage 7 Question 45. Physics Question Pack Question 57. Physics Question Pack Question 74. Physics Question Pack Passage 13 Question 79. Physics Question Pack Passage 13 Question 82. Official Guide C/P Section Passage 2 Question 7. Practice Exam 3 B/B Section Passage 1 Question 1. Practice Exam 3 B/B Section Passage 1 Question 4. Key Points. • Work is the transfer of energy by a force acting on an object as it is displaced.. • The work done by a force is zero if the displacement is either zero or perpendicular to the force.. • The work done is positive if the force and displacement have the same direction, and the work done is negative if they have opposite direction.. Key Terms. Energy: A quantity that denotes the ability to do work and is measured in a unit dimensioned in mass × distance²/time² (ML²/T²) or the equivalent. Force-displacement graph: A graph of the force applied on the y-axis, and distance moved on the x-axis. The area under the curve between two points is the work done.. Billing Information.
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https://www.physicsforums.com/threads/why-does-an-element-in-a-circuit-cause-a-voltage-drop.823346/
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# Why does an element in a circuit cause a "voltage drop"
1. Jul 14, 2015
### Mr Davis 97
I have seen this topic in other threads before, but I have not found an answer that eliminates my confusion. I know that electric potential is defined so that only the position of a charge with respect to a field determines the quantity of the electric potential (voltage). This allows us to quantify, generally, how spacial orientation with respect to a field affects the specific potential energy of a specific charge with some amount of coulombs. This all being said, I don't understand why there is a "voltage drop" after charged particles enter an element (such as a resistor) in a circuit. If we say that a battery creates an electric field and thus well-defined electric potentials around the field, then how come there is a drop in electric potential after a particle enters a a resistor, if that charged particle is roughly the same distance from the battery which the electric field comes from? I've heard people say that there is a voltage drop because the charged particles lose energy in the resistor, but if voltage is electric potential and not electric potential energy, then how does the fact that they lose energy affect electric potential (voltage), which is defined as the amount of potential energy per coulomb? I think my confusion lies in the concept of voltage. Basically I don't know how voltage is supposed to drop when there is a definite, constant amount of voltage for the battery.
2. Jul 14, 2015
### phinds
Suppose you have three resistors of equal value hooked in series with each other and with the battery. Would you expect the voltage across the middle resistor to be identical to the voltage of the battery? This would imply that the other two resistors are acting not like resistors but like wires. Ohm would be upset.
3. Jul 14, 2015
### Staff: Mentor
My understanding, with some not-very-accurate terminology:
Remember that the charges all interact with each other. Charges moving through the resistor require more energy to move than charges in the conductor. You could say that the charges 'pile up' on one side of the resistor until the current flow through the conductor and the resistor are equal, and that the current is mostly limited by how many charges per second can pass through the resistor.
So the distance from the terminals doesn't matter since the charges just move around in response to each other's electric fields as well as the battery's.
4. Jul 14, 2015
### stedwards
You have constant water pressure delivered to you house from the regulator provided by the water company. The pressure is all the same, throughout your house--34 PSI, until someone turns on the water.
You're taking a shower, and someone turns on the dishwater. The water flow drops because the pressure at the nozzle decreases. (It is an unstated universal law, that the water always gets colder, but that's another story.)
This is because there there is a restistance, from the sides of the pipes, to impede the water from flowing. The more water that is in demand, the lower the flow rate, and the less pressure.
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# Why does an element in a circuit cause a "voltage drop". 1. Jul 14, 2015. ### Mr Davis 97. I have seen this topic in other threads before, but I have not found an answer that eliminates my confusion. I know that electric potential is defined so that only the position of a charge with respect to a field determines the quantity of the electric potential (voltage). This allows us to quantify, generally, how spacial orientation with respect to a field affects the specific potential energy of a specific charge with some amount of coulombs. This all being said, I don't understand why there is a "voltage drop" after charged particles enter an element (such as a resistor) in a circuit. If we say that a battery creates an electric field and thus well-defined electric potentials around the field, then how come there is a drop in electric potential after a particle enters a a resistor, if that charged particle is roughly the same distance from the battery which the electric field comes from? I've heard people say that there is a voltage drop because the charged particles lose energy in the resistor, but if voltage is electric potential and not electric potential energy, then how does the fact that they lose energy affect electric potential (voltage), which is defined as the amount of potential energy per coulomb? I think my confusion lies in the concept of voltage. Basically I don't know how voltage is supposed to drop when there is a definite, constant amount of voltage for the battery.. 2. Jul 14, 2015. ### phinds. Suppose you have three resistors of equal value hooked in series with each other and with the battery. Would you expect the voltage across the middle resistor to be identical to the voltage of the battery? This would imply that the other two resistors are acting not like resistors but like wires. Ohm would be upset.. 3. Jul 14, 2015.
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### Staff: Mentor. My understanding, with some not-very-accurate terminology:. Remember that the charges all interact with each other. Charges moving through the resistor require more energy to move than charges in the conductor. You could say that the charges 'pile up' on one side of the resistor until the current flow through the conductor and the resistor are equal, and that the current is mostly limited by how many charges per second can pass through the resistor.. So the distance from the terminals doesn't matter since the charges just move around in response to each other's electric fields as well as the battery's.. 4. Jul 14, 2015. ### stedwards. You have constant water pressure delivered to you house from the regulator provided by the water company. The pressure is all the same, throughout your house--34 PSI, until someone turns on the water.. You're taking a shower, and someone turns on the dishwater. The water flow drops because the pressure at the nozzle decreases. (It is an unstated universal law, that the water always gets colder, but that's another story.). This is because there there is a restistance, from the sides of the pipes, to impede the water from flowing. The more water that is in demand, the lower the flow rate, and the less pressure.
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# Gravitation Quiz
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This is gravitation quiz.
No searchs from google
• 1.
### What quantity is gravitational force ?
• A.
Scalar quantity
• B.
Vector quantity
• C.
A and B both
• D.
None
B. Vector quantity
Explanation
Gravitational force is a vector quantity because it has both magnitude and direction. The magnitude of gravitational force is determined by the masses of the objects involved, and the direction is always towards the center of mass of the objects. This means that gravitational force can be represented by an arrow, indicating both the strength and direction of the force. Therefore, the correct answer is "Vector quantity".
Rate this question:
• 2.
### Which one of the following is true ?
• A.
Gravitational force is indirectly proportional to masses of two objects
• B.
Gravitational force is directly proportional to square of distance between two bodies
• C.
Gravitational force is directly proportional to masses of two bodies
• D.
None
• E.
A , b , c all
C. Gravitational force is directly proportional to masses of two bodies
Explanation
The correct answer is that gravitational force is directly proportional to the masses of two bodies. This means that as the masses of the two objects increase, the gravitational force between them also increases. The greater the mass, the stronger the gravitational force.
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• 3.
### Unit of G is ?
• A.
N m2 kg2
• B.
N m/kg
• C.
N m2 / kg2
• D.
N kg2/m2
C. N m2 / kg2
Explanation
The unit of G is N m2 / kg2 because G represents the gravitational constant, which is a physical constant that is used to calculate the force of gravity between two objects. The unit N represents Newton, which is the unit of force. The unit m2 represents square meters, which is the unit of area. The unit kg2 represents kilograms squared, which is the unit of mass squared. Therefore, the correct unit for G is N m2 / kg2.
Rate this question:
• 4.
### Gravitational acceleration does not depend on ?
• A.
Mass of the object
• B.
Weight of the object
• C.
Gravitation constant
• D.
Distance from earth surface
A. Mass of the object
Explanation
Gravitational acceleration does not depend on the mass of the object. This is because gravitational acceleration is a constant value for a given location and is independent of the mass of the object experiencing the gravitational force. The force of gravity experienced by an object depends on its mass, but the acceleration due to gravity remains the same. This means that objects of different masses will fall at the same rate in a vacuum near the Earth's surface.
Rate this question:
• 5.
### Where is the value of Gravitational acceleration is minimum ?
• A.
Alps - Europe
• B.
Equator
• C.
Poles
• D.
Tropic of cancer
• E.
Tropic of capricorn
B. Equator
Explanation
The value of gravitational acceleration is minimum at the Equator. This is because the Earth is not a perfect sphere, but rather an oblate spheroid, with the equatorial diameter being larger than the polar diameter. As a result, the centrifugal force due to the Earth's rotation is greater at the equator, counteracting some of the gravitational force. Therefore, the net gravitational acceleration is slightly lower at the equator compared to the poles or other latitudes.
Rate this question:
• 6.
### What is the value of Gravitational acceleration on the surface of the moon ?
• A.
9.8 m/s2
• B.
33 cm/s2
• C.
1.608 m/s2
• D.
15 m/s2
C. 1.608 m/s2
Explanation
The value of gravitational acceleration on the surface of the moon is 1.608 m/s2. This value is significantly lower than the gravitational acceleration on Earth, which is 9.8 m/s2. The moon has less mass than Earth, so its gravitational pull is weaker. This means that objects on the moon weigh less and fall more slowly compared to Earth.
Rate this question:
• 7.
### What will be the mass of the object on moon surface whose mass is 60 kg on earths surface
• A.
60 kg
• B.
10 kg
• C.
60 N
• D.
10 N
A. 60 kg
Explanation
The mass of an object remains the same regardless of its location. Therefore, if the mass of the object is 60 kg on Earth's surface, it will still be 60 kg on the moon's surface. The force acting on the object may change due to the difference in gravitational pull, but the mass itself remains constant.
Rate this question:
• 8.
### What is the unit gravitational force ?
• A.
Newton
• B.
Kelvin
• C.
Kg
• D.
Metre
A. Newton
Explanation
The unit gravitational force is measured in Newtons. Newton is the standard unit of force in the International System of Units (SI). It is named after Sir Isaac Newton, who formulated the laws of motion and universal gravitation. The unit represents the force required to accelerate a one-kilogram mass by one meter per second squared. Therefore, Newton is the appropriate unit for measuring gravitational force, as it quantifies the force exerted by gravity on an object.
Rate this question:
• 9.
### What is the weight of an object on earths surfacewhich weighs 60 N on moon's surface ?
• A.
360 N
• B.
10 N
• C.
60 kg
• D.
2 kg
A. 360 N
Explanation
The weight of an object on Earth's surface is greater than its weight on the moon's surface due to the difference in gravitational force between the two celestial bodies. Since the object weighs 60 N on the moon, it must weigh more on Earth. Therefore, the correct answer is 360 N.
Rate this question:
• 10.
### Weight depends on which two factors ?
• A.
Mass and energy
• B.
Mass and gravitation constant
• C.
Mass and Gravitational acceleration
• D.
Above a, b,c
• E.
None
C. Mass and Gravitational acceleration
Explanation
The weight of an object depends on two factors: mass and gravitational acceleration. Mass refers to the amount of matter in an object, while gravitational acceleration refers to the acceleration due to gravity. The weight of an object is calculated by multiplying its mass by the gravitational acceleration. Therefore, both mass and gravitational acceleration play a crucial role in determining the weight of an object.
Rate this question:
Quiz Review Timeline +
Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.
• Current Version
• Mar 22, 2023
Quiz Edited by
ProProfs Editorial Team
• Mar 12, 2010
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# Gravitation Quiz. Approved & Edited by ProProfs Editorial Team. The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes.. Learn about Our Editorial Process. | By Jenil3988. J. Jenil3988. Community Contributor. Quizzes Created: 1 | Total Attempts: 6,983. Questions: 10 | Attempts: 7,019. Settings. This is gravitation quiz.. No searchs from google. • 1.. ### What quantity is gravitational force ?. • A.. Scalar quantity. • B.. Vector quantity. • C.. A and B both. • D.. None. B. Vector quantity. Explanation. Gravitational force is a vector quantity because it has both magnitude and direction. The magnitude of gravitational force is determined by the masses of the objects involved, and the direction is always towards the center of mass of the objects. This means that gravitational force can be represented by an arrow, indicating both the strength and direction of the force. Therefore, the correct answer is "Vector quantity".. Rate this question:. • 2.. ### Which one of the following is true ?. • A.. Gravitational force is indirectly proportional to masses of two objects. • B.. Gravitational force is directly proportional to square of distance between two bodies. • C.. Gravitational force is directly proportional to masses of two bodies. • D.. None. • E.. A , b , c all. C. Gravitational force is directly proportional to masses of two bodies. Explanation. The correct answer is that gravitational force is directly proportional to the masses of two bodies. This means that as the masses of the two objects increase, the gravitational force between them also increases. The greater the mass, the stronger the gravitational force.. Rate this question:. • 3.. ### Unit of G is ?. • A.. N m2 kg2. • B.. N m/kg. • C.. N m2 / kg2. • D.. N kg2/m2. C. N m2 / kg2. Explanation. The unit of G is N m2 / kg2 because G represents the gravitational constant, which is a physical constant that is used to calculate the force of gravity between two objects. The unit N represents Newton, which is the unit of force. The unit m2 represents square meters, which is the unit of area. The unit kg2 represents kilograms squared, which is the unit of mass squared. Therefore, the correct unit for G is N m2 / kg2.. Rate this question:. • 4.. ### Gravitational acceleration does not depend on ?. • A.. Mass of the object. • B.. Weight of the object. • C.. Gravitation constant. • D.. Distance from earth surface. A. Mass of the object. Explanation. Gravitational acceleration does not depend on the mass of the object. This is because gravitational acceleration is a constant value for a given location and is independent of the mass of the object experiencing the gravitational force. The force of gravity experienced by an object depends on its mass, but the acceleration due to gravity remains the same. This means that objects of different masses will fall at the same rate in a vacuum near the Earth's surface.. Rate this question:. • 5.. ### Where is the value of Gravitational acceleration is minimum ?. • A.. Alps - Europe. • B.. Equator. • C.. Poles. • D.. Tropic of cancer. • E.. Tropic of capricorn. B. Equator. Explanation. The value of gravitational acceleration is minimum at the Equator.
|
This is because the Earth is not a perfect sphere, but rather an oblate spheroid, with the equatorial diameter being larger than the polar diameter. As a result, the centrifugal force due to the Earth's rotation is greater at the equator, counteracting some of the gravitational force. Therefore, the net gravitational acceleration is slightly lower at the equator compared to the poles or other latitudes.. Rate this question:. • 6.. ### What is the value of Gravitational acceleration on the surface of the moon ?. • A.. 9.8 m/s2. • B.. 33 cm/s2. • C.. 1.608 m/s2. • D.. 15 m/s2. C. 1.608 m/s2. Explanation. The value of gravitational acceleration on the surface of the moon is 1.608 m/s2. This value is significantly lower than the gravitational acceleration on Earth, which is 9.8 m/s2. The moon has less mass than Earth, so its gravitational pull is weaker. This means that objects on the moon weigh less and fall more slowly compared to Earth.. Rate this question:. • 7.. ### What will be the mass of the object on moon surface whose mass is 60 kg on earths surface. • A.. 60 kg. • B.. 10 kg. • C.. 60 N. • D.. 10 N. A. 60 kg. Explanation. The mass of an object remains the same regardless of its location. Therefore, if the mass of the object is 60 kg on Earth's surface, it will still be 60 kg on the moon's surface. The force acting on the object may change due to the difference in gravitational pull, but the mass itself remains constant.. Rate this question:. • 8.. ### What is the unit gravitational force ?. • A.. Newton. • B.. Kelvin. • C.. Kg. • D.. Metre. A. Newton. Explanation. The unit gravitational force is measured in Newtons. Newton is the standard unit of force in the International System of Units (SI). It is named after Sir Isaac Newton, who formulated the laws of motion and universal gravitation. The unit represents the force required to accelerate a one-kilogram mass by one meter per second squared. Therefore, Newton is the appropriate unit for measuring gravitational force, as it quantifies the force exerted by gravity on an object.. Rate this question:. • 9.. ### What is the weight of an object on earths surfacewhich weighs 60 N on moon's surface ?. • A.. 360 N. • B.. 10 N. • C.. 60 kg. • D.. 2 kg. A. 360 N. Explanation. The weight of an object on Earth's surface is greater than its weight on the moon's surface due to the difference in gravitational force between the two celestial bodies. Since the object weighs 60 N on the moon, it must weigh more on Earth. Therefore, the correct answer is 360 N.. Rate this question:. • 10.. ### Weight depends on which two factors ?. • A.. Mass and energy. • B.. Mass and gravitation constant. • C.. Mass and Gravitational acceleration. • D.. Above a, b,c. • E.. None. C. Mass and Gravitational acceleration. Explanation. The weight of an object depends on two factors: mass and gravitational acceleration. Mass refers to the amount of matter in an object, while gravitational acceleration refers to the acceleration due to gravity. The weight of an object is calculated by multiplying its mass by the gravitational acceleration. Therefore, both mass and gravitational acceleration play a crucial role in determining the weight of an object.. Rate this question:. Quiz Review Timeline +. Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.. • Current Version. • Mar 22, 2023. Quiz Edited by. ProProfs Editorial Team. • Mar 12, 2010. Quiz Created by. Jenil3988. Related Topics.
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https://www.booksamillion.com/p/Trigonometry-Essentials-Practice-Workbook-Answers/Chris-McMullen-Ph-D/9781477497784
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## Trigonometry Essentials Practice Workbook with Answers : Master Basic Trig Skills: Improve Your Math Fluency Series by Chris McMullen
Overview -
AUTHOR Chris McMullen earned his Ph.D. in physics from Oklahoma State University and currently teaches physics at Northwestern State University of Louisiana. He developed the Improve Your Math Fluency series of workbooks to help students become more fluent in basic math skills.
WHAT TO EXPECT This is a workbook designed to offer plenty of practice with essential skills. It is not a textbook designed to teach trigonometry, but a workbook designed to supplement a student's instruction in trigonometry. Examples and a brief description of the concepts are included to serve as a quick refresher and a guide. If you need more instruction, you should use this workbook in combination with a textbook. The last chapter involves applications of trig identities, which is a challenging topic that will mostly interest more advanced students. A few chapters are intended to help students memorize the trig functions of common angles like 30, 150, or 315 degrees, which will be especially useful for students who may take exams without a calculator (that's the case with the MCAT and many math and science courses). There may be more practice than many students need, which is much better than having too little practice. Any extra pages may be helpful for teachers or parents with multiple children.
DESCRIPTION This Trigonometry Essentials Practice Workbook with Answers provides ample practice for developing fluency in very fundamental trigonometry skills. Every problem can be answered without a calculator, which is very helpful for students who aren't allowed to use a calculator. This is the case in some trig and physics courses, as well as some standardized exams (like the MCAT).
CONTENTS This workbook is conveniently divided up into 11 chapters so that students can focus on one trigonometry skill at a time. Skills include the following:
• converting between degrees and radians;
• expressing sine, cosine, tangent, secant, cosecant, and cotangent as fractions by looking at right triangles;
• solving for unknown sides and angles in 45 -45 -90 and 30 -60 -90 right triangles;
• determining the sine, cosine, tangent, secant, cosecant, and cotangent of multiples of 30 and 45 up to 360 (working with both degrees and radians);
• practice finding the reference angle for angles in Quadrants II, III, and IV;
• finding the inverse trig functions;
• applying the law of sines and the law of cosines to solve for unknown sides and angles in acute and obtuse triangles;
• solving problems with trig identities (like the angle sum and difference formulas);
• and solving algebraic equations that feature basic trig functions.
EXAMPLES Each section begins with a few pages of instructions for how to solve the problems followed by some examples. These examples should serve as a useful guide until students are able to solve the problems independently.
ANSWERS Answers to exercises are tabulated at the back of the book. This helps students develop confidence and ensures that students practice correct techniques, rather than practice making mistakes.
PHOTOCOPIES The copyright notice permits parents/teachers who purchase one copy or borrow one copy from a library to make photocopies for their own children/students only. This is very convenient if you have multiple children/students or if a child/student needs additional practice.
INTRODUCTION An introduction describes how parents and teachers can help students make the most of this workbook. Students are encouraged to time and score each page. In this way, they can try to have fun improving on their records, which can help lend them confidence in their math skills.
local_shippingFor Delivery
In Stock.
This item is Non-Returnable.
FREE Shipping for Club Members help
New & Used Marketplace 11 copies from \$6.91
## More About Trigonometry Essentials Practice Workbook with Answers by Chris McMullen
### Overview
AUTHOR Chris McMullen earned his Ph.D. in physics from Oklahoma State University and currently teaches physics at Northwestern State University of Louisiana. He developed the Improve Your Math Fluency series of workbooks to help students become more fluent in basic math skills.
WHAT TO EXPECT This is a workbook designed to offer plenty of practice with essential skills. It is not a textbook designed to teach trigonometry, but a workbook designed to supplement a student's instruction in trigonometry. Examples and a brief description of the concepts are included to serve as a quick refresher and a guide. If you need more instruction, you should use this workbook in combination with a textbook. The last chapter involves applications of trig identities, which is a challenging topic that will mostly interest more advanced students. A few chapters are intended to help students memorize the trig functions of common angles like 30, 150, or 315 degrees, which will be especially useful for students who may take exams without a calculator (that's the case with the MCAT and many math and science courses). There may be more practice than many students need, which is much better than having too little practice. Any extra pages may be helpful for teachers or parents with multiple children.
DESCRIPTION This Trigonometry Essentials Practice Workbook with Answers provides ample practice for developing fluency in very fundamental trigonometry skills. Every problem can be answered without a calculator, which is very helpful for students who aren't allowed to use a calculator. This is the case in some trig and physics courses, as well as some standardized exams (like the MCAT).
CONTENTS This workbook is conveniently divided up into 11 chapters so that students can focus on one trigonometry skill at a time. Skills include the following:
• converting between degrees and radians;
• expressing sine, cosine, tangent, secant, cosecant, and cotangent as fractions by looking at right triangles;
• solving for unknown sides and angles in 45 -45 -90 and 30 -60 -90 right triangles;
• determining the sine, cosine, tangent, secant, cosecant, and cotangent of multiples of 30 and 45 up to 360 (working with both degrees and radians);
• practice finding the reference angle for angles in Quadrants II, III, and IV;
• finding the inverse trig functions;
• applying the law of sines and the law of cosines to solve for unknown sides and angles in acute and obtuse triangles;
• solving problems with trig identities (like the angle sum and difference formulas);
• and solving algebraic equations that feature basic trig functions.
EXAMPLES Each section begins with a few pages of instructions for how to solve the problems followed by some examples. These examples should serve as a useful guide until students are able to solve the problems independently.
ANSWERS Answers to exercises are tabulated at the back of the book. This helps students develop confidence and ensures that students practice correct techniques, rather than practice making mistakes.
PHOTOCOPIES The copyright notice permits parents/teachers who purchase one copy or borrow one copy from a library to make photocopies for their own children/students only. This is very convenient if you have multiple children/students or if a child/student needs additional practice.
INTRODUCTION An introduction describes how parents and teachers can help students make the most of this workbook. Students are encouraged to time and score each page. In this way, they can try to have fun improving on their records, which can help lend them confidence in their math skills.
This item is Non-Returnable.
### Details
• ISBN-13: 9781477497784
• ISBN-10: 1477497781
• Publisher: Createspace Independent Publishing Platform
• Publish Date: May 2012
• Page Count: 186
• Dimensions: 10 x 7.99 x 0.4 inches
• Shipping Weight: 0.84 pounds
Related Categories
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## Trigonometry Essentials Practice Workbook with Answers : Master Basic Trig Skills: Improve Your Math Fluency Series by Chris McMullen. Overview -. AUTHOR Chris McMullen earned his Ph.D. in physics from Oklahoma State University and currently teaches physics at Northwestern State University of Louisiana. He developed the Improve Your Math Fluency series of workbooks to help students become more fluent in basic math skills.. WHAT TO EXPECT This is a workbook designed to offer plenty of practice with essential skills. It is not a textbook designed to teach trigonometry, but a workbook designed to supplement a student's instruction in trigonometry. Examples and a brief description of the concepts are included to serve as a quick refresher and a guide. If you need more instruction, you should use this workbook in combination with a textbook. The last chapter involves applications of trig identities, which is a challenging topic that will mostly interest more advanced students. A few chapters are intended to help students memorize the trig functions of common angles like 30, 150, or 315 degrees, which will be especially useful for students who may take exams without a calculator (that's the case with the MCAT and many math and science courses). There may be more practice than many students need, which is much better than having too little practice. Any extra pages may be helpful for teachers or parents with multiple children.. DESCRIPTION This Trigonometry Essentials Practice Workbook with Answers provides ample practice for developing fluency in very fundamental trigonometry skills. Every problem can be answered without a calculator, which is very helpful for students who aren't allowed to use a calculator. This is the case in some trig and physics courses, as well as some standardized exams (like the MCAT).. CONTENTS This workbook is conveniently divided up into 11 chapters so that students can focus on one trigonometry skill at a time. Skills include the following:. • converting between degrees and radians;. • expressing sine, cosine, tangent, secant, cosecant, and cotangent as fractions by looking at right triangles;. • solving for unknown sides and angles in 45 -45 -90 and 30 -60 -90 right triangles;. • determining the sine, cosine, tangent, secant, cosecant, and cotangent of multiples of 30 and 45 up to 360 (working with both degrees and radians);. • practice finding the reference angle for angles in Quadrants II, III, and IV;. • finding the inverse trig functions;. • applying the law of sines and the law of cosines to solve for unknown sides and angles in acute and obtuse triangles;. • solving problems with trig identities (like the angle sum and difference formulas);. • and solving algebraic equations that feature basic trig functions.. EXAMPLES Each section begins with a few pages of instructions for how to solve the problems followed by some examples. These examples should serve as a useful guide until students are able to solve the problems independently.. ANSWERS Answers to exercises are tabulated at the back of the book. This helps students develop confidence and ensures that students practice correct techniques, rather than practice making mistakes.. PHOTOCOPIES The copyright notice permits parents/teachers who purchase one copy or borrow one copy from a library to make photocopies for their own children/students only. This is very convenient if you have multiple children/students or if a child/student needs additional practice.. INTRODUCTION An introduction describes how parents and teachers can help students make the most of this workbook. Students are encouraged to time and score each page. In this way, they can try to have fun improving on their records, which can help lend them confidence in their math skills.. local_shippingFor Delivery. In Stock.. This item is Non-Returnable.. FREE Shipping for Club Members help. New & Used Marketplace 11 copies from \$6.91. ## More About Trigonometry Essentials Practice Workbook with Answers by Chris McMullen. ### Overview. AUTHOR Chris McMullen earned his Ph.D.
|
in physics from Oklahoma State University and currently teaches physics at Northwestern State University of Louisiana. He developed the Improve Your Math Fluency series of workbooks to help students become more fluent in basic math skills.. WHAT TO EXPECT This is a workbook designed to offer plenty of practice with essential skills. It is not a textbook designed to teach trigonometry, but a workbook designed to supplement a student's instruction in trigonometry. Examples and a brief description of the concepts are included to serve as a quick refresher and a guide. If you need more instruction, you should use this workbook in combination with a textbook. The last chapter involves applications of trig identities, which is a challenging topic that will mostly interest more advanced students. A few chapters are intended to help students memorize the trig functions of common angles like 30, 150, or 315 degrees, which will be especially useful for students who may take exams without a calculator (that's the case with the MCAT and many math and science courses). There may be more practice than many students need, which is much better than having too little practice. Any extra pages may be helpful for teachers or parents with multiple children.. DESCRIPTION This Trigonometry Essentials Practice Workbook with Answers provides ample practice for developing fluency in very fundamental trigonometry skills. Every problem can be answered without a calculator, which is very helpful for students who aren't allowed to use a calculator. This is the case in some trig and physics courses, as well as some standardized exams (like the MCAT).. CONTENTS This workbook is conveniently divided up into 11 chapters so that students can focus on one trigonometry skill at a time. Skills include the following:. • converting between degrees and radians;. • expressing sine, cosine, tangent, secant, cosecant, and cotangent as fractions by looking at right triangles;. • solving for unknown sides and angles in 45 -45 -90 and 30 -60 -90 right triangles;. • determining the sine, cosine, tangent, secant, cosecant, and cotangent of multiples of 30 and 45 up to 360 (working with both degrees and radians);. • practice finding the reference angle for angles in Quadrants II, III, and IV;. • finding the inverse trig functions;. • applying the law of sines and the law of cosines to solve for unknown sides and angles in acute and obtuse triangles;. • solving problems with trig identities (like the angle sum and difference formulas);. • and solving algebraic equations that feature basic trig functions.. EXAMPLES Each section begins with a few pages of instructions for how to solve the problems followed by some examples. These examples should serve as a useful guide until students are able to solve the problems independently.. ANSWERS Answers to exercises are tabulated at the back of the book. This helps students develop confidence and ensures that students practice correct techniques, rather than practice making mistakes.. PHOTOCOPIES The copyright notice permits parents/teachers who purchase one copy or borrow one copy from a library to make photocopies for their own children/students only. This is very convenient if you have multiple children/students or if a child/student needs additional practice.. INTRODUCTION An introduction describes how parents and teachers can help students make the most of this workbook. Students are encouraged to time and score each page. In this way, they can try to have fun improving on their records, which can help lend them confidence in their math skills.. This item is Non-Returnable.. ### Details. • ISBN-13: 9781477497784. • ISBN-10: 1477497781. • Publisher: Createspace Independent Publishing Platform. • Publish Date: May 2012. • Page Count: 186. • Dimensions: 10 x 7.99 x 0.4 inches. • Shipping Weight: 0.84 pounds. Related Categories.
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Attribute VB_Name = "Module2" '*************************************************************************** '* THE BOUNCING BALL * '* ----------------------------------------------------------------------- * '* Explanations: * '* * '* Let us consider an elastic ball dropped with a horizontal speed Vx from * '* a height H. Here are the main equations to calculate the successive * '* rebounds on the ground: * '* * '* Initializing: * '* The first part of the trajectory is a parabolic section, upwards orien- * '* ted. From the physical laws for a falling body, we know that: * '* Vy = sqrt(2GH) and V0 = sqrt(Vx² + Vy²) with theta=arctan(Vx/Vy), * '* where V0 is the speed at ground level and theta = pi - angle of V0 with * '* the Ox axis. * '* The ball moves with the horizontal speed Vx and the vertical speed -Vy * '* during time t=Vy/G until it reaches the ground (G=gravity acceleration).* '* Hence the horizontal distance run is x2=Vx*t=Vx*Vy/G. * '* The equation of the parabolic section is: y = -(H/x2²)*x + H from x=0 * '* to x=x2. * '* * '* Rebounds: * '* For each rebound, the initial speed V0 is multiplied by a damping * '* coefficient, called amort (<1), but the theta initial angle is kept the * '* same. The previous x2 value becomes the next x1 value. The parabolic * '* equation of the next trajectory section has the complete form: * '* y = a*x² + b*x + c (1) * '* P=2*Vx*Vy/G is the correct length for all the rebounds. Hence the next * '* x2 = x1 + P. The parabola summit coordinates are the following: * '* xs = (x1+x2)/2 and ys = (V0²*sin²(theta))/(2G) * '* So to draw the trajectory section of a rebound, we have to know the * '* equation of a parabola passing through the 3 points: (x1,0), (x2,0) and * '* (xs,ys). This leads to the following linear system to solve of order 3: * '* a*x1² + b*x1 + c = 0 * '* a*xs² + b*xs + c = 0 * '* a*x2² + b*x2 + c = 0 * '* where the unknowns are a, b, c, the coefficients of the parabola (1). * '* Using the Cramer's method, we calculate the main determinant Dp and the * '* three sub-determinants, Da, Db, Dc. We find: * '* Dp = x1²*(xs-x2) + x2²*(x1-xs) + xs²*(x2-x1) * '* Da = ys*(x2-x1) Db = ys*(x1²-x2²) Dc = ys*x1*x2*(x2-x1) * '* Hence: a = Da/Dp b = Db/Dp c = Dc/Dp * '* Now we can draw the parabolic section of the next rebound in [x1,x2]. * '* We continue the same process until the required number of rebounds is * '* reached. * '* ----------------------------------------------------------------------- * '* Reference: * '* From "Graphisme dans le plan et dans l'espace avec Turbo Pascal 4.0 de * '* R. Dony - MASSON 1990 page 113" [BIBLI 12]. * '* * '* Visual Basic Release By J-P Moreau, Paris. * '* (www.jpmoreau.fr) * '*************************************************************************** 'Program Rebounds DefInt I-N Const Height = 12 Const G = 9.81 Const xstep = 0.005 Dim Length As Single Dim bounds As Integer Public nrebounds As Integer Public amort As Single Dim x1 As Single Dim x2 As Single Public vx As Single Dim vy As Single Dim v0 As Single Dim angle As Single Dim a As Single Dim b As Single Dim c As Single Public X(10), Y(10) 'dummy arguments for Polygon Sub Data(vx, amort, nrebounds) Form1.Cls vx = Val(Form2.Text1): amort = Val(Form2.Text2) nrebounds = Val(Form2.Text3) End Sub Sub Init() bounds = 0 x1 = 0 vy = Sqr(2 * G * Height) angle = Atn(vy / vx) v0 = Sqr(vx ^ 2 + vy ^ 2) x2 = vx * vy / G a = -Height / x2 ^ 2 b = 0 c = Height End Sub Sub DrawParabol(pt1, pt2) Dim X As Single Dim Y As Single X = pt1 MoveXY X, 0# While X < pt2 Y = a * X ^ 2 + b * X + c LineXY X, Y X = X + xstep Wend bounds = bounds + 1 End Sub Sub CoeffParabol() ys = (v0 * Sin(angle)) ^ 2 / (2 * G) vy = Sqr(2 * G * ys) Length = 2 * vx * vy / G x2 = x1 + Length xs = (x1 + x2) / 2 detprinc = x1 ^ 2 * (xs - x2) + x2 ^ 2 * (x1 - xs) + xs ^ 2 * (x2 - x1) deta = ys * (x2 - x1) a = deta / detprinc '1st coefficient of parabola detb = ys * (x1 ^ 2 - x2 ^ 2) b = detb / detprinc '2nd coefficient detc = ys * x1 * x2 * (x2 - x1) c = detc / detprinc '3rd coefficient End Sub Sub PrintData() Dim s1 As String Dim s2 As String Dim s3 As String Dim s4 As String Dim ch As String s1 = Str\$(Height) s2 = Str\$(vx) s3 = Str\$(x2) s4 = Str\$(amort) ch = "Height=" + s1 + " Speed=" + s2 + " Total length=" + s3 + " Damping=" + s4 Display 700, 60, ch End Sub Sub Exec_Rebounds() 'Data vx, amort, nrebounds Init Form1.Cls Fenetre 0#, 20, 0#, Height Cloture 400, MaxX - 300, 1800, MaxY - 400 Axes Grid 1#, 1# Graduate 2#, 2# Bordure DrawParabol x1, x2 x1 = x2 While bounds < nrebounds v0 = v0 * amort CoeffParabol DrawParabol x1, x2 x1 = x2 Wend PrintData End Sub
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Attribute VB_Name = "Module2" '*************************************************************************** '* THE BOUNCING BALL * '* ----------------------------------------------------------------------- * '* Explanations: * '* * '* Let us consider an elastic ball dropped with a horizontal speed Vx from * '* a height H. Here are the main equations to calculate the successive * '* rebounds on the ground: * '* * '* Initializing: * '* The first part of the trajectory is a parabolic section, upwards orien- * '* ted. From the physical laws for a falling body, we know that: * '* Vy = sqrt(2GH) and V0 = sqrt(Vx² + Vy²) with theta=arctan(Vx/Vy), * '* where V0 is the speed at ground level and theta = pi - angle of V0 with * '* the Ox axis. * '* The ball moves with the horizontal speed Vx and the vertical speed -Vy * '* during time t=Vy/G until it reaches the ground (G=gravity acceleration).* '* Hence the horizontal distance run is x2=Vx*t=Vx*Vy/G. * '* The equation of the parabolic section is: y = -(H/x2²)*x + H from x=0 * '* to x=x2. * '* * '* Rebounds: * '* For each rebound, the initial speed V0 is multiplied by a damping * '* coefficient, called amort (<1), but the theta initial angle is kept the * '* same. The previous x2 value becomes the next x1 value. The parabolic * '* equation of the next trajectory section has the complete form: * '* y = a*x² + b*x + c (1) * '* P=2*Vx*Vy/G is the correct length for all the rebounds.
|
Hence the next * '* x2 = x1 + P. The parabola summit coordinates are the following: * '* xs = (x1+x2)/2 and ys = (V0²*sin²(theta))/(2G) * '* So to draw the trajectory section of a rebound, we have to know the * '* equation of a parabola passing through the 3 points: (x1,0), (x2,0) and * '* (xs,ys). This leads to the following linear system to solve of order 3: * '* a*x1² + b*x1 + c = 0 * '* a*xs² + b*xs + c = 0 * '* a*x2² + b*x2 + c = 0 * '* where the unknowns are a, b, c, the coefficients of the parabola (1). * '* Using the Cramer's method, we calculate the main determinant Dp and the * '* three sub-determinants, Da, Db, Dc. We find: * '* Dp = x1²*(xs-x2) + x2²*(x1-xs) + xs²*(x2-x1) * '* Da = ys*(x2-x1) Db = ys*(x1²-x2²) Dc = ys*x1*x2*(x2-x1) * '* Hence: a = Da/Dp b = Db/Dp c = Dc/Dp * '* Now we can draw the parabolic section of the next rebound in [x1,x2]. * '* We continue the same process until the required number of rebounds is * '* reached. * '* ----------------------------------------------------------------------- * '* Reference: * '* From "Graphisme dans le plan et dans l'espace avec Turbo Pascal 4.0 de * '* R. Dony - MASSON 1990 page 113" [BIBLI 12]. * '* * '* Visual Basic Release By J-P Moreau, Paris. * '* (www.jpmoreau.fr) * '*************************************************************************** 'Program Rebounds DefInt I-N Const Height = 12 Const G = 9.81 Const xstep = 0.005 Dim Length As Single Dim bounds As Integer Public nrebounds As Integer Public amort As Single Dim x1 As Single Dim x2 As Single Public vx As Single Dim vy As Single Dim v0 As Single Dim angle As Single Dim a As Single Dim b As Single Dim c As Single Public X(10), Y(10) 'dummy arguments for Polygon Sub Data(vx, amort, nrebounds) Form1.Cls vx = Val(Form2.Text1): amort = Val(Form2.Text2) nrebounds = Val(Form2.Text3) End Sub Sub Init() bounds = 0 x1 = 0 vy = Sqr(2 * G * Height) angle = Atn(vy / vx) v0 = Sqr(vx ^ 2 + vy ^ 2) x2 = vx * vy / G a = -Height / x2 ^ 2 b = 0 c = Height End Sub Sub DrawParabol(pt1, pt2) Dim X As Single Dim Y As Single X = pt1 MoveXY X, 0# While X < pt2 Y = a * X ^ 2 + b * X + c LineXY X, Y X = X + xstep Wend bounds = bounds + 1 End Sub Sub CoeffParabol() ys = (v0 * Sin(angle)) ^ 2 / (2 * G) vy = Sqr(2 * G * ys) Length = 2 * vx * vy / G x2 = x1 + Length xs = (x1 + x2) / 2 detprinc = x1 ^ 2 * (xs - x2) + x2 ^ 2 * (x1 - xs) + xs ^ 2 * (x2 - x1) deta = ys * (x2 - x1) a = deta / detprinc '1st coefficient of parabola detb = ys * (x1 ^ 2 - x2 ^ 2) b = detb / detprinc '2nd coefficient detc = ys * x1 * x2 * (x2 - x1) c = detc / detprinc '3rd coefficient End Sub Sub PrintData() Dim s1 As String Dim s2 As String Dim s3 As String Dim s4 As String Dim ch As String s1 = Str\$(Height) s2 = Str\$(vx) s3 = Str\$(x2) s4 = Str\$(amort) ch = "Height=" + s1 + " Speed=" + s2 + " Total length=" + s3 + " Damping=" + s4 Display 700, 60, ch End Sub Sub Exec_Rebounds() 'Data vx, amort, nrebounds Init Form1.Cls Fenetre 0#, 20, 0#, Height Cloture 400, MaxX - 300, 1800, MaxY - 400 Axes Grid 1#, 1# Graduate 2#, 2# Bordure DrawParabol x1, x2 x1 = x2 While bounds < nrebounds v0 = v0 * amort CoeffParabol DrawParabol x1, x2 x1 = x2 Wend PrintData End Sub.
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# Pre calculus help
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# Pre calculus help. Best of all, Pre calculus help is free to use, so there's no reason not to give it a try! We will give you answers to homework.. ## The Best Pre calculus help. This Pre calculus help supplies step-by-step instructions for solving all math troubles. This question examines the solution of irregular figure area, with medium difficulty. Solving the area of irregular figures by cutting or complementing is a common question type in IMIS, which must be mastered. Finally, there are three special segments of the triangle, namely the centerline, bisector and height. These three line segments play an important role in triangles. The middle line of a triangle.. In short, this means that some independent third-party providers host the application and provide it to users online without installation. Examples of SaaS applications include image editing tools, e-mail, office applications, calendars, and so on. Application information service refers to the activity of providing users with text, picture, voice, video and other information production, copying, publishing, dissemination and other services through the application, including instant messaging, news information, Knowledge Q & A, forum community, online live broadcasting, e-commerce, online audio and video, life services and other types. This definition is very broad, covering almost all types of services that may be provided based on applications.. This is the power of dimension reduction strike. 3. When doing big math problems, don't be too nervous.
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Just get all the basic scores, and you can get a good score.. The shared area covers a wide range of areas, and its cancellation is not a day's work. Regular geometry solution sets input parameters according to some features of the drawing, and calculates the area and perimeter of the corresponding drawing according to these parameters. In drawing a graph, the coordinates are determined according to the input parameters, and then the coordinates are connected to form a graph. In terms of changing graphics, use the type of drawing to change graphics. In 2018, the Xinhua Daily Telegraph published the contradiction of public sharing hurts the people which needs to be solved.. ## We solve all types of math troubles. This app helps me with every sum that I struggle with. The app explains everything step by step. But it uses data so unfortunately, I don't get to use it all the time to help me with my studies. Alana Powell. I like it if I need to get homework done fast, I know it’s kind of cheating but it tells you the steps to solve the problem then after you catch on to the steps you don't need it This app is very smart in all mathematical expressions. I'm madly in love with this app.. Ireland Rogers. Math tutor online chat Check my math Math photo answer Maths man expert Math homework answers free Pay someone to do your math homework.
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https://www.studypool.com/documents/2234369/1-find-the-expected-return-on-stock-a2-find-the-expected-return-o
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search
# 1 Find the expected return on stock A2 Find the expected return o
Content type
User Generated
Rating
Showing Page:
1/2
1.Find the expected return on stock A
2. Find the expected return on stock B
3. If the T-bill rate is 2 percent and market risk premium is
6 percent, what does CAPM say about the fair expected
rate of return on stock A? Assume that the beta of A is 1.5,
and beta of B is 1.1.
4. If the T-bill rate is 2 percent and market risk premium is
6 percent, what does CAPM say about the fair expected
rate of return on Stock B? Assume that the beta of A is
1.5, and beta of B is 1.1.
Solution
Expected Return = Returns xprobability
1. Expected Return on Stock A = (-5*0.4) + (15*0.4) +
(32*0.2)
= 10.4%
2. Expected Return on Stock B = (1*0.4) +(8*0.4) +
(20*0.2)
= 7.6%
3. Expected Rate of Return using CAPM = Risk free rate +
Beta x (Market risk premium)
Fair Expected Return on Stock A = 2 + 1.5 ( 6) = 11%
4. Fair Expected Return on Stock B = 2 + 1.1 ( 6) = 8.6%
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1.Find the expected return on stock A 2. Find the expected return on stock B 3. If the T-bill rate is 2 percent and market risk premium is 6 percent, what does CAPM say about the fair expected rate of return on stock A? Assume that the beta of A is 1.5, and beta of B is 1.1. 4. If the T-bill rate is 2 percent and market risk premium is 6 percent, what does CAPM say about the fair expected rate of return on Stock B? Assume that the beta of A is 1.5, and beta of B is 1.1. Solution Expected Return = Returns xprobability 1. Expected Return on Stock A = ( -5*0.4) + (15*0.4) + (32*0.2) = 10.4% 2. ...
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### Review
Anonymous
This is great! Exactly what I wanted.
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search. # 1 Find the expected return on stock A2 Find the expected return o. Content type. User Generated. Rating. Showing Page:. 1/2. 1.Find the expected return on stock A. 2. Find the expected return on stock B. 3. If the T-bill rate is 2 percent and market risk premium is. 6 percent, what does CAPM say about the fair expected. rate of return on stock A? Assume that the beta of A is 1.5,. and beta of B is 1.1.. 4. If the T-bill rate is 2 percent and market risk premium is. 6 percent, what does CAPM say about the fair expected. rate of return on Stock B? Assume that the beta of A is. 1.5, and beta of B is 1.1.. Solution. Expected Return = Returns xprobability. 1. Expected Return on Stock A = (-5*0.4) + (15*0.4) +. (32*0.2). = 10.4%. 2. Expected Return on Stock B = (1*0.4) +(8*0.4) +. (20*0.2). = 7.6%. 3.
|
Expected Rate of Return using CAPM = Risk free rate +. Beta x (Market risk premium). Fair Expected Return on Stock A = 2 + 1.5 ( 6) = 11%. 4. Fair Expected Return on Stock B = 2 + 1.1 ( 6) = 8.6%. Sign up to view the full document!. Showing Page:. 2/2. Sign up to view the full document!. Unformatted Attachment Preview. 1.Find the expected return on stock A 2. Find the expected return on stock B 3. If the T-bill rate is 2 percent and market risk premium is 6 percent, what does CAPM say about the fair expected rate of return on stock A? Assume that the beta of A is 1.5, and beta of B is 1.1. 4. If the T-bill rate is 2 percent and market risk premium is 6 percent, what does CAPM say about the fair expected rate of return on Stock B? Assume that the beta of A is 1.5, and beta of B is 1.1. Solution Expected Return = Returns xprobability 1. Expected Return on Stock A = ( -5*0.4) + (15*0.4) + (32*0.2) = 10.4% 2. .... Purchase document to see full attachment. User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.. ### Review. Anonymous. This is great! Exactly what I wanted.. Studypool. 4.7. Trustpilot. 4.5. Sitejabber. 4.4.
|
https://www.educator.com/mathematics/differential-equations/murray/separation-of-variables.php
| 1,725,921,597,000,000,000
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Professor Murray
Separation of Variables
Slide Duration:
Section 1: First-Order Equations
Linear Equations
1h 7m 21s
Intro
0:00
Lesson Objectives
0:19
How to Solve Linear Equations
2:54
Calculate the Integrating Factor
2:58
Changes the Left Side so We Can Integrate Both Sides
3:27
Solving Linear Equations
5:32
Further Notes
6:10
If P(x) is Negative
6:26
Leave Off the Constant
9:38
The C Is Important When Integrating Both Sides of the Equation
9:55
Example 1
10:29
Example 2
22:56
Example 3
36:12
Example 4
39:24
Example 5
44:10
Example 6
56:42
Separable Equations
35m 11s
Intro
0:00
Lesson Objectives
0:19
Some Equations Are Both Linear and Separable So You Can Use Either Technique to Solve Them
1:33
Important to Add C When You Do the Integration
2:27
Example 1
4:28
Example 2
10:45
Example 3
14:43
Example 4
19:21
Example 5
27:23
Slope & Direction Fields
1h 11m 36s
Intro
0:00
Lesson Objectives
0:20
If You Can Manipulate a Differential Equation Into a Certain Form, You Can Draw a Slope Field Also Known as a Direction Field
0:23
How You Do This
0:45
Solution Trajectories
2:49
Never Cross Each Other
3:44
General Solution to the Differential Equation
4:03
Use an Initial Condition to Find Which Solution Trajectory You Want
4:59
Example 1
6:52
Example 2
14:20
Example 3
26:36
Example 4
34:21
Example 5
46:09
Example 6
59:51
Applications, Modeling, & Word Problems of First-Order Equations
1h 5m 19s
Intro
0:00
Lesson Overview
0:38
Mixing
1:00
Population
2:49
Finance
3:22
Set Variables
4:39
Write Differential Equation
6:29
Solve It
10:54
11:47
Example 1
13:29
Example 2
24:53
Example 3
32:13
Example 4
42:46
Example 5
55:05
Autonomous Equations & Phase Plane Analysis
1h 1m 20s
Intro
0:00
Lesson Overview
0:18
Autonomous Differential Equations Have the Form y' = f(x)
0:21
Phase Plane Analysis
0:48
y' < 0
2:56
y' > 0
3:04
If we Perturb the Equilibrium Solutions
5:51
Equilibrium Solutions
7:44
8:06
Solutions Will Tend Away From Unstable Equilibria
9:32
Semistable Equilibria
10:59
Example 1
11:43
Example 2
15:50
Example 3
28:27
Example 4
31:35
Example 5
43:03
Example 6
49:01
Section 2: Second-Order Equations
Distinct Roots of Second Order Equations
28m 44s
Intro
0:00
Lesson Overview
0:36
Linear Means
0:50
Second-Order
1:15
Homogeneous
1:30
Constant Coefficient
1:55
Solve the Characteristic Equation
2:33
Roots r1 and r2
3:43
To Find c1 and c2, Use Initial Conditions
4:50
Example 1
5:46
Example 2
8:20
Example 3
16:20
Example 4
18:26
Example 5
23:52
Complex Roots of Second Order Equations
31m 49s
Intro
0:00
Lesson Overview
0:15
Sometimes The Characteristic Equation Has Complex Roots
1:12
Example 1
3:21
Example 2
7:42
Example 3
15:25
Example 4
18:59
Example 5
27:52
Repeated Roots & Reduction of Order
43m 2s
Intro
0:00
Lesson Overview
0:23
If the Characteristic Equation Has a Double Root
1:46
Reduction of Order
3:10
Example 1
7:23
Example 2
9:20
Example 3
14:12
Example 4
31:49
Example 5
33:21
Undetermined Coefficients of Inhomogeneous Equations
50m 1s
Intro
0:00
Lesson Overview
0:11
Inhomogeneous Equation Means the Right Hand Side is Not 0 Anymore
0:21
First Solve the Homogeneous Equation
1:04
Find a Particular Solution to the Inhomogeneous Equation Using Undetermined Coefficients
2:03
g(t) vs. Guess for ypar
2:42
If Any Term of Your Guess for ypar Looks Like Any Term of yhom
5:07
Example 1
7:54
Example 2
15:25
Example 3
23:45
Example 4
33:35
Example 5
42:57
Inhomogeneous Equations: Variation of Parameters
49m 22s
Intro
0:00
Lesson Overview
0:31
Inhomogeneous vs. Homogeneous
0:47
First Solve the Homogeneous Equation
1:17
Notice There is No Coefficient in Front of y''
1:27
Find a Particular Solution to the Inhomogeneous Equation Using Variation of Parameters
2:32
How to Solve
4:33
Hint on Solving the System
5:23
Example 1
7:27
Example 2
17:46
Example 3
23:14
Example 4
31:49
Example 5
36:00
Section 3: Series Solutions
Review of Power Series
57m 38s
Intro
0:00
Lesson Overview
0:36
Taylor Series Expansion
0:37
Maclaurin Series
2:36
Common Maclaurin Series to Remember From Calculus
3:35
7:58
Ratio Test
12:05
Example 1
15:18
Example 2
20:02
Example 3
27:32
Example 4
39:33
Example 5
45:42
Series Solutions Near an Ordinary Point
1h 20m 28s
Intro
0:00
Lesson Overview
0:49
Guess a Power Series Solution and Calculate Its Derivatives, Example 1
1:03
Guess a Power Series Solution and Calculate Its Derivatives, Example 2
3:14
Combine the Series
5:00
Match Exponents on x By Shifting Indices
5:11
Match Starting Indices By Pulling Out Initial Terms
5:51
Find a Recurrence Relation on the Coefficients
7:09
Example 1
7:46
Example 2
19:10
Example 3
29:57
Example 4
41:46
Example 5
57:23
Example 6
1:09:12
Euler Equations
24m 42s
Intro
0:00
Lesson Overview
0:11
Euler Equation
0:15
Real, Distinct Roots
2:22
Real, Repeated Roots
2:37
Complex Roots
2:49
Example 1
3:51
Example 2
6:20
Example 3
8:27
Example 4
13:04
Example 5
15:31
Example 6
18:31
Series Solutions
1h 26m 17s
Intro
0:00
Lesson Overview
0:13
Singular Point
1:17
Definition: Pole of Order n
1:58
Pole Of Order n
2:04
Regular Singular Point
3:25
Solving Around Regular Singular Points
7:08
Indical Equation
7:30
If the Difference Between the Roots is An Integer
8:06
If the Difference Between the Roots is Not An Integer
8:29
Example 1
8:47
Example 2
14:57
Example 3
25:40
Example 4
47:23
Example 5
1:09:01
Section 4: Laplace Transform
Laplace Transforms
41m 52s
Intro
0:00
Lesson Overview
0:09
Laplace Transform of a Function f(t)
0:18
Laplace Transform is Linear
1:04
Example 1
1:43
Example 2
18:30
Example 3
22:06
Example 4
28:27
Example 5
33:54
Inverse Laplace Transforms
47m 5s
Intro
0:00
Lesson Overview
0:09
Laplace Transform L{f}
0:13
Run Partial Fractions
0:24
Common Laplace Transforms
1:20
Example 1
3:24
Example 2
9:55
Example 3
14:49
Example 4
22:03
Example 5
33:51
Laplace Transform Initial Value Problems
45m 15s
Intro
0:00
Lesson Overview
0:12
0:14
Take the Laplace Transform of Both Sides of the Differential Equation
0:37
Plug in the Identities
1:20
Take the Inverse Laplace Transform to Find y
2:40
Example 1
4:15
Example 2
11:30
Example 3
17:59
Example 4
24:51
Example 5
36:05
Section 5: Review of Linear Algebra
Review of Linear Algebra
57m 30s
Intro
0:00
Lesson Overview
0:41
Matrix
0:54
Determinants
4:45
3x3 Determinants
5:08
Eigenvalues and Eigenvectors
7:01
Eigenvector
7:48
Eigenvalue
7:54
Lesson Overview
8:17
Characteristic Polynomial
8:47
Find Corresponding Eigenvector
9:03
Example 1
10:19
Example 2
16:49
Example 3
20:52
Example 4
25:34
Example 5
35:05
Section 6: Systems of Equations
Distinct Real Eigenvalues
59m 26s
Intro
0:00
Lesson Overview
1:11
How to Solve Systems
2:48
Find the Eigenvalues and Their Corresponding Eigenvectors
2:50
General Solution
4:30
Use Initial Conditions to Find c1 and c2
4:57
Graphing the Solutions
5:20
Solution Trajectories Tend Towards 0 or ∞ Depending on Whether r1 or r2 are Positive or Negative
6:35
Solution Trajectories Tend Towards the Axis Spanned by the Eigenvector Corresponding to the Larger Eigenvalue
7:27
Example 1
9:05
Example 2
21:06
Example 3
26:38
Example 4
36:40
Example 5
43:26
Example 6
51:33
Complex Eigenvalues
1h 3m 54s
Intro
0:00
Lesson Overview
0:47
Recall That to Solve the System of Linear Differential Equations, We find the Eigenvalues and Eigenvectors
0:52
If the Eigenvalues are Complex, Then They Will Occur in Conjugate Pairs
1:13
Expanding Complex Solutions
2:55
Euler's Formula
2:56
Multiply This Into the Eigenvector, and Separate Into Real and Imaginary Parts
1:18
Graphing Solutions From Complex Eigenvalues
5:34
Example 1
9:03
Example 2
20:48
Example 3
28:34
Example 4
41:28
Example 5
51:21
Repeated Eigenvalues
45m 17s
Intro
0:00
Lesson Overview
0:44
If the Characteristic Equation Has a Repeated Root, Then We First Find the Corresponding Eigenvector
1:14
Find the Generalized Eigenvector
1:25
Solutions from Repeated Eigenvalues
2:22
Form the Two Principal Solutions and the Two General Solution
2:23
Use Initial Conditions to Solve for c1 and c2
3:41
Graphing the Solutions
3:53
Example 1
8:10
Example 2
16:24
Example 3
23:25
Example 4
31:04
Example 5
38:17
Section 7: Inhomogeneous Systems
Undetermined Coefficients for Inhomogeneous Systems
43m 37s
Intro
0:00
Lesson Overview
0:35
First Solve the Corresponding Homogeneous System x'=Ax
0:37
Solving the Inhomogeneous System
2:32
Look for a Single Particular Solution xpar to the Inhomogeneous System
2:36
Plug the Guess Into the System and Solve for the Coefficients
3:27
Add the Homogeneous Solution and the Particular Solution to Get the General Solution
3:52
Example 1
4:49
Example 2
9:30
Example 3
15:54
Example 4
20:39
Example 5
29:43
Example 6
37:41
Variation of Parameters for Inhomogeneous Systems
1h 8m 12s
Intro
0:00
Lesson Overview
0:37
Find Two Solutions to the Homogeneous System
2:04
Look for a Single Particular Solution xpar to the inhomogeneous system as follows
2:59
Solutions by Variation of Parameters
3:35
General Solution and Matrix Inversion
6:35
General Solution
6:41
Hint for Finding Ψ-1
6:58
Example 1
8:13
Example 2
16:23
Example 3
32:23
Example 4
37:34
Example 5
49:00
Section 8: Numerical Techniques
Euler's Method
45m 30s
Intro
0:00
Lesson Overview
0:32
Euler's Method is a Way to Find Numerical Approximations for Initial Value Problems That We Cannot Solve Analytically
0:34
Based on Drawing Lines Along Slopes in a Direction Field
1:18
Formulas for Euler's Method
1:57
Example 1
4:47
Example 2
14:45
Example 3
24:03
Example 4
33:01
Example 5
37:55
Runge-Kutta & The Improved Euler Method
41m 4s
Intro
0:00
Lesson Overview
0:43
Runge-Kutta is Know as the Improved Euler Method
0:46
More Sophisticated Than Euler's Method
1:09
It is the Fundamental Algorithm Used in Most Professional Software to Solve Differential Equations
1:16
Order 2 Runge-Kutta Algorithm
1:45
Runge-Kutta Order 2 Algorithm
2:09
Example 1
4:57
Example 2
10:57
Example 3
19:45
Example 4
24:35
Example 5
31:39
Section 9: Partial Differential Equations
Review of Partial Derivatives
38m 22s
Intro
0:00
Lesson Overview
1:04
Partial Derivative of u with respect to x
1:37
Geometrically, ux Represents the Slope As You Walk in the x-direction on the Surface
2:47
Computing Partial Derivatives
3:46
Algebraically, to Find ux You Treat The Other Variable t as a Constant and Take the Derivative with Respect to x
3:49
Second Partial Derivatives
4:16
Clairaut's Theorem Says that the Two 'Mixed Partials' Are Always Equal
5:21
Example 1
5:34
Example 2
7:40
Example 3
11:17
Example 4
14:23
Example 5
31:55
The Heat Equation
44m 40s
Intro
0:00
Lesson Overview
0:28
Partial Differential Equation
0:33
Most Common Ones
1:17
Boundary Value Problem
1:41
Common Partial Differential Equations
3:41
Heat Equation
4:04
Wave Equation
5:44
Laplace's Equation
7:50
Example 1
8:35
Example 2
14:21
Example 3
21:04
Example 4
25:54
Example 5
35:12
Separation of Variables
57m 44s
Intro
0:00
Lesson Overview
0:26
Separation of Variables is a Technique for Solving Some Partial Differential Equations
0:29
Separation of Variables
2:35
Try to Separate the Variables
2:38
If You Can, Then Both Sides Must Be Constant
2:52
Reorganize These Intro Two Ordinary Differential Equations
3:05
Example 1
4:41
Example 2
11:06
Example 3
18:30
Example 4
25:49
Example 5
32:53
Fourier Series
1h 24m 33s
Intro
0:00
Lesson Overview
0:38
Fourier Series
0:42
Find the Fourier Coefficients by the Formulas
2:05
Notes on Fourier Series
3:34
Formula Simplifies
3:35
Function Must be Periodic
4:23
Even and Odd Functions
5:37
Definition
5:45
Examples
6:03
Even and Odd Functions and Fourier Series
9:47
If f is Even
9:52
If f is Odd
11:29
Extending Functions
12:46
If We Want a Cosine Series
14:13
If We Wants a Sine Series
15:20
Example 1
17:39
Example 2
43:23
Example 3
51:14
Example 4
1:01:52
Example 5
1:11:53
Solution of the Heat Equation
47m 41s
Intro
0:00
Lesson Overview
0:22
Solving the Heat Equation
1:03
Procedure for the Heat Equation
3:29
Extend So That its Fourier Series Will Have Only Sines
3:57
Find the Fourier Series for f(x)
4:19
Example 1
5:21
Example 2
8:08
Example 3
17:42
Example 4
25:13
Example 5
28:53
Example 6
42:22
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1 answerLast reply by: Dr. William MurrayFri Sep 27, 2013 5:29 PMPost by michael morris on September 26, 2013i do not see where you cover exact equations. are you going to cover that ?
### Separation of Variables
Separation of Variables (PDF)
### Separation of Variables
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Lesson Overview 0:26
• Separation of Variables is a Technique for Solving Some Partial Differential Equations
• Separation of Variables 2:35
• Try to Separate the Variables
• If You Can, Then Both Sides Must Be Constant
• Reorganize These Intro Two Ordinary Differential Equations
• Example 1 4:41
• Example 2 11:06
• Example 3 18:30
• Example 4 25:49
• Example 5 32:53
### Transcription: Separation of Variables
Welcome back to the differential equations lecture here on educator.com.0000
My name is will Murray, we are studying partial differential equations and were starting to learn how to solve them we already had a couple lectures sort of warm you up to the idea of partial differential equations.0003
We have not actually solved that yet so, today were going to learn how to start to solve them using a technique known as separation of variables so, see what that is all about.0016
The separation of variables like as I said , it is a technique for solving some partial differential equations and the idea is that you assume that the function you are looking for remember work in a call that function U of X of T can be written as a product of a function of X only and a function of T only.0027
So, we assume that the function were looking for can be written as were to call it X of X of X and T of T by the way were to have lots of equations today where were using X is a little x and T little t .0048
Be very careful to keep those straight the little X and little T are the variables X and T are the functions so, let me emphasize that right now, those are functions and the little X and little T are the variables try to be clear when I am writing my own notes which 1 is I am talking about.0066
And so, the point of assuming that you have a product of 2 functions like that is it is really easy to take derivatives because of your taking the partial derivative with respect to X that means that all the Ts are constant so, T of T just comes down as a constant we think of that as being a constant .0091
So, you just take the derivative of capital X with respect to little X X′ of X and then when you take the second derivative with respect to X your taking capital X″ of X and again the T is just a constant when you are looking at derivatives with respect to T is just the other way around .0111
The capital X you think of as being a constant and your taking derivatives of T with respect to little T so, here is the first derivative is T′ and in the second derivative is to″ .0134
So, you take these derivatives and you plug them into the partial differential equation and then let us see what happens with that what you trying to do is get all the X is on 1 side and all the Ts on the other.0149
It does not always work but it is it works for a lot of these differential equations will see some example lots of examples where does if you can do that then a very interesting thing happens you have 1 function that is only dependent on X 1 function that is only dependent on T and are equal to each other.0162
That means if you change T nothing happens on the left because X has not changed so, if you change T nothing happens which means it must be constant so, both of these functions must be constant so, and I call that constant .0184
And so, were to end up with a function of X = λ and independently a function of T = λ so, will reorganize these into 2 ordinary different differential equations a function of X = λ and a function of T = λ.0202
And hopefully you can solve these ordinary differential equations separately for capital X and T remember X that is a function of little X the variable little X T is a function of little T.0220
If you can solve both of those then if you rewrite our original guess remember was new of X and T = X of X and Multiply by T a T so, if you can find X X of X and T of T and you can put them back together and you can compile your solution to the original partial differential equation by multiplying together the solutions to these 2 ordinary differential equations.0237
So, that is the idea of separation of variables it is definitely something that will make more sense after we get some practice so, let us try that out on some examples.0271
So, in example 1 we are going to separation of variables to convert the following partial differential equation into 2 ordinary differential equations so, remember our guess for all of these for all of these separable partial differential equations is U of X T = capital X of X x T of T .0280
It looks like really need to know, some partial derivatives there so, U of X remember the T will be a constant so, that is X′ of X x T of T and U of X X will be X″ of X x T a T and it looks like really need U of T so, that would be now, holding the X is constant so, X of X x T sub' of little T.0308
To plug those into my differential equation so, plug-in plugged in to the pit the partial differential equation that we been given so, U of X X is X″ of X T of T + this is a little X now, U of T is X of X and T′ of T = 0 .0343
The idea remember is to try to solve this equation in such a way that you get all the X is on 1 side and all the Ts on the other so, I am in a move these terms over to the other side so, I get X″ of little X T of little T = - little X x X of X T′ of T .0376
Now, meant across the divide to get all my X on the left and all the T on the right so, capital X″ of X divided by up the - on the left - X X of X = π crust divide T′ of T divided by T of T .0406
Notice now, that I got all X is on 1 side all Ts on the other so, got a function of X = to a function of T and so, this must be constant that is the whole idea of separation of variables this must be constant.0429
So, then assented = to recall my constant λ and down to split that apart into 2 separate equations so, my - X″ of little X over X X of X X of X must be = to λ on 1 side and T′ of T divided by T of T = λ on the other side.0448
Each 1 of those I am not do a little bit of work to make him into a and ordinary differential equation so, X″ of X = a multiply the - over the other side - λ X x X of X and if I just pull the right-hand side over again X″ of X + λ X x X of X = 0 .0483
On the other side with the Ts and got T′ of T = λ x T of T and if I move that over T′ of T - λ x T of T = 0 .0518
So, I am not actually going to solve these differential equations the important thing is that I have done with the problem asked me which is to separate the partial differential equation into 2 ordinary differential equations 1 for in terms of X and 1 in terms of T so, let me summarize that here .0536
We got 2 ordinary differential equations 1 for X and 1 for T of T so, that is all we need to do for that problem me recap how we how we want about it so, we assume that U of XT = X of X x QT that is of an assumption we use for every single problem in this lecture.0559
That is always how you start out with separable with trying to separate partial differential equations than the X derivative the all the T constant take the X derivative second derivative is the same way holding T constant the T derivative you hold the X constant and take the derivative of T and you plug them back into the partial differential equation.0594
CNX″ of X x T + X x X x T′ = 0 down the work to get all my X on 1 side and all my Ts on the other that is the key part of separating a partial differential equation.0614
So, I did some algebra there and got only X is 1 place = to all my Ts but since I have a function of X = to a function of T it must be constant and so, I call that constant λ and then I just separated each of those equations the X = λ the Ts = to λ and each 1 of those I reorganized into a nice ordinary differential equation which you can then solve for X of X for T of T .0631
So, let us try out another example in example 2 were going you separation of variables on the following differential equation U of TT + U of XT + U of X = 0 so, let me start out the exact same guess that were going to make every single time for partial differential equations U of XT = X of X x T of T .0663
Then let me go ahead and find you what X is can be I Treating the T is constant so, X′ of X x T of T looking at the differential equation I am seeing which derivatives of that have to plug-in.0691
U of TT them in a skip a step here and skip right to the second derivative the X will come down is a constant every time and then will have the second derivative of T with respect to T now, U of XT that means you take the T derivative of the X derivative.0709
So, the X relative is a peer X′ of X x T of T the T derivative of that the X is now, constitutes X′ of X x T′ of T so, and take each 1 of those 3 derivatives and plug them into the partial differential equation and will see what happens there.0728
U TT is X of X T″ of T + U of XT is X′ of X x T′ of T + U of X is X′ of X x T of T = 0 .0749
Now, my goal here is to solve this in such a way that I get all the X is on 1 side and all the Ts on the other so, if I look at this is you got X′ here and here them in a factor that out so, X of X T″ of T + X′ of X T′ of T + T of T = 0.0774
Remove that term over to the other side so, X of X T″ of T = - X′ of X x that whole stuff in parentheses there in brackets + T of T and now, I am trying to get all my X on 1 side and all my Ts on the other so, I think I am going to cross multiply and divide .0807
On the left of me get keep double prime of T divided by that compound term T′ of T + T of T on the right I am getting at - X′ of X divided by X of X and I succeeded in getting all my X alongside only Ts on the other so, this again must be constant must be constant and so, it is got a be = to a constant that I am in a call λ.0841
Separate these 2 equations and solve each 1 separately so, my - X′ of X over X of X = λ and on the other side will have T″ of T x T′ of T + T of no divided by T′ of T + T a T is also, = to λ and each 1 of those I am going to sort out into an ordinary differential equation.0879
So, here I am going to get X′ of X move the - over the other side - λ X = - λ X of X and if I move that back I get X′ of X + λ X of X = 0 that is a nice ordinary differential equation there is only 1 variable in their which is little X .0912
On the right I see I got T″ of T = λ x T′ of T + T of T and if I move those terms over to the left I got T″ of T - λ T′ of T - λ T of T = 0 that is second-order ordinary differential equation in terms of T T of T.0949
And so, again I got 2 ordinary differential equations for X of X 1 for X of X and 1 for T of T so, I am done there let me go back over the steps in case you have any lingering questions.0985
U of XT were assuming that to be = to X of X x T of T that is the running assumption for all of these differential equations that were going to try to separate it is always separating it into a function of X x a function of T and the point of that is when you take derivatives U of X you just treating the T is a constant CX′ of X and in U of XT down here we take the T derivative of that so, we get X′ of X x T′ of T and U of TT is X of X x T″ of T.1010
The reason I took those 3 partial derivatives was because I was looking at the partial differential equation and so, I was planning ahead to see what lives can have to plug-in to the partial differential equation and so, I plugged each 1 of those in and then I am trying to solve it in such a way to get all the X over here all the Ts over there.1047
So, I see that I have not X′ of X on 2 terms here so, I factored that out and then if I move that over to the other side then I can cross multiply across divide to get all the Ts on the left and all the X on the right.1067
Which tells me that it must be = to some constant which am calling λ so, I separate that out into 2 separate equations = λ and then it is really 2 parallel tracks here 1 ends up with the differential equation in terms of X and 1 ends up with an ordinary differential equation in terms of T.1086
In example 3 were to use separation of variables to convert another differential equation into 2 ordinary differential equations so, let me remind you that our guests for every partial differential equation of forgot a use separation of variables is U of XT = X of X x T of T.1107
Now, looks like I am going to need the derivatives U X X and U TT so, U X X means were holding T constant and taking the second derivative of X should be to derivatives there x T of T and U TT means were holding X constant so, the X just comes down T″ of T and so, I am in a plugged each 1 of those into my differential equation there. 1930 It looks like I am not the plug-in you by itself as well so, U X X is X″ of X x T of T + U TT is X of X T″ of T + little T x the original U so, that is capital X of X x T of T = 0.1138
Now, I want to I I want to try and separate all the X from all the Ts and it looks like I got an X here and it X of X here thing a factor that term out of whenever I cancel got X″ of X x T of T + X of X x the quantity T″ of T + little T x T of T T T = 0.1198
Remove that term over to the other side so, I get X″ of X x T of T = - X of X x that compound term T″ of T + T x the original TFT and again I am going to cross multiply and divide so, thing the keep that - sin with the X so, get T1 - X″ of X divided by X so, it really divided both sides way X by - X of X. 2124 On the right-hand side I got T″ of T + little T x T of T divided by T of T and now, I successfully gotten all the X on 1 side all the Ts on the other so, again this thing must be constant and I am a call my constant λ that is kind of the universal and in the when you are using separation of variables.1242
And so, I got to a separate into 2 equations - X″ of X over X of X = λ on the 1 side and T″ of T + little T x TT of T over T a T = λ over here on the left and then I multiply my little my X of X out of the denominator and if you bring the - along with the with with the a X of X = - λ x X of X .1314
So, if I move that back X″ of X + λ x capital X of X = 0 on the right multiply my T out of the denominator T″ of T + little T x T of T = λ x T of T and if I move that term over the other side the other side of get T″ of T .1355
Now, I see about a T in each term here so, right this is T little T - λ just factoring their factoring out a T = 0 and so, what I discovered here is that my original partial differential equation just separated into 2 ordinary differential equations.1391
So, 2 ordinary differential equations for X of X and T of T remember were not solving these ordinary differential equations yet will do that in example little later.1416
We are just showing how you can start out with the partial differential equation make a good assumption of the beginning and hopefully separated into to ordinary differential equations so, we go back and recap that for you.1442
Start out with the original assumption U of XT = X of X x TT a T that is a running assumption we use that in every single example then where you go and that starts to vary on looking over the differential equation I see a many U of XX and U of TT .1455
So, for each 1 I hold the other variable constant and just take the second derivative with respect to the corresponding function and then I plugged each 1 of those into the partial differential equation that is what I did here plugged each 1 in.1473
And I want to separate the X for the Ts from each other and I noticed I got it X of X and in each of these terms so, I factor that out here and that enables me to pull that hold term that compound turned over to the other side and then we can cross multiply and divide and get all my X is on 1 side and all my T on the other.1490
So, since I got a function of X = to a function of T means of them both = constant animate a call that constant λ as usual and so, I separate those into 2 equations in terms of λ and 1 involves X and 1 involves Ts but there is no mixing between the 2 .1514
So, each 1 of those I can do a little algebra and rearrange it into a traditional ordinary differential equation 1 in terms of X and 1 in terms of T.1535
So, in example 4 were in you separate separation of variables on the heat equation which I said is out 1 of the most important partial differential equations and it is can be the 1 that were to spend a lot of time solving later we still have not completely solve the heat equation so, to take us a couple lectures to catch there.1545
So, were going to start solving that right now, are going to apply separation of variables to heat equation and there is a little note here that says will we get the step of using λ we should use - λ instead of λ for the separation constant that really make sense why we can do that at this stage but hopefully in a couple problems down the road from now, you will start to see why we worry is - λ instead of λ.1568
So, let us go ahead and try her separation of variables on this 1 starts out just like all the others where you start out with U of XT you assume that it can be written as X of X x T of T and then were to need some derivatives here U so, T X constant so, get X of X x T′ of T and will also, need U X X which means withhold T constant and will get the second derivative of X and then just T of T .1597
So, were to take these and plugged them in were to plug them in to our partial differential equation so, what we get here is U of T is capital X of X x T′ of T = α ² on a really care what out is at this point X″ of X x T of T and my goal here is a separate the X on 1 side the Ts on the other.1634
It is pretty easy for this 1 just get down X of X divided by X X everyone separate alone across multiplied by the other way around so, get my keys on let T′ of T divided by TT a T = anything to bring my α ² along with my T up at that down there and on the other side of got X″ of X divided by X of X .1671
By the way it is not so, obvious at this point y-axis but the α ² with 1 terms or the other there is a lot of different places you put it I am putting it here because I am kind of looking forward to a solution were going to be studying later for the heat equation.1706
So, do not worry about it right now, while putting out for ² in this term as opposed to over here with the X just bare with me and you will see will we get to problem later that it is it is a nice convention to have the α ² with the Ts.1719
Now, this is a constant because precisely because we got all X is on 1 side we got all Ts on the other and it says were going to use - λ instead of the usual λ th= - λ .1733
Remember we can untangle each 1 of those and try to I get some nice differential equations on the X side we get X″ of X divided by X of X = - λ.1750
So, if I clear my denominator there I get X″ of X = - λ x X of X and if I move it over to the other side I get X″ of X + λ x X of X = 0.1767
On the other side I am going to just keep my T′ of T divided by α ² x T of T this can only that = to - λ I think that is actually going to be the format there is going to be useful to solve later so, just can leave it in that form and what that means is that I have successfully separated my partial differential equation into 2 ordinary differential equations .1792
1 in terms of X and 1 in terms of T. I see I left out a′ on my X here; that is X″ of X + λ x X of little X = 0 and here is my differential equation in terms of T. So, it reduced my partial differential equation into two ordinary differential equations .1822
X and T now, you want to hang onto these ordinary differential equations because were going use them in the next example let us go ahead and try to solve these ordinary differential equations in the next example so, make sure you understand where these differential equations are coming for the so, the to be ready to understand them and be ready to solve them and use them in the next example.1845
So, let me go back and remind you where they came from we start with this generic assumption U of XT is a product of a function of X and a function of T and then we take its derivative with respect to T and with respect to X twice so, each time you take a partial derivative it means hold the other variable constant and you take a derivative with respect whichever variable that is.1871
So, U of T hold the X constant and you take T′ of T U of double X that means you hold the T constant and you take the second derivative with respect to X so, we plugged those back into the differential equation and then we try to sort out the X on 1 side and sort out the Ts on the other side.1896
So, we do sort those out X it turns out to be fairly easy get them sorted then we can say that the whole thing there both = to the same constant and for reasons that will be useful later it does not really make sense at this point recalling that constant - λ .1916
So, this separates out into a function of X = - λ and was we solve that into a regular ordinary differential equation and then a function of T is = λ got a little squished function of T = - λ and just do not need that in that form because that is the form that were getting used to solve later .1934
So, the next example you are going solve these 2 ordinary differential equations and were to see what kind of answer we might get to the original partial differential equation.1960
So, let us go ahead and start working on that so, were going to solve the 2 ordinary differential equations below that came from the heat equation from the partial differential equation that was the heat equation.1969
I gives us a little hint here we want to assume that λ is + to try to find solutions that satisfy the boundary conditions U of 0T = U of LT = 0 so, were to try find solutions that also, satisfy that.1985
So, let me go ahead and try to solve this first 1 this is in terms of T this is actually really nicety separable first order differential equation will my α ² over to the other side so, I got T′ of T divided by T of T = - λ α ² .2003
Now, I can integrate both sides can integrate both sides here now, the integral because I have T′ of T over T of T the integral that is exactly natural log of T of T and that is both the = to integrate the right-hand side it is were integrating with respect to T so, is just - λ α ² T + a constant .2027
I want to solve for T here so, when I raised each of both sides so, get T = E to the - λ α ² T + a constant but remember this is the same as E - λ α ² T x EC and that EC were just think of that as being a K being another constant K and so, my function T actually pretty easy to find here was a constant x E - λ α ² T.2056
Let us hang onto that were to be using it later so, before we go on to the next slide let me recap what happened on this slide to try to solve each 1 of these ordinary differential equations that come up in the heat equation and so, far just try to solve the 1 for T try to solve 1 for X on the next slide .2097
I am just trying to solve the 1 for T so, I move the α ² over to the other side key observation here is that the derivative of T of T of natural log of T would be 1 over T x its derivative so, the integral of T priority over T of T is very easy it is just natural log of T of T.2117
Integrate the right-hand side since were integrating with respect to T I just get - λ α ² x T + a constant remember it is very important to add a constant when you are integrating.2141
By the way if it is been a while since you watch the initial lectures here on educator on differential equations might be good to go back and check those out because I am using stuff that we learned back in those early lectures on differential equations .2154
So, to get rid of the natural log I raised E of both sides here and now, by laws of exponents remember E of the X + Y = E X x E Y so, since I got a + C here that is been like multiplied by E the C is just another constant so, I called a K and now, if you solve for T we get K x E - λ α ² x T.2168
So, we still have not looked at the equation for X at all yet so, that is organized go ahead and do on the next slide so, in were still working on example 5 here we solve the equation for T so, that that is looking good we have to solve the equation for X .2198
Now, this is a second order linear ordinary differential equation we had a whole lecture X a 3 lectures on how to solve these earlier on in the differential equations lecture series here on educator.com so, if you do not number how to solve those here is a really quick review.2217
You look at the characteristic equation R² + λ = 0 and using of this X of X you thing that is being X to the 0 derivative so, put R the 0s that is why there is no R right here .2235
So, were solving trying to solve for λ so, R ²s = to - λ and R = the ² root of - λ the + or - there let me remind you that we were given an assumption here which was that λ that was bigger than 0 .2254
For that means that - λ is less than 0 and so, that means the ² root of - λ is complex so, were dealing with complex numbers here it is I x the ² root of + λ.2275
So, we can have to deal with complex solutions here now, we did talk about what happens when you get complex solutions the characteristic equation that was back in 1 of earlier lectures here on educator.com back and check it out if you do not remember but the short version is that your answers look like sins and cosine.2297
So, our R here is + or - I x root λ and so, my solution X of X is I think in our earlier lectures in C1 and C2 C1 x the cosine + C2 x sin call it A and B for my constants this time so, it is a x well remember if you have a + b I then you go E AT x cosine of B T + C2 η T x sin of BT.2317
Sum up all that format simply in a change around what I am calling my constants little bit instead of C1 I am going to use a and B for C1 and C2 this a is just 0 so, you get E 0T which is just 1 so, those terms are both 1 and I get cosine of well my B is the ² root of λ so, I may go constant x quickly go ahead and call that C1 and C2 so, will be too confusing.2369
C1 x cosine of the ² root of λ and that is my variable here is X not T so, put it X in there + C2 x sin of the ² root of λ X and let me go ahead and change that C1 and C2is and call them a and B.2400
So, a cosine the ² root of λ X + Bsin the ² root of λ X so, that is my X of X and my U my my original function that I was looking for U of XT let me remind you what format that had U of XT was = to X of X this is our assumption.2424
We try to separate the variables x T of T and I have already figured out my T of T there is right there and I figured out my X of X so, let me combine those so, that = a cosine the ² of λ X + B x the sin of the ² root of λ X all x K x E - λ α ² T .2459
By the way that K drop it in the reason going to drop it is because if I put a K right here it could be absorbed right into the A and B so, just a cancel that K out and put E- λ α ² x T that is my U of XT .2491
But I am trying to satisfy my 2 boundary conditions so, let me remind you what those were my first boundary condition was U of XT are sorry U of 0T we label as a boundary condition is probably pin a while since you have seen it is a boundary condition this is given to us as part of the differential equation well as part of the package along with the differential equation was that U of 0T had to be = to 0 .2510
So, let us see what we get number that means your plug-in X = 0 let us see what we get if we plug-in X = 0 into our U of XT here so, U of 0T = a x cosine of 0+ B x sin of 0 and all of that gets multiplied by E of the - λ α ² T and that is what is to come out to be 0.2547
So, cosine a 0 is just 1 sin of 0 is 0 and so, what we get is a x E - λ α ² T = 0 on the strength of that we can say that a must be = to 0 .2576
So, we will not be needing to use that a term now, let us look at how were going to incorporate the other boundary condition and let me remind you what that was so, my U of XT simplifies down a bit when U of XT since our a term is gone away it is just B x the sin of root λ X all of that multiplied by E - λ α ² T .2596
And were told by our other boundary condition let me remind you this came as part of the package when we originally got the differential equation so, our other boundary condition was that U of L to had to be 0 remember that represented plug-in X = L which was the right-hand end of the Rod in the heat equation.2636
So, let us plug-in X = L we get B x sin of root λ L x E - λ α ² T = 0 now, we do not want it said B = 0 because otherwise our entire solution disappears .2670
Also, our E - λ α ² T we do not want that to be 0 because and even function can never be 0s that is never 0 I do not want to set B to be = to 0 because otherwise my whole solution disappears and so, what I am forced to look at is the idea that sin of root λ L = 0 .2692
So, I have to figure out what my possibilities there are to make sin of root λ L = 0 so, sin of what = 0 will I know, that sin of any multiple of π in the integer multiple of pies = to 0 so, this will work if I take root λ L = an integer multiple of π where N is an integer .2719
And so, the only thing remember L was the length of the Rod in the heat equation so, that is a constant it is established all the way through we can change that so, do not think I can solve for his is λ and so, if I will be L to the other side and ² both sides 4618 I get λ = N² π ² over L ² .2750
This will work for any value of N we choose so, we get 1 solution for each N so, I am and I write my X sub and of X remember I had an X for what I am discovering is that for each value of NI get a different solution .2784
So, and there is a B there the B could be anything so, do not know, yet what I want B to B but BN sin I have a root λ L there so, remember root λ is N π over L sin of root λ X so, N π X over L and my T = E - now, λ this is rather messy N² π ² over L ² .2813
I still have not α ² T in my exponent for E so, let me extend that α ² T such my T of T and so, ultimately get 1 solution for each end my U of an of XT is remember you multiply the X of X and the T of T so, multiply these 2 things together it is B sub N sin of N π X over L E - N² π ² α ² T all divided by L ² .2859
Pretty messy that is my solution to the heat equation that is still compatible with both of the boundary conditions that you tend to get package along with the heat equation so, that was quite a mouthful of me go back and recap everything came from.2914
We started out with well we had to solve these 2 ordinary differential equations for T of T and X of X so, we solved T of T on the previous slide previous slide so, on the previous side we figured out a form for T of T and we figured out that it was a constant x E - λ α ² T that constant I am X the going to omit that constant and the reason is because down at this step and gets multiplied by X which has Constance attach to it.2937
Anyway so, I let that K be absorbed into the Constance for the the the X functions here so, not to worry about that constant what I do have to focus on next is solving this ordinary differential equation for X and I learned how to do that back in our lessons on second-order differential equations we have an earlier lecture on that read here in the differential equations lecture series on educator.com .2980
So, if you do not remember how to solve those just check back the earlier lecture and you will see that we did a bunch of second-order differential equations and we had the characteristic equation we had cases where there were real roots where there were complex roots where there were repeated roots so, were through all those different equated different cases in different lectures.3012
What happens with this 1 is we get R ² + λ = 0 so, R ² is - λ and so, R turns into the ² root of - λ and were given that λ that was greater than 0 so, - λ is less than 0 so, really trying to here is take the ² of a - number which means we have complex solutions .3039
We learned before that will we have complex solutions A+ B I this is the format of our solution except word I using slightly different variables here so, you translate a little bit this C1 and C2 I a relabeled a and B and that is a little confusing as it is not the same as this a and B.3064
This a and B will be a was just 0 because it is 0+ or - I root λ and the B this be was root λ so, that is where I got to cosine e of that is the B there and that is the B there also, are variable now, is X instead of T we use T before but now, are using axis and so, and I also, wanted to change the Constance C1 and C2 a want to call them a and B a new a and B .3088
So, it is a little confusing with a different variables I apologize for that there is nothing very deeply mathematical going on I am just kind according the solution from the earlier lecture earlier lecture on complex roots on complex roots.3121
So, we have now, found our solution for X of X but we want to match the boundary conditions so, we want to compile our solution together with our solution from T of T so, remember this is our guess with separation of variables.3141
U is the product of X of X and T of T so, here on multiplying them together and are boundary condition was the U 0T = 0 which means you plug-in X = 0 so, I plug-in X = 0 here and that when the plug-in X = 0 you get the cosine a 0 is just 1 the sin of 0 is just 0 and so, you get a x E the something = 0.3162
What you do something this term is never = to 0 so, we must have a = 0 which means or hold term here drops out of our solution so, were just down to the simpler solution Bx the sin term x that exponential term .3191
Now, we bring in our second boundary condition which says that when you plug-in X = L then you must be getting 0 so, if we plug-in X = L here then we get complicated expression = 0 now, the E term again can not be 0 that that can not be 0.3213
I do not want B term to be 0 because it B = 0 in my entire solution disappears and I just get 0 so, I do not want to be the 0 which means I must have sin = to 0 .3214
So, if sin = 0 then I have to think about what angles have sin 0 and I remember that sin of any multiple or any integer multiple of π = 0.3248
That means root λ L must be N π for some integer and an exit is works no matter what and is as long as it is an integer sorted get a whole family of solutions with different integers if I saw for λ I get λ is N² π ² over L ² and I am just going to plug those back into my solution here back in for root λ here and back in for λ here.3261
So, the X part gives us n root λ is N π X over L the E part that the T part gives us E - λ which is - N ² π ² over L ² and we what we multiply those together and putting subscript and 0 because remember getting different solutions for each value of that so, were to have a whole bunch of solutions here .3289
So, we have BNx the X part x the T part so, that is our solution to the partial differential equation that also, satisfies both boundary conditions that is actually officially the end of this lecture here but let me go ahead and give you little Teaser for the next lecture.3318
We write down what we just figured out U of an of XT = BN x sin of N π X over L just figure this out E - N ² π ² α ² T over L ².3339
Now, let me remind you what we did there were trying to solve the heat equation and the original heat equation me make sure I write it down correctly was U of T = α ² U of X X what we have done now, as we solve that using separation of variables.3366
We also, had 2 boundary conditions U of 0T = 0 and U of LT = 0 and what were doing in this example was confirming a solution that satisfied both of those boundary conditions.3395
Now, there was 1 more condition which was the initial condition, U of X 0 = F of X and what we have not done yet is satisfied that initial condition so, that take quite a bit of work that is going to be the content of the next couple of lectures here on educator.com .3415
Were to finish off the solution of the heat equation , we are to try to make it match that initial condition so, hope you stick around and watch those lecture so, we can finish solving the equation together.3436
In the meantime, this is the end of our lecture on separation of variables, and I will just remind you that this is part of the differential equations lecture series here on educator.com. My name is Will Murray, thanks very much for joining us, bye bye.3449
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Professor Murray. Separation of Variables. Slide Duration:. Section 1: First-Order Equations. Linear Equations. 1h 7m 21s. Intro. 0:00. Lesson Objectives. 0:19. How to Solve Linear Equations. 2:54. Calculate the Integrating Factor. 2:58. Changes the Left Side so We Can Integrate Both Sides. 3:27. Solving Linear Equations. 5:32. Further Notes. 6:10. If P(x) is Negative. 6:26. Leave Off the Constant. 9:38. The C Is Important When Integrating Both Sides of the Equation. 9:55. Example 1. 10:29. Example 2. 22:56. Example 3. 36:12. Example 4. 39:24. Example 5. 44:10. Example 6. 56:42. Separable Equations. 35m 11s. Intro. 0:00. Lesson Objectives. 0:19. Some Equations Are Both Linear and Separable So You Can Use Either Technique to Solve Them. 1:33. Important to Add C When You Do the Integration. 2:27. Example 1. 4:28. Example 2. 10:45. Example 3. 14:43. Example 4. 19:21. Example 5. 27:23. Slope & Direction Fields. 1h 11m 36s. Intro. 0:00. Lesson Objectives. 0:20. If You Can Manipulate a Differential Equation Into a Certain Form, You Can Draw a Slope Field Also Known as a Direction Field. 0:23. How You Do This. 0:45. Solution Trajectories. 2:49. Never Cross Each Other. 3:44. General Solution to the Differential Equation. 4:03. Use an Initial Condition to Find Which Solution Trajectory You Want. 4:59. Example 1. 6:52. Example 2. 14:20. Example 3. 26:36. Example 4. 34:21. Example 5. 46:09. Example 6. 59:51. Applications, Modeling, & Word Problems of First-Order Equations. 1h 5m 19s. Intro. 0:00. Lesson Overview. 0:38. Mixing. 1:00. Population. 2:49. Finance. 3:22. Set Variables. 4:39. Write Differential Equation. 6:29. Solve It. 10:54. 11:47. Example 1. 13:29. Example 2. 24:53. Example 3. 32:13. Example 4. 42:46. Example 5. 55:05. Autonomous Equations & Phase Plane Analysis. 1h 1m 20s. Intro. 0:00. Lesson Overview. 0:18. Autonomous Differential Equations Have the Form y' = f(x). 0:21. Phase Plane Analysis. 0:48. y' < 0. 2:56. y' > 0. 3:04. If we Perturb the Equilibrium Solutions. 5:51. Equilibrium Solutions. 7:44. 8:06. Solutions Will Tend Away From Unstable Equilibria. 9:32. Semistable Equilibria. 10:59. Example 1. 11:43. Example 2. 15:50. Example 3. 28:27. Example 4. 31:35. Example 5. 43:03. Example 6. 49:01. Section 2: Second-Order Equations. Distinct Roots of Second Order Equations. 28m 44s. Intro. 0:00. Lesson Overview. 0:36. Linear Means. 0:50. Second-Order. 1:15. Homogeneous. 1:30. Constant Coefficient. 1:55. Solve the Characteristic Equation. 2:33. Roots r1 and r2. 3:43. To Find c1 and c2, Use Initial Conditions. 4:50. Example 1. 5:46. Example 2. 8:20. Example 3. 16:20. Example 4. 18:26. Example 5. 23:52. Complex Roots of Second Order Equations. 31m 49s. Intro. 0:00. Lesson Overview. 0:15. Sometimes The Characteristic Equation Has Complex Roots. 1:12. Example 1. 3:21. Example 2. 7:42. Example 3. 15:25. Example 4. 18:59. Example 5. 27:52. Repeated Roots & Reduction of Order. 43m 2s. Intro. 0:00. Lesson Overview. 0:23. If the Characteristic Equation Has a Double Root. 1:46. Reduction of Order. 3:10. Example 1. 7:23. Example 2. 9:20. Example 3. 14:12. Example 4. 31:49. Example 5. 33:21. Undetermined Coefficients of Inhomogeneous Equations. 50m 1s. Intro. 0:00. Lesson Overview. 0:11. Inhomogeneous Equation Means the Right Hand Side is Not 0 Anymore. 0:21. First Solve the Homogeneous Equation. 1:04. Find a Particular Solution to the Inhomogeneous Equation Using Undetermined Coefficients. 2:03. g(t) vs. Guess for ypar. 2:42. If Any Term of Your Guess for ypar Looks Like Any Term of yhom. 5:07. Example 1. 7:54. Example 2. 15:25. Example 3. 23:45. Example 4. 33:35. Example 5. 42:57. Inhomogeneous Equations: Variation of Parameters. 49m 22s. Intro. 0:00. Lesson Overview. 0:31. Inhomogeneous vs. Homogeneous. 0:47. First Solve the Homogeneous Equation. 1:17. Notice There is No Coefficient in Front of y''. 1:27. Find a Particular Solution to the Inhomogeneous Equation Using Variation of Parameters. 2:32. How to Solve. 4:33. Hint on Solving the System. 5:23. Example 1. 7:27. Example 2. 17:46. Example 3. 23:14. Example 4. 31:49. Example 5. 36:00. Section 3: Series Solutions. Review of Power Series. 57m 38s. Intro. 0:00. Lesson Overview. 0:36. Taylor Series Expansion. 0:37. Maclaurin Series. 2:36. Common Maclaurin Series to Remember From Calculus. 3:35. 7:58. Ratio Test. 12:05. Example 1. 15:18. Example 2. 20:02. Example 3. 27:32. Example 4. 39:33. Example 5. 45:42. Series Solutions Near an Ordinary Point. 1h 20m 28s. Intro. 0:00. Lesson Overview. 0:49. Guess a Power Series Solution and Calculate Its Derivatives, Example 1. 1:03. Guess a Power Series Solution and Calculate Its Derivatives, Example 2. 3:14. Combine the Series. 5:00. Match Exponents on x By Shifting Indices. 5:11. Match Starting Indices By Pulling Out Initial Terms. 5:51. Find a Recurrence Relation on the Coefficients. 7:09. Example 1. 7:46. Example 2. 19:10. Example 3. 29:57. Example 4. 41:46. Example 5. 57:23. Example 6. 1:09:12. Euler Equations. 24m 42s. Intro. 0:00. Lesson Overview. 0:11. Euler Equation. 0:15. Real, Distinct Roots. 2:22. Real, Repeated Roots. 2:37. Complex Roots. 2:49. Example 1. 3:51. Example 2. 6:20. Example 3. 8:27. Example 4. 13:04. Example 5. 15:31. Example 6. 18:31. Series Solutions. 1h 26m 17s. Intro. 0:00. Lesson Overview. 0:13. Singular Point. 1:17. Definition: Pole of Order n. 1:58. Pole Of Order n. 2:04. Regular Singular Point. 3:25. Solving Around Regular Singular Points. 7:08. Indical Equation. 7:30. If the Difference Between the Roots is An Integer. 8:06. If the Difference Between the Roots is Not An Integer. 8:29. Example 1. 8:47. Example 2. 14:57. Example 3. 25:40. Example 4. 47:23. Example 5. 1:09:01. Section 4: Laplace Transform. Laplace Transforms. 41m 52s. Intro. 0:00. Lesson Overview. 0:09. Laplace Transform of a Function f(t). 0:18. Laplace Transform is Linear. 1:04. Example 1. 1:43. Example 2. 18:30. Example 3. 22:06. Example 4. 28:27. Example 5. 33:54. Inverse Laplace Transforms. 47m 5s. Intro. 0:00. Lesson Overview. 0:09. Laplace Transform L{f}. 0:13. Run Partial Fractions. 0:24. Common Laplace Transforms. 1:20. Example 1. 3:24. Example 2. 9:55. Example 3. 14:49. Example 4. 22:03. Example 5. 33:51. Laplace Transform Initial Value Problems. 45m 15s. Intro. 0:00. Lesson Overview. 0:12. 0:14. Take the Laplace Transform of Both Sides of the Differential Equation. 0:37. Plug in the Identities. 1:20. Take the Inverse Laplace Transform to Find y. 2:40. Example 1. 4:15. Example 2. 11:30. Example 3. 17:59. Example 4. 24:51. Example 5. 36:05. Section 5: Review of Linear Algebra. Review of Linear Algebra. 57m 30s. Intro. 0:00. Lesson Overview. 0:41. Matrix. 0:54. Determinants. 4:45. 3x3 Determinants. 5:08. Eigenvalues and Eigenvectors. 7:01. Eigenvector. 7:48. Eigenvalue. 7:54. Lesson Overview. 8:17. Characteristic Polynomial. 8:47. Find Corresponding Eigenvector. 9:03. Example 1. 10:19. Example 2. 16:49. Example 3. 20:52. Example 4. 25:34. Example 5. 35:05. Section 6: Systems of Equations. Distinct Real Eigenvalues. 59m 26s. Intro. 0:00. Lesson Overview. 1:11. How to Solve Systems. 2:48. Find the Eigenvalues and Their Corresponding Eigenvectors. 2:50. General Solution. 4:30. Use Initial Conditions to Find c1 and c2. 4:57. Graphing the Solutions.
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5:20. Solution Trajectories Tend Towards 0 or ∞ Depending on Whether r1 or r2 are Positive or Negative. 6:35. Solution Trajectories Tend Towards the Axis Spanned by the Eigenvector Corresponding to the Larger Eigenvalue. 7:27. Example 1. 9:05. Example 2. 21:06. Example 3. 26:38. Example 4. 36:40. Example 5. 43:26. Example 6. 51:33. Complex Eigenvalues. 1h 3m 54s. Intro. 0:00. Lesson Overview. 0:47. Recall That to Solve the System of Linear Differential Equations, We find the Eigenvalues and Eigenvectors. 0:52. If the Eigenvalues are Complex, Then They Will Occur in Conjugate Pairs. 1:13. Expanding Complex Solutions. 2:55. Euler's Formula. 2:56. Multiply This Into the Eigenvector, and Separate Into Real and Imaginary Parts. 1:18. Graphing Solutions From Complex Eigenvalues. 5:34. Example 1. 9:03. Example 2. 20:48. Example 3. 28:34. Example 4. 41:28. Example 5. 51:21. Repeated Eigenvalues. 45m 17s. Intro. 0:00. Lesson Overview. 0:44. If the Characteristic Equation Has a Repeated Root, Then We First Find the Corresponding Eigenvector. 1:14. Find the Generalized Eigenvector. 1:25. Solutions from Repeated Eigenvalues. 2:22. Form the Two Principal Solutions and the Two General Solution. 2:23. Use Initial Conditions to Solve for c1 and c2. 3:41. Graphing the Solutions. 3:53. Example 1. 8:10. Example 2. 16:24. Example 3. 23:25. Example 4. 31:04. Example 5. 38:17. Section 7: Inhomogeneous Systems. Undetermined Coefficients for Inhomogeneous Systems. 43m 37s. Intro. 0:00. Lesson Overview. 0:35. First Solve the Corresponding Homogeneous System x'=Ax. 0:37. Solving the Inhomogeneous System. 2:32. Look for a Single Particular Solution xpar to the Inhomogeneous System. 2:36. Plug the Guess Into the System and Solve for the Coefficients. 3:27. Add the Homogeneous Solution and the Particular Solution to Get the General Solution. 3:52. Example 1. 4:49. Example 2. 9:30. Example 3. 15:54. Example 4. 20:39. Example 5. 29:43. Example 6. 37:41. Variation of Parameters for Inhomogeneous Systems. 1h 8m 12s. Intro. 0:00. Lesson Overview. 0:37. Find Two Solutions to the Homogeneous System. 2:04. Look for a Single Particular Solution xpar to the inhomogeneous system as follows. 2:59. Solutions by Variation of Parameters. 3:35. General Solution and Matrix Inversion. 6:35. General Solution. 6:41. Hint for Finding Ψ-1. 6:58. Example 1. 8:13. Example 2. 16:23. Example 3. 32:23. Example 4. 37:34. Example 5. 49:00. Section 8: Numerical Techniques. Euler's Method. 45m 30s. Intro. 0:00. Lesson Overview. 0:32. Euler's Method is a Way to Find Numerical Approximations for Initial Value Problems That We Cannot Solve Analytically. 0:34. Based on Drawing Lines Along Slopes in a Direction Field. 1:18. Formulas for Euler's Method. 1:57. Example 1. 4:47. Example 2. 14:45. Example 3. 24:03. Example 4. 33:01. Example 5. 37:55. Runge-Kutta & The Improved Euler Method. 41m 4s. Intro. 0:00. Lesson Overview. 0:43. Runge-Kutta is Know as the Improved Euler Method. 0:46. More Sophisticated Than Euler's Method. 1:09. It is the Fundamental Algorithm Used in Most Professional Software to Solve Differential Equations. 1:16. Order 2 Runge-Kutta Algorithm. 1:45. Runge-Kutta Order 2 Algorithm. 2:09. Example 1. 4:57. Example 2. 10:57. Example 3. 19:45. Example 4. 24:35. Example 5. 31:39. Section 9: Partial Differential Equations. Review of Partial Derivatives. 38m 22s. Intro. 0:00. Lesson Overview. 1:04. Partial Derivative of u with respect to x. 1:37. Geometrically, ux Represents the Slope As You Walk in the x-direction on the Surface. 2:47. Computing Partial Derivatives. 3:46. Algebraically, to Find ux You Treat The Other Variable t as a Constant and Take the Derivative with Respect to x. 3:49. Second Partial Derivatives. 4:16. Clairaut's Theorem Says that the Two 'Mixed Partials' Are Always Equal. 5:21. Example 1. 5:34. Example 2. 7:40. Example 3. 11:17. Example 4. 14:23. Example 5. 31:55. The Heat Equation. 44m 40s. Intro. 0:00. Lesson Overview. 0:28. Partial Differential Equation. 0:33. Most Common Ones. 1:17. Boundary Value Problem. 1:41. Common Partial Differential Equations. 3:41. Heat Equation. 4:04. Wave Equation. 5:44. Laplace's Equation. 7:50. Example 1. 8:35. Example 2. 14:21. Example 3. 21:04. Example 4. 25:54. Example 5. 35:12. Separation of Variables. 57m 44s. Intro. 0:00. Lesson Overview. 0:26. Separation of Variables is a Technique for Solving Some Partial Differential Equations. 0:29. Separation of Variables. 2:35. Try to Separate the Variables. 2:38. If You Can, Then Both Sides Must Be Constant. 2:52. Reorganize These Intro Two Ordinary Differential Equations. 3:05. Example 1. 4:41. Example 2. 11:06. Example 3. 18:30. Example 4. 25:49. Example 5. 32:53. Fourier Series. 1h 24m 33s. Intro. 0:00. Lesson Overview. 0:38. Fourier Series. 0:42. Find the Fourier Coefficients by the Formulas. 2:05. Notes on Fourier Series. 3:34. Formula Simplifies. 3:35. Function Must be Periodic. 4:23. Even and Odd Functions. 5:37. Definition. 5:45. Examples. 6:03. Even and Odd Functions and Fourier Series. 9:47. If f is Even. 9:52. If f is Odd. 11:29. Extending Functions. 12:46. If We Want a Cosine Series. 14:13. If We Wants a Sine Series. 15:20. Example 1. 17:39. Example 2. 43:23. Example 3. 51:14. Example 4. 1:01:52. Example 5. 1:11:53. Solution of the Heat Equation. 47m 41s. Intro. 0:00. Lesson Overview. 0:22. Solving the Heat Equation. 1:03. Procedure for the Heat Equation. 3:29. Extend So That its Fourier Series Will Have Only Sines. 3:57. Find the Fourier Series for f(x). 4:19. Example 1. 5:21. Example 2. 8:08. Example 3. 17:42. Example 4. 25:13. Example 5. 28:53. Example 6. 42:22. Bookmark & Share Embed. ## Copy & Paste this embed code into your website’s HTML. Please ensure that your website editor is in text mode when you paste the code.. (In Wordpress, the mode button is on the top right corner.). ×. • - Allow users to view the embedded video in full-size.. Since this lesson is not free, only the preview will appear on your website.. • ## Related Books. 1 answerLast reply by: Dr. William MurrayFri Sep 27, 2013 5:29 PMPost by michael morris on September 26, 2013i do not see where you cover exact equations. are you going to cover that ?. ### Separation of Variables. Separation of Variables (PDF). ### Separation of Variables. Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.. • Intro 0:00. • Lesson Overview 0:26. • Separation of Variables is a Technique for Solving Some Partial Differential Equations. • Separation of Variables 2:35. • Try to Separate the Variables. • If You Can, Then Both Sides Must Be Constant. • Reorganize These Intro Two Ordinary Differential Equations. • Example 1 4:41. • Example 2 11:06. • Example 3 18:30. • Example 4 25:49. • Example 5 32:53. ### Transcription: Separation of Variables. Welcome back to the differential equations lecture here on educator.com.0000. My name is will Murray, we are studying partial differential equations and were starting to learn how to solve them we already had a couple lectures sort of warm you up to the idea of partial differential equations.0003. We have not actually solved that yet so, today were going to learn how to start to solve them using a technique known as separation of variables so, see what that is all about.0016. The separation of variables like as I said , it is a technique for solving some partial differential equations and the idea is that you assume that the function you are looking for remember work in a call that function U of X of T can be written as a product of a function of X only and a function of T only.0027. So, we assume that the function were looking for can be written as were to call it X of X of X and T of T by the way were to have lots of equations today where were using X is a little x and T little t .0048. Be very careful to keep those straight the little X and little T are the variables X and T are the functions so, let me emphasize that right now, those are functions and the little X and little T are the variables try to be clear when I am writing my own notes which 1 is I am talking about.0066. And so, the point of assuming that you have a product of 2 functions like that is it is really easy to take derivatives because of your taking the partial derivative with respect to X that means that all the Ts are constant so, T of T just comes down as a constant we think of that as being a constant .0091. So, you just take the derivative of capital X with respect to little X X′ of X and then when you take the second derivative with respect to X your taking capital X″ of X and again the T is just a constant when you are looking at derivatives with respect to T is just the other way around .0111. The capital X you think of as being a constant and your taking derivatives of T with respect to little T so, here is the first derivative is T′ and in the second derivative is to″ .0134. So, you take these derivatives and you plug them into the partial differential equation and then let us see what happens with that what you trying to do is get all the X is on 1 side and all the Ts on the other.0149. It does not always work but it is it works for a lot of these differential equations will see some example lots of examples where does if you can do that then a very interesting thing happens you have 1 function that is only dependent on X 1 function that is only dependent on T and are equal to each other.0162. That means if you change T nothing happens on the left because X has not changed so, if you change T nothing happens which means it must be constant so, both of these functions must be constant so, and I call that constant .0184. And so, were to end up with a function of X = λ and independently a function of T = λ so, will reorganize these into 2 ordinary different differential equations a function of X = λ and a function of T = λ.0202. And hopefully you can solve these ordinary differential equations separately for capital X and T remember X that is a function of little X the variable little X T is a function of little T.0220. If you can solve both of those then if you rewrite our original guess remember was new of X and T = X of X and Multiply by T a T so, if you can find X X of X and T of T and you can put them back together and you can compile your solution to the original partial differential equation by multiplying together the solutions to these 2 ordinary differential equations.0237. So, that is the idea of separation of variables it is definitely something that will make more sense after we get some practice so, let us try that out on some examples.0271. So, in example 1 we are going to separation of variables to convert the following partial differential equation into 2 ordinary differential equations so, remember our guess for all of these for all of these separable partial differential equations is U of X T = capital X of X x T of T .0280. It looks like really need to know, some partial derivatives there so, U of X remember the T will be a constant so, that is X′ of X x T of T and U of X X will be X″ of X x T a T and it looks like really need U of T so, that would be now, holding the X is constant so, X of X x T sub' of little T.0308. To plug those into my differential equation so, plug-in plugged in to the pit the partial differential equation that we been given so, U of X X is X″ of X T of T + this is a little X now, U of T is X of X and T′ of T = 0 .0343. The idea remember is to try to solve this equation in such a way that you get all the X is on 1 side and all the Ts on the other so, I am in a move these terms over to the other side so, I get X″ of little X T of little T = - little X x X of X T′ of T .0376. Now, meant across the divide to get all my X on the left and all the T on the right so, capital X″ of X divided by up the - on the left - X X of X = π crust divide T′ of T divided by T of T .0406. Notice now, that I got all X is on 1 side all Ts on the other so, got a function of X = to a function of T and so, this must be constant that is the whole idea of separation of variables this must be constant.0429. So, then assented = to recall my constant λ and down to split that apart into 2 separate equations so, my - X″ of little X over X X of X X of X must be = to λ on 1 side and T′ of T divided by T of T = λ on the other side.0448. Each 1 of those I am not do a little bit of work to make him into a and ordinary differential equation so, X″ of X = a multiply the - over the other side - λ X x X of X and if I just pull the right-hand side over again X″ of X + λ X x X of X = 0 .0483. On the other side with the Ts and got T′ of T = λ x T of T and if I move that over T′ of T - λ x T of T = 0 .0518. So, I am not actually going to solve these differential equations the important thing is that I have done with the problem asked me which is to separate the partial differential equation into 2 ordinary differential equations 1 for in terms of X and 1 in terms of T so, let me summarize that here .0536. We got 2 ordinary differential equations 1 for X and 1 for T of T so, that is all we need to do for that problem me recap how we how we want about it so, we assume that U of XT = X of X x QT that is of an assumption we use for every single problem in this lecture.0559. That is always how you start out with separable with trying to separate partial differential equations than the X derivative the all the T constant take the X derivative second derivative is the same way holding T constant the T derivative you hold the X constant and take the derivative of T and you plug them back into the partial differential equation.0594. CNX″ of X x T + X x X x T′ = 0 down the work to get all my X on 1 side and all my Ts on the other that is the key part of separating a partial differential equation.0614. So, I did some algebra there and got only X is 1 place = to all my Ts but since I have a function of X = to a function of T it must be constant and so, I call that constant λ and then I just separated each of those equations the X = λ the Ts = to λ and each 1 of those I reorganized into a nice ordinary differential equation which you can then solve for X of X for T of T .0631. So, let us try out another example in example 2 were going you separation of variables on the following differential equation U of TT + U of XT + U of X = 0 so, let me start out the exact same guess that were going to make every single time for partial differential equations U of XT = X of X x T of T .0663. Then let me go ahead and find you what X is can be I Treating the T is constant so, X′ of X x T of T looking at the differential equation I am seeing which derivatives of that have to plug-in.0691. U of TT them in a skip a step here and skip right to the second derivative the X will come down is a constant every time and then will have the second derivative of T with respect to T now, U of XT that means you take the T derivative of the X derivative.0709. So, the X relative is a peer X′ of X x T of T the T derivative of that the X is now, constitutes X′ of X x T′ of T so, and take each 1 of those 3 derivatives and plug them into the partial differential equation and will see what happens there.0728. U TT is X of X T″ of T + U of XT is X′ of X x T′ of T + U of X is X′ of X x T of T = 0 .0749. Now, my goal here is to solve this in such a way that I get all the X is on 1 side and all the Ts on the other so, if I look at this is you got X′ here and here them in a factor that out so, X of X T″ of T + X′ of X T′ of T + T of T = 0.0774. Remove that term over to the other side so, X of X T″ of T = - X′ of X x that whole stuff in parentheses there in brackets + T of T and now, I am trying to get all my X on 1 side and all my Ts on the other so, I think I am going to cross multiply and divide .0807. On the left of me get keep double prime of T divided by that compound term T′ of T + T of T on the right I am getting at - X′ of X divided by X of X and I succeeded in getting all my X alongside only Ts on the other so, this again must be constant must be constant and so, it is got a be = to a constant that I am in a call λ.0841. Separate these 2 equations and solve each 1 separately so, my - X′ of X over X of X = λ and on the other side will have T″ of T x T′ of T + T of no divided by T′ of T + T a T is also, = to λ and each 1 of those I am going to sort out into an ordinary differential equation.0879. So, here I am going to get X′ of X move the - over the other side - λ X = - λ X of X and if I move that back I get X′ of X + λ X of X = 0 that is a nice ordinary differential equation there is only 1 variable in their which is little X .0912. On the right I see I got T″ of T = λ x T′ of T + T of T and if I move those terms over to the left I got T″ of T - λ T′ of T - λ T of T = 0 that is second-order ordinary differential equation in terms of T T of T.0949. And so, again I got 2 ordinary differential equations for X of X 1 for X of X and 1 for T of T so, I am done there let me go back over the steps in case you have any lingering questions.0985. U of XT were assuming that to be = to X of X x T of T that is the running assumption for all of these differential equations that were going to try to separate it is always separating it into a function of X x a function of T and the point of that is when you take derivatives U of X you just treating the T is a constant CX′ of X and in U of XT down here we take the T derivative of that so, we get X′ of X x T′ of T and U of TT is X of X x T″ of T.1010. The reason I took those 3 partial derivatives was because I was looking at the partial differential equation and so, I was planning ahead to see what lives can have to plug-in to the partial differential equation and so, I plugged each 1 of those in and then I am trying to solve it in such a way to get all the X over here all the Ts over there.1047. So, I see that I have not X′ of X on 2 terms here so, I factored that out and then if I move that over to the other side then I can cross multiply across divide to get all the Ts on the left and all the X on the right.1067. Which tells me that it must be = to some constant which am calling λ so, I separate that out into 2 separate equations = λ and then it is really 2 parallel tracks here 1 ends up with the differential equation in terms of X and 1 ends up with an ordinary differential equation in terms of T.1086. In example 3 were to use separation of variables to convert another differential equation into 2 ordinary differential equations so, let me remind you that our guests for every partial differential equation of forgot a use separation of variables is U of XT = X of X x T of T.1107. Now, looks like I am going to need the derivatives U X X and U TT so, U X X means were holding T constant and taking the second derivative of X should be to derivatives there x T of T and U TT means were holding X constant so, the X just comes down T″ of T and so, I am in a plugged each 1 of those into my differential equation there. 1930 It looks like I am not the plug-in you by itself as well so, U X X is X″ of X x T of T + U TT is X of X T″ of T + little T x the original U so, that is capital X of X x T of T = 0.1138. Now, I want to I I want to try and separate all the X from all the Ts and it looks like I got an X here and it X of X here thing a factor that term out of whenever I cancel got X″ of X x T of T + X of X x the quantity T″ of T + little T x T of T T T = 0.1198. Remove that term over to the other side so, I get X″ of X x T of T = - X of X x that compound term T″ of T + T x the original TFT and again I am going to cross multiply and divide so, thing the keep that - sin with the X so, get T1 - X″ of X divided by X so, it really divided both sides way X by - X of X. 2124 On the right-hand side I got T″ of T + little T x T of T divided by T of T and now, I successfully gotten all the X on 1 side all the Ts on the other so, again this thing must be constant and I am a call my constant λ that is kind of the universal and in the when you are using separation of variables.1242. And so, I got to a separate into 2 equations - X″ of X over X of X = λ on the 1 side and T″ of T + little T x TT of T over T a T = λ over here on the left and then I multiply my little my X of X out of the denominator and if you bring the - along with the with with the a X of X = - λ x X of X .1314. So, if I move that back X″ of X + λ x capital X of X = 0 on the right multiply my T out of the denominator T″ of T + little T x T of T = λ x T of T and if I move that term over the other side the other side of get T″ of T .1355. Now, I see about a T in each term here so, right this is T little T - λ just factoring their factoring out a T = 0 and so, what I discovered here is that my original partial differential equation just separated into 2 ordinary differential equations.1391. So, 2 ordinary differential equations for X of X and T of T remember were not solving these ordinary differential equations yet will do that in example little later.1416. We are just showing how you can start out with the partial differential equation make a good assumption of the beginning and hopefully separated into to ordinary differential equations so, we go back and recap that for you.1442. Start out with the original assumption U of XT = X of X x TT a T that is a running assumption we use that in every single example then where you go and that starts to vary on looking over the differential equation I see a many U of XX and U of TT .1455. So, for each 1 I hold the other variable constant and just take the second derivative with respect to the corresponding function and then I plugged each 1 of those into the partial differential equation that is what I did here plugged each 1 in.1473. And I want to separate the X for the Ts from each other and I noticed I got it X of X and in each of these terms so, I factor that out here and that enables me to pull that hold term that compound turned over to the other side and then we can cross multiply and divide and get all my X is on 1 side and all my T on the other.1490. So, since I got a function of X = to a function of T means of them both = constant animate a call that constant λ as usual and so, I separate those into 2 equations in terms of λ and 1 involves X and 1 involves Ts but there is no mixing between the 2 .1514. So, each 1 of those I can do a little algebra and rearrange it into a traditional ordinary differential equation 1 in terms of X and 1 in terms of T.1535. So, in example 4 were in you separate separation of variables on the heat equation which I said is out 1 of the most important partial differential equations and it is can be the 1 that were to spend a lot of time solving later we still have not completely solve the heat equation so, to take us a couple lectures to catch there.1545. So, were going to start solving that right now, are going to apply separation of variables to heat equation and there is a little note here that says will we get the step of using λ we should use - λ instead of λ for the separation constant that really make sense why we can do that at this stage but hopefully in a couple problems down the road from now, you will start to see why we worry is - λ instead of λ.1568. So, let us go ahead and try her separation of variables on this 1 starts out just like all the others where you start out with U of XT you assume that it can be written as X of X x T of T and then were to need some derivatives here U so, T X constant so, get X of X x T′ of T and will also, need U X X which means withhold T constant and will get the second derivative of X and then just T of T .1597. So, were to take these and plugged them in were to plug them in to our partial differential equation so, what we get here is U of T is capital X of X x T′ of T = α ² on a really care what out is at this point X″ of X x T of T and my goal here is a separate the X on 1 side the Ts on the other.1634. It is pretty easy for this 1 just get down X of X divided by X X everyone separate alone across multiplied by the other way around so, get my keys on let T′ of T divided by TT a T = anything to bring my α ² along with my T up at that down there and on the other side of got X″ of X divided by X of X .1671. By the way it is not so, obvious at this point y-axis but the α ² with 1 terms or the other there is a lot of different places you put it I am putting it here because I am kind of looking forward to a solution were going to be studying later for the heat equation.1706. So, do not worry about it right now, while putting out for ² in this term as opposed to over here with the X just bare with me and you will see will we get to problem later that it is it is a nice convention to have the α ² with the Ts.1719. Now, this is a constant because precisely because we got all X is on 1 side we got all Ts on the other and it says were going to use - λ instead of the usual λ th= - λ .1733. Remember we can untangle each 1 of those and try to I get some nice differential equations on the X side we get X″ of X divided by X of X = - λ.1750. So, if I clear my denominator there I get X″ of X = - λ x X of X and if I move it over to the other side I get X″ of X + λ x X of X = 0.1767. On the other side I am going to just keep my T′ of T divided by α ² x T of T this can only that = to - λ I think that is actually going to be the format there is going to be useful to solve later so, just can leave it in that form and what that means is that I have successfully separated my partial differential equation into 2 ordinary differential equations .1792. 1 in terms of X and 1 in terms of T. I see I left out a′ on my X here; that is X″ of X + λ x X of little X = 0 and here is my differential equation in terms of T. So, it reduced my partial differential equation into two ordinary differential equations .1822. X and T now, you want to hang onto these ordinary differential equations because were going use them in the next example let us go ahead and try to solve these ordinary differential equations in the next example so, make sure you understand where these differential equations are coming for the so, the to be ready to understand them and be ready to solve them and use them in the next example.1845. So, let me go back and remind you where they came from we start with this generic assumption U of XT is a product of a function of X and a function of T and then we take its derivative with respect to T and with respect to X twice so, each time you take a partial derivative it means hold the other variable constant and you take a derivative with respect whichever variable that is.1871. So, U of T hold the X constant and you take T′ of T U of double X that means you hold the T constant and you take the second derivative with respect to X so, we plugged those back into the differential equation and then we try to sort out the X on 1 side and sort out the Ts on the other side.1896. So, we do sort those out X it turns out to be fairly easy get them sorted then we can say that the whole thing there both = to the same constant and for reasons that will be useful later it does not really make sense at this point recalling that constant - λ .1916. So, this separates out into a function of X = - λ and was we solve that into a regular ordinary differential equation and then a function of T is = λ got a little squished function of T = - λ and just do not need that in that form because that is the form that were getting used to solve later .1934. So, the next example you are going solve these 2 ordinary differential equations and were to see what kind of answer we might get to the original partial differential equation.1960. So, let us go ahead and start working on that so, were going to solve the 2 ordinary differential equations below that came from the heat equation from the partial differential equation that was the heat equation.1969. I gives us a little hint here we want to assume that λ is + to try to find solutions that satisfy the boundary conditions U of 0T = U of LT = 0 so, were to try find solutions that also, satisfy that.1985. So, let me go ahead and try to solve this first 1 this is in terms of T this is actually really nicety separable first order differential equation will my α ² over to the other side so, I got T′ of T divided by T of T = - λ α ² .2003. Now, I can integrate both sides can integrate both sides here now, the integral because I have T′ of T over T of T the integral that is exactly natural log of T of T and that is both the = to integrate the right-hand side it is were integrating with respect to T so, is just - λ α ² T + a constant .2027. I want to solve for T here so, when I raised each of both sides so, get T = E to the - λ α ² T + a constant but remember this is the same as E - λ α ² T x EC and that EC were just think of that as being a K being another constant K and so, my function T actually pretty easy to find here was a constant x E - λ α ² T.2056. Let us hang onto that were to be using it later so, before we go on to the next slide let me recap what happened on this slide to try to solve each 1 of these ordinary differential equations that come up in the heat equation and so, far just try to solve the 1 for T try to solve 1 for X on the next slide .2097. I am just trying to solve the 1 for T so, I move the α ² over to the other side key observation here is that the derivative of T of T of natural log of T would be 1 over T x its derivative so, the integral of T priority over T of T is very easy it is just natural log of T of T.2117. Integrate the right-hand side since were integrating with respect to T I just get - λ α ² x T + a constant remember it is very important to add a constant when you are integrating.2141. By the way if it is been a while since you watch the initial lectures here on educator on differential equations might be good to go back and check those out because I am using stuff that we learned back in those early lectures on differential equations .2154. So, to get rid of the natural log I raised E of both sides here and now, by laws of exponents remember E of the X + Y = E X x E Y so, since I got a + C here that is been like multiplied by E the C is just another constant so, I called a K and now, if you solve for T we get K x E - λ α ² x T.2168. So, we still have not looked at the equation for X at all yet so, that is organized go ahead and do on the next slide so, in were still working on example 5 here we solve the equation for T so, that that is looking good we have to solve the equation for X .2198. Now, this is a second order linear ordinary differential equation we had a whole lecture X a 3 lectures on how to solve these earlier on in the differential equations lecture series here on educator.com so, if you do not number how to solve those here is a really quick review.2217. You look at the characteristic equation R² + λ = 0 and using of this X of X you thing that is being X to the 0 derivative so, put R the 0s that is why there is no R right here .2235. So, were solving trying to solve for λ so, R ²s = to - λ and R = the ² root of - λ the + or - there let me remind you that we were given an assumption here which was that λ that was bigger than 0 .2254. For that means that - λ is less than 0 and so, that means the ² root of - λ is complex so, were dealing with complex numbers here it is I x the ² root of + λ.2275. So, we can have to deal with complex solutions here now, we did talk about what happens when you get complex solutions the characteristic equation that was back in 1 of earlier lectures here on educator.com back and check it out if you do not remember but the short version is that your answers look like sins and cosine.2297. So, our R here is + or - I x root λ and so, my solution X of X is I think in our earlier lectures in C1 and C2 C1 x the cosine + C2 x sin call it A and B for my constants this time so, it is a x well remember if you have a + b I then you go E AT x cosine of B T + C2 η T x sin of BT.2317. Sum up all that format simply in a change around what I am calling my constants little bit instead of C1 I am going to use a and B for C1 and C2 this a is just 0 so, you get E 0T which is just 1 so, those terms are both 1 and I get cosine of well my B is the ² root of λ so, I may go constant x quickly go ahead and call that C1 and C2 so, will be too confusing.2369. C1 x cosine of the ² root of λ and that is my variable here is X not T so, put it X in there + C2 x sin of the ² root of λ X and let me go ahead and change that C1 and C2is and call them a and B.2400. So, a cosine the ² root of λ X + Bsin the ² root of λ X so, that is my X of X and my U my my original function that I was looking for U of XT let me remind you what format that had U of XT was = to X of X this is our assumption.2424. We try to separate the variables x T of T and I have already figured out my T of T there is right there and I figured out my X of X so, let me combine those so, that = a cosine the ² of λ X + B x the sin of the ² root of λ X all x K x E - λ α ² T .2459. By the way that K drop it in the reason going to drop it is because if I put a K right here it could be absorbed right into the A and B so, just a cancel that K out and put E- λ α ² x T that is my U of XT .2491. But I am trying to satisfy my 2 boundary conditions so, let me remind you what those were my first boundary condition was U of XT are sorry U of 0T we label as a boundary condition is probably pin a while since you have seen it is a boundary condition this is given to us as part of the differential equation well as part of the package along with the differential equation was that U of 0T had to be = to 0 .2510. So, let us see what we get number that means your plug-in X = 0 let us see what we get if we plug-in X = 0 into our U of XT here so, U of 0T = a x cosine of 0+ B x sin of 0 and all of that gets multiplied by E of the - λ α ² T and that is what is to come out to be 0.2547. So, cosine a 0 is just 1 sin of 0 is 0 and so, what we get is a x E - λ α ² T = 0 on the strength of that we can say that a must be = to 0 .2576. So, we will not be needing to use that a term now, let us look at how were going to incorporate the other boundary condition and let me remind you what that was so, my U of XT simplifies down a bit when U of XT since our a term is gone away it is just B x the sin of root λ X all of that multiplied by E - λ α ² T .2596. And were told by our other boundary condition let me remind you this came as part of the package when we originally got the differential equation so, our other boundary condition was that U of L to had to be 0 remember that represented plug-in X = L which was the right-hand end of the Rod in the heat equation.2636. So, let us plug-in X = L we get B x sin of root λ L x E - λ α ² T = 0 now, we do not want it said B = 0 because otherwise our entire solution disappears .2670. Also, our E - λ α ² T we do not want that to be 0 because and even function can never be 0s that is never 0 I do not want to set B to be = to 0 because otherwise my whole solution disappears and so, what I am forced to look at is the idea that sin of root λ L = 0 .2692. So, I have to figure out what my possibilities there are to make sin of root λ L = 0 so, sin of what = 0 will I know, that sin of any multiple of π in the integer multiple of pies = to 0 so, this will work if I take root λ L = an integer multiple of π where N is an integer .2719. And so, the only thing remember L was the length of the Rod in the heat equation so, that is a constant it is established all the way through we can change that so, do not think I can solve for his is λ and so, if I will be L to the other side and ² both sides 4618 I get λ = N² π ² over L ² .2750. This will work for any value of N we choose so, we get 1 solution for each N so, I am and I write my X sub and of X remember I had an X for what I am discovering is that for each value of NI get a different solution .2784. So, and there is a B there the B could be anything so, do not know, yet what I want B to B but BN sin I have a root λ L there so, remember root λ is N π over L sin of root λ X so, N π X over L and my T = E - now, λ this is rather messy N² π ² over L ² .2813. I still have not α ² T in my exponent for E so, let me extend that α ² T such my T of T and so, ultimately get 1 solution for each end my U of an of XT is remember you multiply the X of X and the T of T so, multiply these 2 things together it is B sub N sin of N π X over L E - N² π ² α ² T all divided by L ² .2859. Pretty messy that is my solution to the heat equation that is still compatible with both of the boundary conditions that you tend to get package along with the heat equation so, that was quite a mouthful of me go back and recap everything came from.2914. We started out with well we had to solve these 2 ordinary differential equations for T of T and X of X so, we solved T of T on the previous slide previous slide so, on the previous side we figured out a form for T of T and we figured out that it was a constant x E - λ α ² T that constant I am X the going to omit that constant and the reason is because down at this step and gets multiplied by X which has Constance attach to it.2937. Anyway so, I let that K be absorbed into the Constance for the the the X functions here so, not to worry about that constant what I do have to focus on next is solving this ordinary differential equation for X and I learned how to do that back in our lessons on second-order differential equations we have an earlier lecture on that read here in the differential equations lecture series on educator.com .2980. So, if you do not remember how to solve those just check back the earlier lecture and you will see that we did a bunch of second-order differential equations and we had the characteristic equation we had cases where there were real roots where there were complex roots where there were repeated roots so, were through all those different equated different cases in different lectures.3012. What happens with this 1 is we get R ² + λ = 0 so, R ² is - λ and so, R turns into the ² root of - λ and were given that λ that was greater than 0 so, - λ is less than 0 so, really trying to here is take the ² of a - number which means we have complex solutions .3039. We learned before that will we have complex solutions A+ B I this is the format of our solution except word I using slightly different variables here so, you translate a little bit this C1 and C2 I a relabeled a and B and that is a little confusing as it is not the same as this a and B.3064. This a and B will be a was just 0 because it is 0+ or - I root λ and the B this be was root λ so, that is where I got to cosine e of that is the B there and that is the B there also, are variable now, is X instead of T we use T before but now, are using axis and so, and I also, wanted to change the Constance C1 and C2 a want to call them a and B a new a and B .3088. So, it is a little confusing with a different variables I apologize for that there is nothing very deeply mathematical going on I am just kind according the solution from the earlier lecture earlier lecture on complex roots on complex roots.3121. So, we have now, found our solution for X of X but we want to match the boundary conditions so, we want to compile our solution together with our solution from T of T so, remember this is our guess with separation of variables.3141. U is the product of X of X and T of T so, here on multiplying them together and are boundary condition was the U 0T = 0 which means you plug-in X = 0 so, I plug-in X = 0 here and that when the plug-in X = 0 you get the cosine a 0 is just 1 the sin of 0 is just 0 and so, you get a x E the something = 0.3162. What you do something this term is never = to 0 so, we must have a = 0 which means or hold term here drops out of our solution so, were just down to the simpler solution Bx the sin term x that exponential term .3191. Now, we bring in our second boundary condition which says that when you plug-in X = L then you must be getting 0 so, if we plug-in X = L here then we get complicated expression = 0 now, the E term again can not be 0 that that can not be 0.3213. I do not want B term to be 0 because it B = 0 in my entire solution disappears and I just get 0 so, I do not want to be the 0 which means I must have sin = to 0 .3214. So, if sin = 0 then I have to think about what angles have sin 0 and I remember that sin of any multiple or any integer multiple of π = 0.3248. That means root λ L must be N π for some integer and an exit is works no matter what and is as long as it is an integer sorted get a whole family of solutions with different integers if I saw for λ I get λ is N² π ² over L ² and I am just going to plug those back into my solution here back in for root λ here and back in for λ here.3261. So, the X part gives us n root λ is N π X over L the E part that the T part gives us E - λ which is - N ² π ² over L ² and we what we multiply those together and putting subscript and 0 because remember getting different solutions for each value of that so, were to have a whole bunch of solutions here .3289. So, we have BNx the X part x the T part so, that is our solution to the partial differential equation that also, satisfies both boundary conditions that is actually officially the end of this lecture here but let me go ahead and give you little Teaser for the next lecture.3318. We write down what we just figured out U of an of XT = BN x sin of N π X over L just figure this out E - N ² π ² α ² T over L ².3339. Now, let me remind you what we did there were trying to solve the heat equation and the original heat equation me make sure I write it down correctly was U of T = α ² U of X X what we have done now, as we solve that using separation of variables.3366. We also, had 2 boundary conditions U of 0T = 0 and U of LT = 0 and what were doing in this example was confirming a solution that satisfied both of those boundary conditions.3395. Now, there was 1 more condition which was the initial condition, U of X 0 = F of X and what we have not done yet is satisfied that initial condition so, that take quite a bit of work that is going to be the content of the next couple of lectures here on educator.com .3415. Were to finish off the solution of the heat equation , we are to try to make it match that initial condition so, hope you stick around and watch those lecture so, we can finish solving the equation together.3436. In the meantime, this is the end of our lecture on separation of variables, and I will just remind you that this is part of the differential equations lecture series here on educator.com. My name is Will Murray, thanks very much for joining us, bye bye.3449. OR. ### Start Learning Now. Our free lessons will get you started (Adobe Flash® required).
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https://www.gradesaver.com/textbooks/math/algebra/college-algebra-10th-edition/chapter-r-section-r-4-polynomials-r-4-assess-your-understanding-page-48/47
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# Chapter R - Section R.4 - Polynomials - R.4 Assess Your Understanding - Page 48: 47
$x^3+3x^2-2x-4$
#### Work Step by Step
Multiply each term in the first parentheses set by each term in the second parentheses set. $(x+1)(x^2+2x-4)$ $x^3+2x^2-4x+x^2+2x-4$ Combine like terms $x^3+3x^2-2x-4$ To put in polynomial standard form, make sure to rearrange the numbers by degree from high to low going left to right
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# Chapter R - Section R.4 - Polynomials - R.4 Assess Your Understanding - Page 48: 47. $x^3+3x^2-2x-4$. #### Work Step by Step. Multiply each term in the first parentheses set by each term in the second parentheses set.
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$(x+1)(x^2+2x-4)$ $x^3+2x^2-4x+x^2+2x-4$ Combine like terms $x^3+3x^2-2x-4$ To put in polynomial standard form, make sure to rearrange the numbers by degree from high to low going left to right. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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PLATE VELOCITY
a math-based problem asking you to calculate plate velocity using a hot spot track.
Name:
Calculating plate velocity using hot spots
To find out the speed (velocity) of motion of plates you will need to use the following equation
t
d
v
=
with v= velocity (or rate of motion), d= distance and t=time. To get full credit (assuming you have done all your work correctly) you need to also show all of your mathematical work, as well as the unit conversions. Your textbook has conversion tables in the Appendix. However to make it easier for you I am including the ones you will need below:1km=1000 m
1m=100 cm
therefore: 1km= 100,000 cm
The diagram below is a series of volcanic islands, with their ages labeled. In the next page, answer the following questions:
1. Provide 2 observable lines of evidence (they must be observations that you can make simply looking at the map) that this is a hot spot chain and not an island arc . These must be observations you can make strictly from looking at the map and not related to any other knowledge about hot spots or island arcs.
2. In the above diagram, in what direction is the plate moving?
3. How fast is the plate moving (in cm/year)? SHOW YOUR WORK
N
Volcano erupting
4 million years old
2 million years old
0 200 km 400 km 600 km 800km
8 million years old
6 million years old
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PLATE VELOCITY. a math-based problem asking you to calculate plate velocity using a hot spot track.. Name:. Calculating plate velocity using hot spots. To find out the speed (velocity) of motion of plates you will need to use the following equation. t. d. v. =. with v= velocity (or rate of motion), d= distance and t=time. To get full credit (assuming you have done all your work correctly) you need to also show all of your mathematical work, as well as the unit conversions. Your textbook has conversion tables in the Appendix. However to make it easier for you I am including the ones you will need below:1km=1000 m. 1m=100 cm. therefore: 1km= 100,000 cm. The diagram below is a series of volcanic islands, with their ages labeled.
|
In the next page, answer the following questions:. 1. Provide 2 observable lines of evidence (they must be observations that you can make simply looking at the map) that this is a hot spot chain and not an island arc . These must be observations you can make strictly from looking at the map and not related to any other knowledge about hot spots or island arcs.. 2. In the above diagram, in what direction is the plate moving?. 3. How fast is the plate moving (in cm/year)? SHOW YOUR WORK. N. Volcano erupting. 4 million years old. 2 million years old. 0 200 km 400 km 600 km 800km. 8 million years old. 6 million years old.
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https://tnboardsolutions.com/samacheer-kalvi-11th-physics-guide-chapter-8/
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Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Physics Guide Pdf Chapter 8 Heat and Thermodynamics Text Book Back Questions and Answers, Notes.
## Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics
### 11th Physics Guide Heat and Thermodynamics Book Back Questions and Answers
Part – I:
I. Multiple choice questions:
Question 1.
In hot summer after a bath, the body’s:
(a) internal energy decreases
(b) internal energy increases
(c) heat decreases
(d) no change in internal energy and heat
(a) internal energy decreases
Question 2.
The graph between volume and temperature in Charle’s law is:
(a) an ellipse
(b) a circle
(c) a straight line
(d) a parabola
(c) a straight line
Question 3.
When a cycle tyre suddenly bursts, the air inside the tyre expands. This process is:
(a) isothermal
(c) isobaric
(d) isochoric
Question 4.
An ideal gas passes from one equilibrium state (P1, V1, T1, N) to another equilibrium state (2P1, 3V1, T2, N). Then:
(a) T1 = T2
(b) T1 = $$\frac{\mathrm{T}_{2}}{6}$$
(c) T1 = 6T2
(d) T1 = 3T2
(b) T1 = $$\frac{\mathrm{T}_{2}}{6}$$
Question 5.
When a uniform rod is heated, which of the following quantity of the rod will increase:
(a) mass
(b) weight
(c) centre of mass
(d) moment of inertia
(d) moment of inertia
Hint:
M.I. = MK²
During heating K would be increased. Hence moment of inertia will increase.
Question 6.
When food is cooked in a vessel by keeping the lid closed, after some time the steam pushes the lid outward. By considering the steam as a thermodynamic system, then in the cooking process:
(a) Q > 0, W > 0
(b) Q < 0, W > 0
(c) Q > 0, W < 0
(d) Q < 0, W < 0
(a) Q > 0, W > 0
Hint:
Q > 0; W > 0
∆Q = ∆U1 + ∆W.
∴ ∆Q ∝ ∆W
When ∆Q > 0
∆W > 0.
Question 7.
When you exercise in the morning, by considering your body as a thermodynamic system, which of the following is true?
(a) ∆U > 0, W > 0
(b) ∆U < 0, W > 0
(c) ∆U< 0, W < 0
(d) ∆U = 0, W > 0
(b) ∆U < 0, W > 0
Hint:
∆Q = ∆U + ∆W
When exercise is performed, internal energy is utilised (∆U < 0). So it will decrease.
But as internal energy is utilised, the exercise (∆W) will be more. So work will be more.
∴ ∆W > 0
∆U < 0 W > 0
Question 8.
A hot cup of coffee is kept on the table. After some time it attains a thermal equilibrium with the surroundings. By considering the air molecules in the room as a thermodynamic system, which of the following is true?
(a) ∆U > 0, Q = 0
(b) ∆U > 0, W < 0
(c) ∆U > 0, Q > 0
(d) ∆U = 0, Q > 0
(c) ∆U > 0, Q > 0
Hint: During the thermal equilibrium with surrounding as heat energy (Q) is increased, internal energy will be increased.
∴ ∆U > 0 Q > 0
Question 9.
An ideal gas is taken from (Pi, Vi) to (Pf, Vf) in three different ways. Identify the process in which the work done on the gas the most.
(a) Process A
(b) Process B
(c) Process C
(d) Equal work is done in Process A, B & C
(b) Process B
Hint:
When work is done on the gas, compression takes place. Final pressure Pf will be increased and remains constant. It is shown by process B.
Question 10.
The V-T diagram of an ideal gas which goes through a reversible cycle A → B → is shown below. (Processes D → A and B → C are adiabatic)
The corresponding PV diagram for the process is (all figures are schematic)
Question 11.
A distant star emits radiation with maximum intensity at 350 nm. The temperature of the star is:
(a) 8280 K
(b) 5000 K
(c) 7260 K
(d) 9044 K
(a) 8280 K
Hint:
Question 12.
Identify the state variables given here.
(a) Q, T, W
(b) P,T,U
(c) Q, W
(d) P,T,Q
(b) P,T,U
Question 13.
In an isochoric process, we have:
(a) W = 0
(b) Q = 0
(c) ∆U = 0
(d) ∆T = 0
(a) W = 0
Hint:
Work done in an isochoric process is zero.
Question 14.
The efficiency of a heat engine working between the freezing point and boiling point of water is: (NEET2018)
(a) 6.25%
(b) 20%
(c) 26.8%
(d) 12.5%
(b) 20%
Hint:
T2 = 0° C = 0 + 273 = 273 K
T1 = 100° C = 100 + 273 = 373 K
Question 15.
An ideal refrigerator has a freezer at a temperature of -12°C. The coefficient of performance of the engine is 5. The temperature of the air (to which the heat ejected) is:
(a) 50°C
(b) 45.2°C
(c) 40.2°C
(d) 37.5°C
(c) 40.2°C
Hint:
Question 1.
‘An object contains more heat’- is it a right statement? If not why?
When heated, an object receives heat from the agency. Now object has more internal energy than before. Heat is the energy in transit and which flows from an object at a higher temperature to an object at lower temperature. Heat is not a quantity. So the statement I would prefer “an object contains more thermal energy”.
Question 2.
Obtain an ideal gas law from Boyle’s and Charles’law.
According to Boyle’s law, Pressure ∝ $$\frac { 1 }{ 2 }$$
i.e., P ∝ $$\frac { 1 }{ V }$$
According to Charle’s law,
Volume ∝ Temperature, i.e., V ∝ T
By combining these two laws, we get
PV= CT … (1)
Where C is a positive constant
But C = k x Number of particles (N)
C = kN
Where k – Boltzmann’s constant.
∴ Equation (1) becomes
PV = NkT
Question 3.
Define one mole.
One mole of any substance is the amount of that substance which contains Avogadro number (NA) of particles (such as atoms or molecules).
Question 4.
Define specific heat capacity and give its unit.
Specific heat capacity of a substance is defined as the amount of heat energy required to raise the temperature of 1 kg of a substance by 1 Kelvin or 1 °C.
The SI unit for specific heat capacity is J kg-1 K-1.
Question 5.
Define molar specific heat capacity.
Heat energy required to increase the temperature of one mole of substance by IK or 1°C.
Question 6.
What is a thermal expansion?
Thermal expansion is the tendency of matter to change in shape, area, and volume due to a change in temperature.
Question 7.
Give the expressions for linear, area and volume thermal expansions.
(i) Linear expansion:
$$\frac { ∆L }{ L }$$ = α ∆T ⇒ αL = $$\frac { ∆L }{ L∆T }$$
(ii) Area expansion:
$$\frac { ∆A }{ A }$$ = 2α ∆T ⇒ αA = $$\frac { ∆A }{ A∆T }$$
(iii) Volume expansion:
$$\frac { ∆V }{ V }$$ = 3α ∆T ⇒ αv = $$\frac { ∆V }{ V∆T }$$
Question 8.
Define latent heat capacity. Give its unit.
Latent heat capacity of a substance is defined as the amount of heat energy required to change the state of a unit mass of the material. The SI unit for latent heat capacity is J kg-1.
Question 9.
State Stefan-Boltzmann law.
The total amount of heat energy radiated per second per unit area of a black body is directly proportional to the fourth power of its absolute temperature.
Question 10.
What is Wien’s law?
The wavelength of maximum intensity of emission of a black body radiation is inversely proportional to the absolute temperature of the black body.
Question 11.
Define thermal conductivity. Give its unit.
The quantity of heat transferred through a unit length of a material in a direction normal to Unit surface area due to a unit temperature difference under steady-state conditions is known as the thermal conductivity of a material.
Question 12.
What is a black body?
A black body is one which neither reflects nor transmits but absorbs whole of the radiation incident on it.
Question 13.
What is a thermodynamic system? Give examples.
Thermodynamic system: A thermodynamic system is a finite part of the universe. It is a collection of a large number of particles (atoms and molecules) specified by certain parameters called pressure (P), Volume (V), and Temperature (T). The remaining part of the universe is called the surrounding. Both are separated by a boundary.
Examples: A thermodynamic system can be liquid, solid, gas, and radiation.
Question 14.
What are the different types of thermodynamic systems?
(i) Open system can exchange both matter and energy with the environment.
(ii) Closed system exchange energy but not matter with the environment.
(iii) Isolated system can exchange neither energy nor matter with the environment.
Question 15.
What is meant by ‘thermal equilibrium’?
Two systems are said to be in thermal equilibrium with each other if they are at the same temperature, which will not change with time.
Question 16.
What is mean by state variable? Give example.
The quantities that are used to describe the equilibrium states of a Thermodynamic system. .
Example: Pressure, Volume, Temperature.
Question 17.
What are intensive and extensive variables? Give examples.
Extensive variable depends on the size or mass of the system.
Example: Volume, total mass, entropy, internal energy, heat capacity etc.
Intensive variables do not depend on the size or mass of the system.
Example: Temperature, pressure, specific heat capacity, density etc.
Question 18.
What is an equation of state? Give an example.
The equation that relates the state variables in a specific manner is called equation of state.
Examples:
(i) Gas equation
PV = NkT
(ii) Vander Waal’s equation, .
(p +$$\frac{a}{v^{2}}$$)(v – b) = RT
Question 19.
State Zeroth law of thermodynamics,
The zeroth law of thermodynamics states that if two systems, A and B, are in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other.
Question 20.
Define the internal energy of the system.
The sum of kinetic and potential energies of all the molecules of the system with respect to the centre of mass of the system.
Question 21.
Are internal energy and heat energy the same? Explain.
Internal energy and thermal energy do not mean the same thing, but they are related. Internal energy is the energy stored in a body. It increases when the temperature of the body rises, or when the body changes from solid to liquid or from liquid to gas.
“Heat is the energy transferred from one body to another as a result of a temperature difference.”
Question 22.
Define one calorie.
To raise 1 g of an object by 1°C , 4.186 J of energy is required. In earlier days the heat energy was measured in calorie.
1 cal = 4.186J
Question 23.
Did joule converted mechanical energy to heat energy? Explain.
In Joule’s experiment, when the masses fall, the paddle wheel turns. The frictional force between paddle wheel and water causes a rise in temperature of water. It is due to the work done by the masses. This infers that gravitational potential energy is transformed into internal energy of water. Thus Joule converted mechanical energy into heat energy.
Question 24.
State the first law of thermodynamics.
‘Change in internal energy (AU) of the system is equal to heat supplied to the system (Q) minus the work done by the system (W) on the surroundings’.
Question 25.
Can we measure the temperature of the object by touching it?
No, we can’t measure the temperature of the object touching it. Because the temperature is the degree of hotness or coolness of a body. Only we can sense the hotness or coolness of the object.
Question 26.
Give the sign convention for Q and W.
Question 27.
Define the quasi-static process.
A quasi-static process is an infinitely slow process in which the system changes its variables (P, V, T) so slowly such that it remains in thermal, mechanical, and chemical equilibrium with its surroundings throughout.
Question 28.
Give the expression for work done by the gas.
In general, the work done by the gas by increasing the volume from Vi to Vf is given by
W = $$\int_{V_{i}}^{V_{f}} \mathrm{P} d \mathrm{~V}$$
Question 29.
What is PV diagram?
PV diagram is a graph between pressure P and volume V of the system. The P-V diagram is used to calculate the amount of work done by the gas during expansion or on the gas during compression.
Question 30.
Explain why the specific heat capacity at constant pressure is greater than the specific heat capacity at constant volume.
When increasing the temperature of a gas at constant pressure, gas expands and consume some heat to do work. It is not happened, while increasing the temperature of a gas at constant volume. Hence less heat energy is required to increase the temperature of the gas at constant volume. Hence CP is always greater than Cv.
Question 31.
State the equation of state for an isothermal process.
The equation of state for isothermal process is given by,
PV = constant
Question 32.
Give an expression for work done in an isothermal process.
W = μRTln $$\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)$$
Question 33.
Express the change in internal energy in terms of molar specific heat capacity.
When the gas is heated at the constant volume the temperature increases by dT. As no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU. dU = µωdT
Question 34.
Apply first law for (i) an isothermal (ii) adiabatic (Hi) isobaric processes.
(i) For an isothermal process temperature is maintained as constant. Hence,
∆V= 0
∴ W = P∆V = 0
According to first law of thermodynamics
∆U = Q – W
= Q – 0
= Q
∴ ∆U = Q
(ii) For adiabatic process, Q = 0
By applying first law of thermodynamics
∆U = Q – W
∆U = |- W| = W
∆U = W = $$\frac{\mu R}{\gamma-1}$$ = (Ti – Tf)
(iii) (a) For isobaric expansion Q > O
According to first law of thermodynamics
∆U = – Q – W
∆U = Q – P∆V
(b) For isobaric compression Q < O.
According to first law of thermodynamics
∆U = – Q – W
= – Q – P∆V
Question 35.
Give the equation of state for an adiabatic process.
The equation of state for an adiabatic proc ess is given by,
PVγ = Constant
Question 36.
State an equation state for an isochoric process.
The equation of state for an isochoric process is given by,
P = ($$\frac { μR }{ V }$$)T
Question 37.
if the piston of a container is pushed fast inward. Will the ideal gas equation be valid in the intermediate stage? ff not, why?
When the piston is compressed so quickly that there is no time to exchange heat to the surrounding, the temperature of the gas increases rapidly. In this intermediate stage the ideal gas equation be not valid. Because this equation can be relates the pressure, volume and temperature of thermodynamic system at equilibrium.
Question 38.
Draw the PV diagram for,
(a) isothermal process
(c) lsobaric process
(d) lsochoric process
(a) isothermal process:
(c) lsobaric process:
(d) lsochoric process:
Question 39.
What is a cyclic process?
It is a process in which the thermodynamic system returns to its initial state after undergoing a series of changes.
Question 40.
What is meant by a reversible and irreversible processes?
Reversible processes: A thermodynamic process can be considered reversible only if it possible to retrace the path in the opposite direction in such a way that the system and surroundings pass through the same states as in the initial, direct process.
Irreversible processes: All natural processes are irreversible. Irreversible process cannot be plotted in a PV diagram, because these processes cannot have unique values of pressure, temperature at every stage of the process.
Question 41.
State Clausius form of the second law of thermodynamics.
“Heat energy always flows from hotter object to colder object spontaneously”. This is known as the Clausius form of second law of thermodynamics.
Question 42.
State Kelvin-Planck statement of second law of thermodynamics.
It is impossible to construct a heat engine that operates in a cycle, whose sole effect is to convert the heat completely into work.
Question 43.
Define heat engine.
Heat engine is a device which takes heat as input and converts this heat in to work by undergoing a cyclic process.
Question 44.
What are processes involves in a Carnot engine?
1. Isothermal expansion
3. Isothermal compression
Question 45.
Can the given heat energy be completely converted to work in a cyclic process? If not, when can the heat can completely converted to work?
No.
According to second law it can’t be completely converted. It is not possible to convert heat completely converted into work, in a cyclic process. But, in Non-cyclic process, heat can be completely converted into work.
Question 46.
State the second law of thermodynamics in terms of entropy.
“For all the processes that occur in nature (irreversible process), the entropy always increases. For reversible process entropy will not change”. Entropy determines the direction in which natural processes should occur.
Question 47.
Why does heat flow from a hot object to a cold object?
Because entropy increases when heat flows from hot object to cold object. If heat were to flow from a cold to a hot object, entropy will be decreased. So second law thermodynamics is violated.
Question 48.
Define the coefficient of performance.
The ratio of heat extracted from the cold body (sink) to the external work done by the compressor W.
COP = ß = $$\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{W}}$$.
Question 1.
Explain the meaning of heat and work with suitable examples.
The spontaneous flow of energy from the object at higher temperature to the one at lower temperature. When they are contact. This energy is called heat. This process of energy transfer from higher temperature object to lower temperature object is called heating.
Example: ‘A hot cup of coffee has more heat’. This statement is correct or wrong?
When heated, a cup of coffee receives heat from the stove. Once the coffee is taken from the stove, the cup of coffee has more internal energy than before. ‘Heat’ is the energy in transit and that flows from a body at higher temperature to an other body at lower temperature. Heat is not a quantity. So the statement ‘A hot cup of coffee has more heat’ is wrong, instead ‘coffee is hot’ will be appropriate.
Meaning of Work: If a body undergoes a displacement then only work is said to be done by the body. Work is also the transfer of energy by other means such as moving a piston of a cylinder containing gas.
Example: When you rub your hands against each other the temperature of the hand is increased. You have done some work on your hands by rubbing. The temperature of the hands increases due to this work. Now when you place your hands on the chin, the temperature of the chin increases.
This is because the hands are at higher temperature than the chin. In the above example, the temperature of hands is increased due to work and temperature of the chin is increased due to heat transfer from the hands to the chin.
The temperature in the system will increase and sometimes may not by doing work on the system. Like heat, work is also not a quantity and through the work energy is transferred to the system.
Question 2.
Discuss the ideal gas laws.
(i) Boyles law: When the gas is kept at constant temperature, the pressure of the gas is inversely proportional to the volume.
P ∝ $$\frac { 1 }{ v }$$
(ii) Charles law: When the gas is kept at constant pressure, the volume of the gas is directly proportional to absolute temperature. V ∝ T
(iii) By combining these two equations we have PV= CT … (1)
(iv) Let the constant C as k times the number of particles N. Here k is the Boltzmann constant and it is found to be a universal constant. So the ideal gas law can be stated as follows,
PV = NkT … (2)
If the gas consists of p mole of particles then total number of particles is
N = μNA … (3)
Substituting the value of N in the equation (2) we get
PV = μNAkT
Here NAk = R
So, the ideal gas law can be written for μ mole of gas is
PV= pRT
It is called equation of state for an ideal gas.
Question 3.
Explain in detail the thermal expansion.
Thermal expansion is the tendency of matter to change in shape, area, and volume due to a change in temperature. When a solid is heated, its atoms vibrate with higher amplitude about their fixed points. At that time, relative change in the size of solids is small.
Liquids expand more than solids because, they have less intermolecular forces than solids. In the case of gas molecules, the intermolecular forces are almost negligible and so they expand much more than solids.
Linear Expansion: In solids, for a small change in temperature ∆T, the fractional change in length $$\frac { ∆L }{ L }$$ is directly proportional to ∆T.
$$\frac { ∆L }{ L }$$ = αL∆T
Therefore, αL = $$\frac { ∆L }{ L∆T }$$
Area Expansion: For an infinitesimal change in temperature ∆T the fractional change in area of a substance is directly proportional to ∆T and it can be written as
$$\frac { ∆A }{ A }$$ = αA∆T
Therefore, αA = $$\frac { ∆A }{ A∆T }$$
Volume Expansion: For a very small change in temperature ∆T the fractional change in volume $$\frac { ∆V }{ V }$$ of a substance is directly proportional to AT.
$$\frac { ∆V }{ V }$$ = αV∆T
Therefore, αV = $$\frac { ∆L }{ V∆T }$$
Question 4.
Describe the anomalous expansion of water. How is it helpful in our lives?
On heating liquids expand and contract on cooling at moderate temperatures. But water exhibits an anomalous behavior. It contracts on heating between 0°C and 4°C. The volume of the given amount of water decreases as . it is cooled from room temperature, until it reach 4°C . Below 4°C the volume increases and so the density decreases. It means that the water has a maximum density at 4°C. This behaviour of water is called anomalous expansion of water.
During winter in cold countries, the surface of the lakes would be at lower temperature than the bottom. Since the solid water (ice) has lower density than its liquid form,’below 4°C, the frozen water will be on the top surface above the liquid water (ice floats). It is due to the anomalous expansion of water. The water in lakes and ponds freeze only at the top. So, the species living in the lakes will be safe at the bottom.
Question 5.
Explain Calorimetry and derive an expression for final temperature when two thermodynamic systems are mixed.
Calorimetry means the measurement of the amount of heat released or absorbed by thermodynamic system during the heating process. When a body at higher temperature is brought in contact with another body at lower temperature, due to transformation of heat. The heat lost by the hot body is equal to the heat gained by the cold body. No heat is allowed to escape to the surroundings. It can be mathematically expressed as
Qgain = – Qlost
Qgain + Qlost = 0
Heat gained or lost can be measured with a calorimeter.
When a sample is heated at high temperature (T1) and immersed into water at room temperature (T2) in the calorimeter. After some time both sample and water reach a final equilibrium temperature Tf. Since the calorimeter is insulated, heat given by the hot sample is equal to heat gained by the water.
Qgain = – Qlost
As per the sign convention, the heat energy lost is denoted by negative sign and heat gained is denoted as positive.
From the definition of specific heat capacity,
Qgain= m2s2(Tf – T2)
Qlost = m1s1(Tf – T1)
Here s1 and s2 specific heat capacity of hot sample and water respectively.
So we can write,
Question 6.
Discuss various modes of heat transfer.
There are three modes of heat transfer: Conduction, Convection and Radiation.
Conduction: Conduction is the process of direct transfer of heat through matter due to temperature difference. If two objects are placed in direct contact with one another, then heat will be transferred from the hotter object to the colder one.
Convection: Convection is the process in which heat energy transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another.
Example: Boiling of water in a pot.
Radiation: Radiation is a form of energy transfer from one body to another by electromagnetic waves.
Example: Solar energy from the Sun.
Question 7.
Explain in detail Newton’s law of cooling.
Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.
$$\frac { dQ }{ dt }$$ ∝ – (T – Ts) … (1)
The negative sign implies that the quantity of heat lost by liquid goes on decreasing with time. Where,
T = Temperature of the object
T = Temperature of the surrounding
From the above graph, it is clear that the rate of cooling is high initially and decreases with falling temperature.
Consider a body of mass m and specific heat capacity s at temperature T. Let T be the temperature of the surroundings. If the temperature falls by a small amount dl in time dt, then the amount of heat lost is,
dQ = msdT
Dividing both sides of equation (2) by dt
$$\frac { dQ }{ dt }$$ ∝ $$\frac { msdT }{ dt }$$ … (3)
From Newton’s law of cooling
$$\frac { dQ }{ dt }$$ ∝ (T – Ts)
$$\frac { dQ }{ dt }$$ ∝ (T – Ts) … (4)
Where a is some positive constant.
From equation (3) and (4)
– a(T – Ts) = ms$$\frac { dT }{ dt }$$
$$\frac{d \mathrm{~T}}{\mathrm{~T}-\mathrm{T}_{s}}$$ = – $$\frac { a }{ ms }$$dt … (5)
Integrating equation (5) on both sides,
Where b1 is the constant of integration. Taking exponential both sides, we get
T = Ts + b2 $$e^{-\frac{a}{m s} t}$$
here b2 = $$e^{b_{1}}$$ = constant.
Question 8.
Explain Wien’s law and why our eyes are sensitive only to visible rays?
Wien’s law states that, ‘the wavelength of maximum intensity of emission of a black body radiation is inversely proportional to the absolute temperature of the black body’.
λm ∝ $$\frac { 1 }{ T }$$
(or)
λm = $$\frac { b }{ T }$$ … (1)
It is implied that if temperature of the body increases, maximal intensity wavelength (λm) shifts towards lower wavelength (higher frequency) of electromagnetic spectrum.
Wien’s law and Vision: Why our eye is sensitive to only visible wavelength (in the range 400 nm to 700nm)?
The Sun is approximately considered as a black body. Since any object above 0 K will emit radiation, Sun also emits radiation. Its surface temperature is about 5700 K. By substituting this value in the equation (1),
It is the wavelength at which maximum intensity is 508nm. Since the Sun’s temperature is around 5700K, the spectrum of radiations emitted by Sun lie between 400 nm to 700 nm which is the visible part of the spectrum.
The humans evolved under the Sun by receiving its radiations. The human eye is sensitive only in the visible.
Question 9.
Discuss the,
(i) Thermal equilibrium
(ii) Mechanical equilibrium
(iii) Chemical equilibrium
(iv) Thermodynamic equilibrium.
(i) Thermal equilibrium: Two systems are said to be in thermal equilibrium with each other if they are at the same temperature, which will not change with time.
(ii) Mechanical equilibrium: A system is said to be in mechanical equilibrium if no unbalanced force acts on the thermodynamic system or on the surrounding by thermodynamic system.
(iii) Chemical equilibrium: If there is no net chemical reaction between two thermodynamic systems in contact with each other then it is said to be in chemical equilibrium.
(iv) Thermodynamic equilibrium: If two systems are set to be in thermodynamic equilibrium, then the systems are at thermal, mechanical and chemical equilibrium with each other. In a state of thermodynamic equilibrium the macroscopic variables such as pressure, volume and temperature will have fixed values and do not change with time.
Question 10.
Explain Joule’s Experiment of the mechanical equivalent of heat.
Joule showed that mechanical energy can be converted into internal energy and vice versa. In his experiment, two masses were tied with a rope and a paddle wheel as shown in Figure. When these masses fall through a distance h due to gravity, both the masses lose potential energy which is equal to 2mgh. When the masses fall, the paddle wheel turns. Due to the turning of wheel inside water, frictional force comes in between the water and the paddle wheel.
This causes a rise in temperature of the water. This confirms that gravitational potential energy is converted to internal energy of water. The temperature of water increases due to the work done by the masses. In fact, Joule was able to show that the mechanical work has the same effect as giving heat energy. He found that to raise lg of an object by 1°C , 4.186J of energy is required. In earlier days the heat was measured in calorie.
1 cal = 4.186 J
This is called Joule’s mechanical equivalent of heat.
Question 11.
Derive the expression for the work done in a volume change in a thermodynamic system.
Let us consider a gas contained in the cylinder fitted with a movable piston. When the gas is expanded quasi-statically by pushing the piston by a small distance dx as shown in Figure. Because the expansion occurs quasi- statically. The pressure, temperature and internal energy will have unique values at every instant.
The small work done by the gas on the piston
dW = F dx … (1)
The force exerted by the gas on the piston F – PA.
A – area of the piston and
P – pressure exerted by the gas on the piston.
Equation (1) can be rewritten as
dW = PA dx … (2)
But A dx = dV change in volume during this expansion process.
Hence the small work done by the gas during the expansion is given by
dW = dV … (3)
It is noted that is positive since the volume is increased. In general the work done by the gas by increasing the volume from Vi to Vf is given by
W = $$\int_{V_{i}}^{\mathrm{V}_{f}} \mathbf{P} d \mathrm{~V}$$
Suppose if the work is done on the system, then Vi > Vf.
Then, W is negative.
Question 12.
Derive Meyer’s relation for an ideal gas.
Let us consider μ mole of an ideal gas in a container with volume V, pressure P and temperature T. When the gas is heated at constant volume the temperature increases by dT. Since no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU.
If Cv is the molar specific heat capacity at constant volume,.
dU= μCvdT … (1)
When the gas is heated at constant pressure so that the temperature increases by dT. If ‘Q’ is the heat supplied in this process and ‘dV’ the change in volume of the gas.
Q = μCpdT … (2)
If W is the workdone by the gas in this process, then
W = P dV … (3)
But from the first law of thermodynamics,
Q = dU + W … (4)
Substituting equations (1), (2) and (3) in (4), we get,
μCpdT = μCvdT + P dV … (5)
For mole of ideal gas, the equation of state is given by,
PV = μRT
⇒ PdV + VdP = μRdT … (6)
Since the pressure is constant, dP = 0
∴ CpdT = CvdT + RdT
∴ Cp = Cv + R
(or) Cp – Cv = R … (7)
This relation is called Meyer’s relation.
Question 13.
Explain in detail the isothermal process.
It is a process in which both the pressure and volume of a thermodynamic system will change at constant temperature. The ideal gas equation is
PV = μRT
Here, T is constant for this process.
So the equation of state for isothermal process is given by,
PV= constant … (1)
It is implied that if the gas goes from one equilibrium state (P1,V1) to another equilibrium state (P2,V2) the following relation holds for this process
P1,V1 = P2,V2 … (2)
Since PV = constant, P is inversely proportional to V(P ∝ $$\frac { 1 }{ V }$$). This implies that PV graph is a hyperbola. The pressure-volume graph for constant temperature is also called isotherm.
The following figure shows the PV diagram for quasi-static isothermal expansion and quasi-static isothermal compression. For an isothermal process since temperature is constant, the internal energy is also constant. It implies that dU or ∆U – 0.
For an isothermal process, the first law of thermodynamics can be written as follows,
Q = W … (3)
From equation (3), it implies that the heat supplied to a gas is used to do only external work. It is a common misconception that when there is flow of heat energy to the system, the temperature will increase. For isothermal process it is not true. When the piston of the cylinder is pushed, the isothermal compression takes place. This will increase the internal energy which will flow out of the system through thermal contact.
Question 14.
Derive the work done in an isothermal process.
Let us consider an ideal gas which is allowed to expand quasi-statically at constant temperature from initial state (Pi,Vi) to the final state (Pf, Vf. Let us calculate the work done by the gas during this process.
gas, W = $$\int_{V_{i}}^{\mathrm{V}_{f}} \mathrm{P} d \mathrm{~V}$$ … (1)
As the process occurs quasi-statically, at every stage the gas is at equilibrium with the surroundings. Since it is in equilibrium at every stage the ideal gas law is valid.
P = $$\frac { μRT }{ V }$$ … (2)
Substituting equation (2) and (1) we get,
In equation (3), μRT is taken out of the integral, since it is constant throughout the isothermal process.
By performing the integration in equation (3), we get
W = μRTln$$\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)$$ … (4)
Since we have an isothermal expansion,
$$\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}$$ > 1, so $$\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)$$ > 0. As a result the work done by the gas during an isothermal expansion is positive.
Equation (4) is true for isothermal compression also. But in an isothermal compression $$\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}$$ < 1. So $$\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)$$ < 0.
Hence the work done on the gas in an isothermal compression is negative. In the PV diagram the work done during the isothermal expansion is equal to the area under the graph as shown in figure.
Similarly for an isothermal compression, the area under the PV graph is equal to the work done on the gas that turns out to be the area with a negative sign.
Question 15.
Explain in detail an adiabatic process.
It is a process in which no heat energy flows into or out of the system (Q = 0). But the gas can expand by spending its internal energy or gas can be compressed through some external work. Hence the pressure, volume and temperature of the system may change in an adiabatic process.
From the first law for an adiabatic process, ∆U = W. This implies that the work is done by the gas at the expense of internal energy or work is done on the system that increases its internal energy. The adiabatic process can be achieved by the following methods
(i) Thermally insulating the system from surroundings. So that no heat energy flows into or out of the system. For instance, when thermally insulated cylinder of gas is compressed (adiabatic compression) or expanded (adiabatic expansion) as shown in the Figure.
(ii) If the process occurs so quickly so that there is no time to exchange heat with surroundings even though there is no thermal insulation
The equation (1). Implies that if the gas goes from an equilibrium state (Pi, Vi) to another equilibrium state (Pf,Vf) adiabatically then it satisfies the relation,
PVγ = constant … (1)
The equation (1) Here γ is called adiabatic exponent ( γ = $$\frac{C_{\mathrm{P}}}{C_{\mathrm{V}}}$$) which depends on the nature of the gas.
The equation (1). Implies that if the gas goes from an equilibrium state (Pi,Vi) to another equilibrium state (Pf, Vf) adiabatically then it satisfies the relation,
PiViγ = PfVfγ … (2)
The PV diagram for an adiabatic process is also called adiabatic. It is noted that the PV diagram for isothermal and adiabatic processes look similar. But actually the adiabatic curve is steeper than isothermal curse.
Let us rewrite the equation (1) in terms of T and V. From ideal gas equation, the pressure P = $$\frac { μRT }{ V }$$. Substituting this equation in the equation (1), we have
$$\frac { μRT }{ V }$$Vγ = constant
(or) $$\frac { T }{ V }$$Vγ = $$\frac { constant }{ μR }$$
Note here that is another constant. So it can
be written as
TVγ-1 = constant … (3)
From the equation (3) it is known that if the gas goes from an initial equilibrium state (Ti,Vi) to final equilibrium state (Tf, Vf) adiabatically then it satisfies the relation
$$\mathrm{T}_{i} \mathrm{~V}_{i}^{\gamma-1}=\mathrm{T}_{f} \mathrm{~V}_{f}^{\gamma-1}$$ … (4)
The equation of state for adiabatic process can also be written in terms of T arid P as
$$\mathrm{T}^{\gamma} \mathrm{P}^{1-\gamma}$$ = constant
Question 16.
Derive the work done in an adiabatic process.
Let us consider p moles of an ideal gas enclosed in a cylinder having perfectly non conducting walls and base. A frictionless and insulating piston A is fitted in the cylinder as shown in Figure. It’s area of cross-section be A.
Let W be the work done when the system goes from the initial state (PiViTi) to the final state (PfVfTf) adiabatically.
W = $$\int_{V_{i}}^{\mathrm{V}_{f}} \mathrm{P} d \mathrm{~V}$$ … (1)
It is assumed that the adiabatic process occurs quasi-statically, at every stage the ideal gas law is valid. Under this condition, the adiabatic equation of state is PVγ = constant (or) P = $$\frac{\text { constant }}{\mathrm{V}^{\gamma}}$$ can be substituted in the equation (1), we get
From ideal gas law,
In adiabatic expansion, work is done by the gas. i.e., Wadia is positive. As Ti > Tf the gas cools during adiabatic expansion.
In adiabatic compression, work is done on the gas. i.e., Wadia is negative. As Ti < Tf hence the temperature of the gas increases during adiabatic compression.
Question 17.
Explain the isobaric process and derive the work done in this process.
Isobaric process is a thermodynamic process that occurs at constant pressure. Inspite of the pressure consistency of this process, temperature, volume and internal energy are not constant. From the ideal gas equation, we have
V = ($$\frac { μR }{ P }$$)T … (1)
Here, $$\frac { μR }{ P }$$ = constant
In an isobaric process the temperature is directly proportional to volume.
V ∝ T(Isobaric process) … (2)
It is implied that for a isobaric process, the V-T graph is a straight line passing through the origin.
Work done by the gas
W= $$\int_{V_{i}}^{\mathrm{V}_{f}} \mathrm{P} d \mathrm{~V}$$
In an isobaric process, the pressure is constant, So P comes out of the integral,
∆V – change in the volume. If AV is negative, W is also negative. It is implied that the work is done on the gas. If ∆V is positive, W is also positive, implying that work is done by the gas. From the ideal gas equation.
equation (4) can be rewritten as
PV = μRT and V = $$\frac { μRT }{ P }$$
Substituting this in equation (4) we get,
W = μRTf(1 – $$\frac{\mathrm{T}_{i}}{\mathrm{~T}_{f}}$$) … (5)
In the PV diagram, area under the isobaric curve is equal to the work done in isobaric process. The shaded area in the following Figure is equal to the work done by the gas.
The first law of thermodynamics for isobaric process is given by,
∆U = Q – P∆V
Question 18.
Explain in detail the isochoric process.
Isochoric process is a thermodynamic process in which the volume of the system is kept constant. But pressure, temperature and internal energy continue to be variables. The pressure-volume graph for an isochoric process is a vertical line parallel to pressure axis as shown in Figure.
The equation of state for an isochoric process is given by,
P = $$\frac { μR }{ V }$$ … (1)
Here $$\frac { μR }{ V }$$ = constant
∴ P ∝ T
It is inferred that the pressure is directly proportional to temperature. This implies that the P-T graph for an isochoric process is a straight line passing through origin. If a gas goes from state (Pi, Ti) to (Pf, Tf) at constant volume, then the system satisfies the following equation,
$$\frac{P_{i}}{T_{i}}=\frac{P_{f}}{T_{f}}$$
For an isochoric processes, AV=0 and W-0. Then the first law of thermodynamics becomes
∆U = Q … (3)
It is implied that the heat supplied is used to increase only the internal energy. As a result the temperature increases and pressure also increases.
Question 19.
What are the limitations of the first law of thermodynamics?
The first law of thermodynamics explains well the inter convertibility of heat and work. But is not indicated the direction of change.
Example:
(i) When a hot object is in contact with a cold object, heat always flows from the hot object to cold object but not in the reverse direction. According to first law of thermodynamics, it is possible for the energy to flow from hot object to cold object and viceversa. But in nature the direction of heat flow is always from higher temperature to lower temperature.
(ii) When brakes are applied, a car stops due to friction and the work done against friction is converted into heat. But this heat is not reconverted into the kinetic energy of the car. Hence the first law is not sufficient to explain many of natural phenomena.
Question 20.
Explain the heat engine and obtain its efficiency.
Heat engine is a device which takes heat as input and converts this heat into work by undergoing a cyclic process.
A heat engine has three parts:
(i) Hot reservoir
(ii) Working substance
(iii) Cold reservoir
A Schematic diagram for heat engine is given below in the figure.
• Hot reservoir (or) Source: It supplies heat to the engine. Which is always maintained at a high temperature TH.
• Working substance: It is a substance like gas or water, that converts the heat supplied into work.
• Cold reservoir (or) Sink: The heat engine ejects some amount of heat (QL) into cold reservoir after it doing work. It is always maintained at a low temperature TL.
The heat engine works in a cyclic process. After a cyclic process it returns to the same state. Since the heat engine returns to the same state after it ejects heat, the change in the internal energy of the heat engine is zero.
The efficiency of the heat engine is defined as the ratio of the work done (output) to the heat absorbed (input) in one cyclic process. Let the working substance absorb heat QH units from the source and reject QL units to the sink after doing work W units.
We can write,
Input heat = Work done + ejected heat
QH = W + QL
W = QH – QL
Then the efficiency of heat engine
Question 21.
Explain in detail Carnot heat engine.
A reversible heat engine operating in a cycle between two temperatures in a particular way is called a Carnot Engine. The carnot engine consists of four parts that are given below.
1. Source: It is the source of heat maintained at constant high temperature TH. Without changing its temperature any amount of heat can be extracted from it.
2. Sink: It is a cold body maintained at a constant low temperature TL. It can absorb any amount of heat.
3. Insulating stand: It is made of perfectly non-conducting material. Heat is not conducted through this stand.
4. Working substance: It is an ideal gas enclosed in a cylinder with perfectly non-conducting walls and perfectly conducting bottom. A non-conducting and frictionless piston is fitted in it.
The four parts are shown in the following Figure.
Carnot’s cycle: In camot’s cycle, the working substance is subjected to four successive reversible processes.
Let the initial pressure, volume of the working substance be P1, V1.
When the cylinder is placed on the source, the heat (QH) flows from source to the working substance (ideal gas) through the bottom of the cylinder.
The input heat increases the volume of the gas.So the piston is allowed to move out very’ slowly.
W1 is the work done by the gas it expands from volume V1 to volume V2 with a decrease of pressure from P1 to P2.
Then the work done by the gas (working substance) is given by
∴ QH = QA→B $$\int_{V_{1}}^{\mathrm{V}_{2}} \mathrm{P} d \mathrm{~V}$$
When the cylinder is placed on the insulating stand and the piston is allowed to move out, the gas expands adiabatically from volume V2 to volume V3 the pressure falls from P2 to P3. The temperature falls to TLK.
The work done by the gas in an adiabatic expansion is given by,
Area under the curve BC.
Isothermal compression from (P3, V3, TL) to (P4, V4, TL).
Let the cylinder is placed on the sink and the gas is isothermally compressed until the pressure and volume become P4 and V4 respectively.
Let the cylinder is placed on the insulating stand again and the gas is compressed adiabatically till it attains the initial pressure P1, volume V2 and temperature TH, Which is shown by the curve DA in the P – V diagram.
= – Area under the curve DA
Let ‘W’ be the net work done by the working substance in one cycle
The net work done by the Carnot engine in one cycle is given by,
W = WA→B – WC→D
Net work done by the working substance in one cycle is equal to the area (enclosed by ABCD) of the P-V diagram.
Question 22.
Derive the expression for Carnot engine efficiency.
Efficiency is defined as the ratio of work done by the working substance in one cycle to the amount of heat extracted from the source.
Let us omit the negative sign. Since we are interested in only the amount of heat (QL) ejected into the sink, we have
By applying adiabatic conditions, we get,
TH V2γ-1 = TL V3γ-1
TH V1γ-1 = TL V4γ-1
By dividing the above two equations, we get
$$\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)$$γ-1 = $$\left(\frac{\mathrm{V}_{3}}{\mathrm{~V}_{4}}\right)$$γ-1
Which implies that
$$\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}=\frac{\mathrm{V}_{3}}{\mathrm{~V}_{4}}$$ … (5)
Substituting equation (5) and (4), we get
$$\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{Q}_{\mathrm{H}}}=\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$$
∴ The efficiency
η = 1 – $$\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$$
Question 23.
Explain the second law of thermodynamics in terms of entropy.
$$\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{Q}_{\mathrm{H}}}$$ = $$\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$$ … (1)
According to efficiency of Carnot’s engine equation, quantity $$\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}$$ = $$\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{L}}}$$.
The quantity $$\frac { Q }{ T }$$ is called entropy. It is a very important thermodynamic property of a System, which is also a state variable.
$$\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}$$ is the entropy received by the Carnot engine from hot reservoir and $$\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{L}}}$$ is entropy given out by the Carnot engineLto the cold reservoir. For reversible engines (Carnot Engine) both entropies should be same, so that the change in entropy of the Carnot engine in one cycle is zero. It is proved in equation (1).
But for all practical engines like diesel and petrol engines which are not reversible engines, they satisfy the relation $$\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{L}}}$$ > $$\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}$$.
In fact the second law of thermodynamics can be reformulated as follows “For all the processes that occur in nature(irreversible process), the entropy always increases. For reversible process entropy will not change”. Entropy determines the direction in which natural process must occur.
Question 24.
Explain in detail the working of a refrigerator.
A refrigerator is a Carnot’s engine working in the reverse order which is shown in the figure.
8.17 Schematic diagram of a refrigerator The working substance (gas) absorbs a quantity of heat QL from the cold body (sink) at a lower temperature TL. A certain amount of work W is done on the working substance by the compressor and a quantity of heat QH is rejected to the hot body (source) i.e., the atmosphere at TH. When we stand beneath of the refrigerator, we can feel warmth air.
From the first law of thermodynamics, we have QL + W = QH
Hence the cold reservoir (refrigerator) further cools down and the surroundings (kitchen or atmosphere) gets hotter.
IV. Numerical problems:
Question 1.
Calculate the number of moles of air is in the inflated balloon at room temperature as shown in the figure.
The radius of the balloon is 10 cm, and pressure inside the balloon is 180 kPa.
The pressure inside the balloon P = 1.8 x 105 P
Room temperature
T = 273 + 30 = 303K.
Let the radius of the balloon be R = 10 x 10-2m.
Volume of the balloon
Question 2.
In the planet Mars, the average temperature is around – 53°C and atmospheric pressure is 0.9 kPa. Calculate the number of moles of
the molecules in unit volume in the planet Mars? Is this greater than that in earth?
= 0.9 x 10³Pa
PV = NkT
Since volume is unity V = 1
Equation (1) becomes
P = NkT
Where k is Boltzmann constant
Question 3.
An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V1 and contains ideal gas at pressure P1 and temperature T1. The other chamber has volume V1 and contains ideal gas at pressure P2 and temperature T2. If the partition is removed without doing any work on the gases, calculate the final equilibrium temperature of the container.
In first chamber let the pressure be P1
In second chamber let the pressure be P2
In first chamber let the volume be V1
In second chamber let the volume be V2
In first chamber let the temperature be T1
In second chamber let the temperature be T2
According to conservation of energy,
Question 4.
The temperature of a uniform rod of length L having a coefficient of linear expansion αL is changed by ∆T. Calculate the new moment of inertia of the uniform rod about axis passing through its centre and perpendicular to an axis of the rod.
Moment of inertia of uniform rod of mass M, and length L about axis passing through its centre and perpendicular to an axis of rod is given by
I = $$\frac{\mathrm{ML}^{2}}{12}$$
The increase in length of the rod
∆l = LαL∆T
αL – Coefficient of linear expansion.
∆T – change in temperature
After the rod is heated,
Moment of inertia,
Question 5.
Draw the TP diagram (P-x axis, T-y axis), VT(T-x axis, V-y axis) diagram for
(a) Isochoric process
(b) Isothermal process
(c) isobaric process.
(a) For isochoric process:
V = V0 = constant
The points a and b are represented as a = (P1V0T0) b = (P2, V0, T2)
(b) For isothermal process:
T = T0 = constant
PV = nRT0
P(T) = multivalue
P(T) = $$\frac{n \mathrm{RT}_{0}}{\mathrm{~V}}$$
The points a and b are represented as
a = (P1V1,T) b = (P2V0T0)
(c) For isobaric process:
P = P0 = constant
T(V) = $$\frac{\mathrm{P}_{0} \mathrm{~V}}{n^{2}}$$
P0V = nRT
P(T) = P0
Question 6.
A man starts bicycling in the morning at a temperature around 25°C, he checked the pressure of tire which is equal to be 500 kPa. Afternoon he found that the absolute pressure in the tyre is increased to 520 kPa. By assuming the expansion of tyre is negligible, what is the temperature of tyre at afternoon?
Question 7.
The temperature of a normal human body is 98.6°F. During high fever if the temperature increases to 104°F, what is the change in peak wavelength that emitted by our body? (Assume human body is a black body)
Normal temperature of human body = 98.6°F
Conversion of F° to C°
C = $$\frac { (F-32) }{ 1.8 }$$
= $$\frac { (98.6-32 }{ 1.8 }$$
= $$\frac { 66.6 }{ 1.8 }$$
= 37°C
Conversion of C° to Kelvin
T = 37 + 273 = 310
Peak wavelength at 98.6°F is
λm = $$\frac { b }{ T }$$
= $$\frac{2.898 \times 10^{-3}}{310}$$
= 0.009348 x 10-3 m
= 9348 x 10-9
= 9348 nm
Final temperature of the human body = 104°F
Conversion of F° to C°
C = $$\frac { F-32 }{ 1.8 }$$
= $$\frac { 104-32 }{ 1.8 }$$
= $$\frac { 72) }{ 1.8 }$$
Conversion of 40° to Kelvin
T = 40+ 273 = 313 K
Peak wavelength at 104°F is
λm = $$\frac { b }{ T }$$
= $$\frac{2.898 \times 10^{3}}{313}$$
= 0.009258 x 10-3 m
= 9258 x 10-9 = 9258 nm
λmax = 9348 nm at 98.6°F
λmax = 9258 nm at 104°F
Question 8.
In an adiabatic expansion of the air, the volume is increased by 4%, what is percentage change in pressure? (For air γ = 1.4)
Percentage of increased volume = 4%
$$\frac { ∆V }{ V }$$ x 100 = 4%
γ = 1.4
PVγ = Constant
Question 9.
In a petrol engine, (internal combustion engine) air at atmospheric pressure and temperature of 20°C is compressed in the cylinder by the piston to 1/8 of its original volume. Calculate the temperature of the compressed air. (For air γ = 1.4)
P1 = 1 atmospheric pressure, V1 = V, V2 = $$\frac { V }{ 8 }$$
T1 = 20 + 273 – 293K, T2 – ?, γ = 1.4.
∴ The temperature of the compressed air = 400°C
Question 10.
Consider the following cyclic process consist of isotherm, isochoric and isobar which is given in the figure.
Draw the same cyclic process qualitatively in the V-T diagram where T is taken along x direction and V is taken along y-direction. Analyze the nature of heat exchange in each process.
Process 1 to 2: increase in volume. So heat must be added.
Process 2 to 3: Volume remains constant. Increase in temperature. The given heat is used to increase the internal energy.
Process 3 to 1: Pressure remains constant. Volume and Temperature are reduced. Heat flows out of the system. It is an isobaric compression where the work is done on the system.
Explanation: In the graph, during the process (1 to 2), the gas undergoes isothermal expansion. It receives certain amount of heat from the surroundings. It uses this heat in doing the work. Hence internal energy of the gas remains unchanged.
During the process represented by (2 to 3) the gas is heated at constant volume. Since no work is done and volume does not change, the process is isochoric process.
Since heat is transferred to the gas from the surroundings, the internal energy of the gas is increased. During the process represented by (3 to 1) the gas is compressed isobarically. Work is done on the gas. Since temperature drops internal energy is reached. Hence the gas gives up heat to the surroundings.
Question 11.
An ideal gas is taken in a cyclic process as shown in the figure. Calculate
(a) work done by the gas.
(b) work done on the gas.
(c) Net work done in the process
(a) Let the work done by the gas along AB be W. From the fig pressure,
P = 600 N/m²
From the fig change in volume,
∆V = 3
∴Work done,
W = PAV
= 600 x 3 = 1800J
= 1.8 kJ
(b) Isobaric compression take place and work is done on the gas along BC.
W = – PAV
∆V = (6 – 3) = 3
P = +400N/m²
∴ W= – 400 x 3
= – 1200
= – 1.2 kJ
(c)
Question 12.
For a given ideal gas 6 x 105 J heat energy is supplied and the volume of gas is increased from 4 m³ to 6 m² at atmospheric pressure. Calculate
(a) the work done by the gas
(b) change in internal energy of the gas
(c) graph this process in PV and TV diagram.
Heat energy supplied to the gas
Q = 6 x 105J
(b) Change in internal energy
∆U = ∆Q – AW
= ∆Q – P∆V
= 6 x 105 – 2.026 x 105
= 3,974 x 103
∆U = 3.974 kJ
(c)
Question 13.
Suppose a person wants to increase the efficiency of the reversible heat engine that is operating between 100°C and 300°C. He had two ways to increase the efficiency.
(a) By decreasing the cold reservoir temperature from 100°C to 50°C and keeping the hot reservoir temperature constant
(b) By increasing the temperature of the hot reservoir from 300°C tc 350°C by keeping the cold reservoir temperature constant. Which is the suitable method?
Temperature of cold reservoir
T1 = 100°C
= 100 + 273 – 373 K
Temperature of hot reservoir
TH = 300 + 273 = 573 K
η = 1 – $$\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$$
η = 1 – $$\frac { 373 }{ 573 }$$
η = 1 – $$\frac { 573 – 373 }{ 573 }$$
= $$\frac { 200 }{ 753 }$$
= 0.349 x 100
Initial efficiency = 34.9%
(a) By decreasing the temperature of the cold reservoir from 100°C to 50PC.
(b) By increasing the temperature of the hot reservoir from 300°C to 350°C,
T1 = 350 + 273 = 623K
T1 = 100 + 273 =373K
∴ Efficiency,
η = 1 – $$\frac{\mathrm{T}_{\mathrm{2}}}{\mathrm{T}_{\mathrm{1}}}$$
η = 1 – $$\frac { 373 }{ 623 }$$
η = $$\frac { 623 – 373 }{ 623 }$$
η = $$\frac { 250 }{ 623 }$$
η = 0.4012
η = 0.4012 x 100 = 40.12%
∴ Method (a) is more efficient than method (b).
Question 14.
A Carnot engine whose efficiency is 45% takes heat from a source maintained atatemperature of 327°C. To have an engine of efficiency 60% what must be the intake temperature for the same exhaust (sink) temperature?
Temperature of a source,
T1 = 327°C = 327 + 273 = 600 k
η = $$\frac { 45 }{ 100 }$$ = 0.45
Efficiency η = 1 – $$\frac{\mathrm{T}_{\mathrm{2}}}{\mathrm{T}_{\mathrm{1}}}$$
∴ 0.45 = 1 – $$\frac{\mathrm{T}_{2}}{600}$$
$$\frac{\mathrm{T}_{2}}{600}$$ = 1 – 0.45 = 0.55
∴ T2 = 0.55 x 600 = 330 K
η = 60% = $$\frac { 60 }{ 100 }$$ = 0.6
0.6 = 1 – $$\frac{\mathrm{T}_{\mathrm{2}}}{\mathrm{T}_{\mathrm{1}}}$$ = 1 – $$\frac{330}{\mathrm{~T}_{1}}$$
$$\frac{330}{\mathrm{~T}_{1}}$$ = 1 – 0.6 = 0.4
In take temperature
T1 = $$\frac { 330 }{ 0.4 }$$ = 825K
= 825 – 273
= 552°C
The intake temperature = 552°C
Question 15.
An ideal refrigerator keeps its content at 0°C while the room temperature is 27°C. Calculate its coefficient of performance.
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Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Physics Guide Pdf Chapter 8 Heat and Thermodynamics Text Book Back Questions and Answers, Notes.. ## Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics. ### 11th Physics Guide Heat and Thermodynamics Book Back Questions and Answers. Part – I:. I. Multiple choice questions:. Question 1.. In hot summer after a bath, the body’s:. (a) internal energy decreases. (b) internal energy increases. (c) heat decreases. (d) no change in internal energy and heat. (a) internal energy decreases. Question 2.. The graph between volume and temperature in Charle’s law is:. (a) an ellipse. (b) a circle. (c) a straight line. (d) a parabola. (c) a straight line. Question 3.. When a cycle tyre suddenly bursts, the air inside the tyre expands. This process is:. (a) isothermal. (c) isobaric. (d) isochoric. Question 4.. An ideal gas passes from one equilibrium state (P1, V1, T1, N) to another equilibrium state (2P1, 3V1, T2, N). Then:. (a) T1 = T2. (b) T1 = $$\frac{\mathrm{T}_{2}}{6}$$. (c) T1 = 6T2. (d) T1 = 3T2. (b) T1 = $$\frac{\mathrm{T}_{2}}{6}$$. Question 5.. When a uniform rod is heated, which of the following quantity of the rod will increase:. (a) mass. (b) weight. (c) centre of mass. (d) moment of inertia. (d) moment of inertia. Hint:. M.I. = MK². During heating K would be increased. Hence moment of inertia will increase.. Question 6.. When food is cooked in a vessel by keeping the lid closed, after some time the steam pushes the lid outward. By considering the steam as a thermodynamic system, then in the cooking process:. (a) Q > 0, W > 0. (b) Q < 0, W > 0. (c) Q > 0, W < 0. (d) Q < 0, W < 0. (a) Q > 0, W > 0. Hint:. Q > 0; W > 0. ∆Q = ∆U1 + ∆W.. ∴ ∆Q ∝ ∆W. When ∆Q > 0. ∆W > 0.. Question 7.. When you exercise in the morning, by considering your body as a thermodynamic system, which of the following is true?. (a) ∆U > 0, W > 0. (b) ∆U < 0, W > 0. (c) ∆U< 0, W < 0. (d) ∆U = 0, W > 0. (b) ∆U < 0, W > 0. Hint:. ∆Q = ∆U + ∆W. When exercise is performed, internal energy is utilised (∆U < 0). So it will decrease.. But as internal energy is utilised, the exercise (∆W) will be more. So work will be more.. ∴ ∆W > 0. ∆U < 0 W > 0. Question 8.. A hot cup of coffee is kept on the table. After some time it attains a thermal equilibrium with the surroundings. By considering the air molecules in the room as a thermodynamic system, which of the following is true?. (a) ∆U > 0, Q = 0. (b) ∆U > 0, W < 0. (c) ∆U > 0, Q > 0. (d) ∆U = 0, Q > 0. (c) ∆U > 0, Q > 0. Hint: During the thermal equilibrium with surrounding as heat energy (Q) is increased, internal energy will be increased.. ∴ ∆U > 0 Q > 0. Question 9.. An ideal gas is taken from (Pi, Vi) to (Pf, Vf) in three different ways. Identify the process in which the work done on the gas the most.. (a) Process A. (b) Process B. (c) Process C. (d) Equal work is done in Process A, B & C. (b) Process B. Hint:. When work is done on the gas, compression takes place. Final pressure Pf will be increased and remains constant. It is shown by process B.. Question 10.. The V-T diagram of an ideal gas which goes through a reversible cycle A → B → is shown below. (Processes D → A and B → C are adiabatic). The corresponding PV diagram for the process is (all figures are schematic). Question 11.. A distant star emits radiation with maximum intensity at 350 nm. The temperature of the star is:. (a) 8280 K. (b) 5000 K. (c) 7260 K. (d) 9044 K. (a) 8280 K. Hint:. Question 12.. Identify the state variables given here.. (a) Q, T, W. (b) P,T,U. (c) Q, W. (d) P,T,Q. (b) P,T,U. Question 13.. In an isochoric process, we have:. (a) W = 0. (b) Q = 0. (c) ∆U = 0. (d) ∆T = 0. (a) W = 0. Hint:. Work done in an isochoric process is zero.. Question 14.. The efficiency of a heat engine working between the freezing point and boiling point of water is: (NEET2018). (a) 6.25%. (b) 20%. (c) 26.8%. (d) 12.5%. (b) 20%. Hint:. T2 = 0° C = 0 + 273 = 273 K. T1 = 100° C = 100 + 273 = 373 K. Question 15.. An ideal refrigerator has a freezer at a temperature of -12°C. The coefficient of performance of the engine is 5. The temperature of the air (to which the heat ejected) is:. (a) 50°C. (b) 45.2°C. (c) 40.2°C. (d) 37.5°C. (c) 40.2°C. Hint:. Question 1.. ‘An object contains more heat’- is it a right statement? If not why?. When heated, an object receives heat from the agency. Now object has more internal energy than before. Heat is the energy in transit and which flows from an object at a higher temperature to an object at lower temperature. Heat is not a quantity. So the statement I would prefer “an object contains more thermal energy”.. Question 2.. Obtain an ideal gas law from Boyle’s and Charles’law.. According to Boyle’s law, Pressure ∝ $$\frac { 1 }{ 2 }$$. i.e., P ∝ $$\frac { 1 }{ V }$$. According to Charle’s law,. Volume ∝ Temperature, i.e., V ∝ T. By combining these two laws, we get. PV= CT … (1). Where C is a positive constant. But C = k x Number of particles (N). C = kN. Where k – Boltzmann’s constant.. ∴ Equation (1) becomes. PV = NkT. Question 3.. Define one mole.. One mole of any substance is the amount of that substance which contains Avogadro number (NA) of particles (such as atoms or molecules).. Question 4.. Define specific heat capacity and give its unit.. Specific heat capacity of a substance is defined as the amount of heat energy required to raise the temperature of 1 kg of a substance by 1 Kelvin or 1 °C.. The SI unit for specific heat capacity is J kg-1 K-1.. Question 5.. Define molar specific heat capacity.. Heat energy required to increase the temperature of one mole of substance by IK or 1°C.. Question 6.. What is a thermal expansion?. Thermal expansion is the tendency of matter to change in shape, area, and volume due to a change in temperature.. Question 7.. Give the expressions for linear, area and volume thermal expansions.. (i) Linear expansion:. $$\frac { ∆L }{ L }$$ = α ∆T ⇒ αL = $$\frac { ∆L }{ L∆T }$$. (ii) Area expansion:. $$\frac { ∆A }{ A }$$ = 2α ∆T ⇒ αA = $$\frac { ∆A }{ A∆T }$$. (iii) Volume expansion:. $$\frac { ∆V }{ V }$$ = 3α ∆T ⇒ αv = $$\frac { ∆V }{ V∆T }$$. Question 8.. Define latent heat capacity. Give its unit.. Latent heat capacity of a substance is defined as the amount of heat energy required to change the state of a unit mass of the material. The SI unit for latent heat capacity is J kg-1.. Question 9.. State Stefan-Boltzmann law.. The total amount of heat energy radiated per second per unit area of a black body is directly proportional to the fourth power of its absolute temperature.. Question 10.. What is Wien’s law?. The wavelength of maximum intensity of emission of a black body radiation is inversely proportional to the absolute temperature of the black body.. Question 11.. Define thermal conductivity. Give its unit.. The quantity of heat transferred through a unit length of a material in a direction normal to Unit surface area due to a unit temperature difference under steady-state conditions is known as the thermal conductivity of a material.. Question 12.. What is a black body?. A black body is one which neither reflects nor transmits but absorbs whole of the radiation incident on it.. Question 13.. What is a thermodynamic system? Give examples.. Thermodynamic system: A thermodynamic system is a finite part of the universe. It is a collection of a large number of particles (atoms and molecules) specified by certain parameters called pressure (P), Volume (V), and Temperature (T). The remaining part of the universe is called the surrounding. Both are separated by a boundary.. Examples: A thermodynamic system can be liquid, solid, gas, and radiation.. Question 14.. What are the different types of thermodynamic systems?. (i) Open system can exchange both matter and energy with the environment.. (ii) Closed system exchange energy but not matter with the environment.. (iii) Isolated system can exchange neither energy nor matter with the environment.. Question 15.. What is meant by ‘thermal equilibrium’?. Two systems are said to be in thermal equilibrium with each other if they are at the same temperature, which will not change with time.. Question 16.. What is mean by state variable? Give example.. The quantities that are used to describe the equilibrium states of a Thermodynamic system. .. Example: Pressure, Volume, Temperature.. Question 17.. What are intensive and extensive variables? Give examples.. Extensive variable depends on the size or mass of the system.. Example: Volume, total mass, entropy, internal energy, heat capacity etc.. Intensive variables do not depend on the size or mass of the system.. Example: Temperature, pressure, specific heat capacity, density etc.. Question 18.. What is an equation of state? Give an example.. The equation that relates the state variables in a specific manner is called equation of state.. Examples:. (i) Gas equation. PV = NkT. (ii) Vander Waal’s equation, .. (p +$$\frac{a}{v^{2}}$$)(v – b) = RT. Question 19.. State Zeroth law of thermodynamics,. The zeroth law of thermodynamics states that if two systems, A and B, are in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other.. Question 20.. Define the internal energy of the system.. The sum of kinetic and potential energies of all the molecules of the system with respect to the centre of mass of the system.. Question 21.. Are internal energy and heat energy the same? Explain.. Internal energy and thermal energy do not mean the same thing, but they are related. Internal energy is the energy stored in a body. It increases when the temperature of the body rises, or when the body changes from solid to liquid or from liquid to gas.. “Heat is the energy transferred from one body to another as a result of a temperature difference.”. Question 22.. Define one calorie.. To raise 1 g of an object by 1°C , 4.186 J of energy is required. In earlier days the heat energy was measured in calorie.. 1 cal = 4.186J. Question 23.. Did joule converted mechanical energy to heat energy? Explain.. In Joule’s experiment, when the masses fall, the paddle wheel turns. The frictional force between paddle wheel and water causes a rise in temperature of water. It is due to the work done by the masses. This infers that gravitational potential energy is transformed into internal energy of water. Thus Joule converted mechanical energy into heat energy.. Question 24.. State the first law of thermodynamics.. ‘Change in internal energy (AU) of the system is equal to heat supplied to the system (Q) minus the work done by the system (W) on the surroundings’.. Question 25.. Can we measure the temperature of the object by touching it?. No, we can’t measure the temperature of the object touching it. Because the temperature is the degree of hotness or coolness of a body. Only we can sense the hotness or coolness of the object.. Question 26.. Give the sign convention for Q and W.. Question 27.. Define the quasi-static process.. A quasi-static process is an infinitely slow process in which the system changes its variables (P, V, T) so slowly such that it remains in thermal, mechanical, and chemical equilibrium with its surroundings throughout.. Question 28.. Give the expression for work done by the gas.. In general, the work done by the gas by increasing the volume from Vi to Vf is given by. W = $$\int_{V_{i}}^{V_{f}} \mathrm{P} d \mathrm{~V}$$. Question 29.. What is PV diagram?. PV diagram is a graph between pressure P and volume V of the system. The P-V diagram is used to calculate the amount of work done by the gas during expansion or on the gas during compression.. Question 30.. Explain why the specific heat capacity at constant pressure is greater than the specific heat capacity at constant volume.. When increasing the temperature of a gas at constant pressure, gas expands and consume some heat to do work. It is not happened, while increasing the temperature of a gas at constant volume. Hence less heat energy is required to increase the temperature of the gas at constant volume. Hence CP is always greater than Cv.. Question 31.. State the equation of state for an isothermal process.. The equation of state for isothermal process is given by,. PV = constant. Question 32.. Give an expression for work done in an isothermal process.. W = μRTln $$\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)$$. Question 33.. Express the change in internal energy in terms of molar specific heat capacity.. When the gas is heated at the constant volume the temperature increases by dT. As no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU. dU = µωdT. Question 34.. Apply first law for (i) an isothermal (ii) adiabatic (Hi) isobaric processes.. (i) For an isothermal process temperature is maintained as constant. Hence,. ∆V= 0. ∴ W = P∆V = 0. According to first law of thermodynamics. ∆U = Q – W. = Q – 0. = Q. ∴ ∆U = Q. (ii) For adiabatic process, Q = 0. By applying first law of thermodynamics. ∆U = Q – W. ∆U = |- W| = W. ∆U = W = $$\frac{\mu R}{\gamma-1}$$ = (Ti – Tf). (iii) (a) For isobaric expansion Q > O. According to first law of thermodynamics. ∆U = – Q – W. ∆U = Q – P∆V. (b) For isobaric compression Q < O.. According to first law of thermodynamics. ∆U = – Q – W. = – Q – P∆V. Question 35.. Give the equation of state for an adiabatic process.. The equation of state for an adiabatic proc ess is given by,. PVγ = Constant. Question 36.. State an equation state for an isochoric process.. The equation of state for an isochoric process is given by,. P = ($$\frac { μR }{ V }$$)T. Question 37.. if the piston of a container is pushed fast inward. Will the ideal gas equation be valid in the intermediate stage? ff not, why?. When the piston is compressed so quickly that there is no time to exchange heat to the surrounding, the temperature of the gas increases rapidly. In this intermediate stage the ideal gas equation be not valid. Because this equation can be relates the pressure, volume and temperature of thermodynamic system at equilibrium.. Question 38.. Draw the PV diagram for,. (a) isothermal process. (c) lsobaric process. (d) lsochoric process. (a) isothermal process:. (c) lsobaric process:. (d) lsochoric process:. Question 39.. What is a cyclic process?. It is a process in which the thermodynamic system returns to its initial state after undergoing a series of changes.. Question 40.. What is meant by a reversible and irreversible processes?. Reversible processes: A thermodynamic process can be considered reversible only if it possible to retrace the path in the opposite direction in such a way that the system and surroundings pass through the same states as in the initial, direct process.. Irreversible processes: All natural processes are irreversible. Irreversible process cannot be plotted in a PV diagram, because these processes cannot have unique values of pressure, temperature at every stage of the process.. Question 41.. State Clausius form of the second law of thermodynamics.. “Heat energy always flows from hotter object to colder object spontaneously”. This is known as the Clausius form of second law of thermodynamics.. Question 42.. State Kelvin-Planck statement of second law of thermodynamics.. It is impossible to construct a heat engine that operates in a cycle, whose sole effect is to convert the heat completely into work.. Question 43.. Define heat engine.. Heat engine is a device which takes heat as input and converts this heat in to work by undergoing a cyclic process.. Question 44.. What are processes involves in a Carnot engine?. 1. Isothermal expansion. 3. Isothermal compression. Question 45.. Can the given heat energy be completely converted to work in a cyclic process? If not, when can the heat can completely converted to work?. No.. According to second law it can’t be completely converted. It is not possible to convert heat completely converted into work, in a cyclic process. But, in Non-cyclic process, heat can be completely converted into work.. Question 46.. State the second law of thermodynamics in terms of entropy.. “For all the processes that occur in nature (irreversible process), the entropy always increases. For reversible process entropy will not change”. Entropy determines the direction in which natural processes should occur.. Question 47.. Why does heat flow from a hot object to a cold object?. Because entropy increases when heat flows from hot object to cold object. If heat were to flow from a cold to a hot object, entropy will be decreased. So second law thermodynamics is violated.. Question 48.. Define the coefficient of performance.. The ratio of heat extracted from the cold body (sink) to the external work done by the compressor W.. COP = ß = $$\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{W}}$$.. Question 1.. Explain the meaning of heat and work with suitable examples.. The spontaneous flow of energy from the object at higher temperature to the one at lower temperature. When they are contact. This energy is called heat. This process of energy transfer from higher temperature object to lower temperature object is called heating.. Example: ‘A hot cup of coffee has more heat’. This statement is correct or wrong?. When heated, a cup of coffee receives heat from the stove. Once the coffee is taken from the stove, the cup of coffee has more internal energy than before. ‘Heat’ is the energy in transit and that flows from a body at higher temperature to an other body at lower temperature. Heat is not a quantity. So the statement ‘A hot cup of coffee has more heat’ is wrong, instead ‘coffee is hot’ will be appropriate.. Meaning of Work: If a body undergoes a displacement then only work is said to be done by the body. Work is also the transfer of energy by other means such as moving a piston of a cylinder containing gas.. Example: When you rub your hands against each other the temperature of the hand is increased. You have done some work on your hands by rubbing. The temperature of the hands increases due to this work. Now when you place your hands on the chin, the temperature of the chin increases.. This is because the hands are at higher temperature than the chin. In the above example, the temperature of hands is increased due to work and temperature of the chin is increased due to heat transfer from the hands to the chin.. The temperature in the system will increase and sometimes may not by doing work on the system. Like heat, work is also not a quantity and through the work energy is transferred to the system.. Question 2.. Discuss the ideal gas laws.. (i) Boyles law: When the gas is kept at constant temperature, the pressure of the gas is inversely proportional to the volume.. P ∝ $$\frac { 1 }{ v }$$. (ii) Charles law: When the gas is kept at constant pressure, the volume of the gas is directly proportional to absolute temperature. V ∝ T. (iii) By combining these two equations we have PV= CT … (1). (iv) Let the constant C as k times the number of particles N. Here k is the Boltzmann constant and it is found to be a universal constant. So the ideal gas law can be stated as follows,. PV = NkT … (2). If the gas consists of p mole of particles then total number of particles is. N = μNA … (3). Substituting the value of N in the equation (2) we get. PV = μNAkT. Here NAk = R. So, the ideal gas law can be written for μ mole of gas is. PV= pRT. It is called equation of state for an ideal gas.. Question 3.. Explain in detail the thermal expansion.. Thermal expansion is the tendency of matter to change in shape, area, and volume due to a change in temperature. When a solid is heated, its atoms vibrate with higher amplitude about their fixed points. At that time, relative change in the size of solids is small.. Liquids expand more than solids because, they have less intermolecular forces than solids. In the case of gas molecules, the intermolecular forces are almost negligible and so they expand much more than solids.. Linear Expansion: In solids, for a small change in temperature ∆T, the fractional change in length $$\frac { ∆L }{ L }$$ is directly proportional to ∆T.. $$\frac { ∆L }{ L }$$ = αL∆T. Therefore, αL = $$\frac { ∆L }{ L∆T }$$. Area Expansion: For an infinitesimal change in temperature ∆T the fractional change in area of a substance is directly proportional to ∆T and it can be written as. $$\frac { ∆A }{ A }$$ = αA∆T. Therefore, αA = $$\frac { ∆A }{ A∆T }$$. Volume Expansion: For a very small change in temperature ∆T the fractional change in volume $$\frac { ∆V }{ V }$$ of a substance is directly proportional to AT.. $$\frac { ∆V }{ V }$$ = αV∆T. Therefore, αV = $$\frac { ∆L }{ V∆T }$$. Question 4.. Describe the anomalous expansion of water. How is it helpful in our lives?. On heating liquids expand and contract on cooling at moderate temperatures. But water exhibits an anomalous behavior. It contracts on heating between 0°C and 4°C. The volume of the given amount of water decreases as . it is cooled from room temperature, until it reach 4°C . Below 4°C the volume increases and so the density decreases. It means that the water has a maximum density at 4°C. This behaviour of water is called anomalous expansion of water.. During winter in cold countries, the surface of the lakes would be at lower temperature than the bottom. Since the solid water (ice) has lower density than its liquid form,’below 4°C, the frozen water will be on the top surface above the liquid water (ice floats). It is due to the anomalous expansion of water. The water in lakes and ponds freeze only at the top. So, the species living in the lakes will be safe at the bottom.. Question 5.. Explain Calorimetry and derive an expression for final temperature when two thermodynamic systems are mixed.. Calorimetry means the measurement of the amount of heat released or absorbed by thermodynamic system during the heating process. When a body at higher temperature is brought in contact with another body at lower temperature, due to transformation of heat. The heat lost by the hot body is equal to the heat gained by the cold body. No heat is allowed to escape to the surroundings. It can be mathematically expressed as. Qgain = – Qlost. Qgain + Qlost = 0. Heat gained or lost can be measured with a calorimeter.. When a sample is heated at high temperature (T1) and immersed into water at room temperature (T2) in the calorimeter. After some time both sample and water reach a final equilibrium temperature Tf. Since the calorimeter is insulated, heat given by the hot sample is equal to heat gained by the water.. Qgain = – Qlost. As per the sign convention, the heat energy lost is denoted by negative sign and heat gained is denoted as positive.. From the definition of specific heat capacity,. Qgain= m2s2(Tf – T2). Qlost = m1s1(Tf – T1). Here s1 and s2 specific heat capacity of hot sample and water respectively.. So we can write,. Question 6.. Discuss various modes of heat transfer.. There are three modes of heat transfer: Conduction, Convection and Radiation.. Conduction: Conduction is the process of direct transfer of heat through matter due to temperature difference. If two objects are placed in direct contact with one another, then heat will be transferred from the hotter object to the colder one.. Convection: Convection is the process in which heat energy transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another.. Example: Boiling of water in a pot.. Radiation: Radiation is a form of energy transfer from one body to another by electromagnetic waves.. Example: Solar energy from the Sun.. Question 7.. Explain in detail Newton’s law of cooling.. Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.. $$\frac { dQ }{ dt }$$ ∝ – (T – Ts) … (1). The negative sign implies that the quantity of heat lost by liquid goes on decreasing with time. Where,. T = Temperature of the object. T = Temperature of the surrounding. From the above graph, it is clear that the rate of cooling is high initially and decreases with falling temperature.. Consider a body of mass m and specific heat capacity s at temperature T. Let T be the temperature of the surroundings. If the temperature falls by a small amount dl in time dt, then the amount of heat lost is,. dQ = msdT. Dividing both sides of equation (2) by dt. $$\frac { dQ }{ dt }$$ ∝ $$\frac { msdT }{ dt }$$ … (3). From Newton’s law of cooling. $$\frac { dQ }{ dt }$$ ∝ (T – Ts). $$\frac { dQ }{ dt }$$ ∝ (T – Ts) … (4). Where a is some positive constant.. From equation (3) and (4). – a(T – Ts) = ms$$\frac { dT }{ dt }$$. $$\frac{d \mathrm{~T}}{\mathrm{~T}-\mathrm{T}_{s}}$$ = – $$\frac { a }{ ms }$$dt … (5). Integrating equation (5) on both sides,. Where b1 is the constant of integration. Taking exponential both sides, we get. T = Ts + b2 $$e^{-\frac{a}{m s} t}$$. here b2 = $$e^{b_{1}}$$ = constant.. Question 8.. Explain Wien’s law and why our eyes are sensitive only to visible rays?. Wien’s law states that, ‘the wavelength of maximum intensity of emission of a black body radiation is inversely proportional to the absolute temperature of the black body’.. λm ∝ $$\frac { 1 }{ T }$$. (or). λm = $$\frac { b }{ T }$$ … (1). It is implied that if temperature of the body increases, maximal intensity wavelength (λm) shifts towards lower wavelength (higher frequency) of electromagnetic spectrum.. Wien’s law and Vision: Why our eye is sensitive to only visible wavelength (in the range 400 nm to 700nm)?. The Sun is approximately considered as a black body. Since any object above 0 K will emit radiation, Sun also emits radiation. Its surface temperature is about 5700 K. By substituting this value in the equation (1),. It is the wavelength at which maximum intensity is 508nm. Since the Sun’s temperature is around 5700K, the spectrum of radiations emitted by Sun lie between 400 nm to 700 nm which is the visible part of the spectrum.. The humans evolved under the Sun by receiving its radiations. The human eye is sensitive only in the visible.. Question 9.. Discuss the,.
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(i) Thermal equilibrium. (ii) Mechanical equilibrium. (iii) Chemical equilibrium. (iv) Thermodynamic equilibrium.. (i) Thermal equilibrium: Two systems are said to be in thermal equilibrium with each other if they are at the same temperature, which will not change with time.. (ii) Mechanical equilibrium: A system is said to be in mechanical equilibrium if no unbalanced force acts on the thermodynamic system or on the surrounding by thermodynamic system.. (iii) Chemical equilibrium: If there is no net chemical reaction between two thermodynamic systems in contact with each other then it is said to be in chemical equilibrium.. (iv) Thermodynamic equilibrium: If two systems are set to be in thermodynamic equilibrium, then the systems are at thermal, mechanical and chemical equilibrium with each other. In a state of thermodynamic equilibrium the macroscopic variables such as pressure, volume and temperature will have fixed values and do not change with time.. Question 10.. Explain Joule’s Experiment of the mechanical equivalent of heat.. Joule showed that mechanical energy can be converted into internal energy and vice versa. In his experiment, two masses were tied with a rope and a paddle wheel as shown in Figure. When these masses fall through a distance h due to gravity, both the masses lose potential energy which is equal to 2mgh. When the masses fall, the paddle wheel turns. Due to the turning of wheel inside water, frictional force comes in between the water and the paddle wheel.. This causes a rise in temperature of the water. This confirms that gravitational potential energy is converted to internal energy of water. The temperature of water increases due to the work done by the masses. In fact, Joule was able to show that the mechanical work has the same effect as giving heat energy. He found that to raise lg of an object by 1°C , 4.186J of energy is required. In earlier days the heat was measured in calorie.. 1 cal = 4.186 J. This is called Joule’s mechanical equivalent of heat.. Question 11.. Derive the expression for the work done in a volume change in a thermodynamic system.. Let us consider a gas contained in the cylinder fitted with a movable piston. When the gas is expanded quasi-statically by pushing the piston by a small distance dx as shown in Figure. Because the expansion occurs quasi- statically. The pressure, temperature and internal energy will have unique values at every instant.. The small work done by the gas on the piston. dW = F dx … (1). The force exerted by the gas on the piston F – PA.. A – area of the piston and. P – pressure exerted by the gas on the piston.. Equation (1) can be rewritten as. dW = PA dx … (2). But A dx = dV change in volume during this expansion process.. Hence the small work done by the gas during the expansion is given by. dW = dV … (3). It is noted that is positive since the volume is increased. In general the work done by the gas by increasing the volume from Vi to Vf is given by. W = $$\int_{V_{i}}^{\mathrm{V}_{f}} \mathbf{P} d \mathrm{~V}$$. Suppose if the work is done on the system, then Vi > Vf.. Then, W is negative.. Question 12.. Derive Meyer’s relation for an ideal gas.. Let us consider μ mole of an ideal gas in a container with volume V, pressure P and temperature T. When the gas is heated at constant volume the temperature increases by dT. Since no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU.. If Cv is the molar specific heat capacity at constant volume,.. dU= μCvdT … (1). When the gas is heated at constant pressure so that the temperature increases by dT. If ‘Q’ is the heat supplied in this process and ‘dV’ the change in volume of the gas.. Q = μCpdT … (2). If W is the workdone by the gas in this process, then. W = P dV … (3). But from the first law of thermodynamics,. Q = dU + W … (4). Substituting equations (1), (2) and (3) in (4), we get,. μCpdT = μCvdT + P dV … (5). For mole of ideal gas, the equation of state is given by,. PV = μRT. ⇒ PdV + VdP = μRdT … (6). Since the pressure is constant, dP = 0. ∴ CpdT = CvdT + RdT. ∴ Cp = Cv + R. (or) Cp – Cv = R … (7). This relation is called Meyer’s relation.. Question 13.. Explain in detail the isothermal process.. It is a process in which both the pressure and volume of a thermodynamic system will change at constant temperature. The ideal gas equation is. PV = μRT. Here, T is constant for this process.. So the equation of state for isothermal process is given by,. PV= constant … (1). It is implied that if the gas goes from one equilibrium state (P1,V1) to another equilibrium state (P2,V2) the following relation holds for this process. P1,V1 = P2,V2 … (2). Since PV = constant, P is inversely proportional to V(P ∝ $$\frac { 1 }{ V }$$). This implies that PV graph is a hyperbola. The pressure-volume graph for constant temperature is also called isotherm.. The following figure shows the PV diagram for quasi-static isothermal expansion and quasi-static isothermal compression. For an isothermal process since temperature is constant, the internal energy is also constant. It implies that dU or ∆U – 0.. For an isothermal process, the first law of thermodynamics can be written as follows,. Q = W … (3). From equation (3), it implies that the heat supplied to a gas is used to do only external work. It is a common misconception that when there is flow of heat energy to the system, the temperature will increase. For isothermal process it is not true. When the piston of the cylinder is pushed, the isothermal compression takes place. This will increase the internal energy which will flow out of the system through thermal contact.. Question 14.. Derive the work done in an isothermal process.. Let us consider an ideal gas which is allowed to expand quasi-statically at constant temperature from initial state (Pi,Vi) to the final state (Pf, Vf. Let us calculate the work done by the gas during this process.. gas, W = $$\int_{V_{i}}^{\mathrm{V}_{f}} \mathrm{P} d \mathrm{~V}$$ … (1). As the process occurs quasi-statically, at every stage the gas is at equilibrium with the surroundings. Since it is in equilibrium at every stage the ideal gas law is valid.. P = $$\frac { μRT }{ V }$$ … (2). Substituting equation (2) and (1) we get,. In equation (3), μRT is taken out of the integral, since it is constant throughout the isothermal process.. By performing the integration in equation (3), we get. W = μRTln$$\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)$$ … (4). Since we have an isothermal expansion,. $$\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}$$ > 1, so $$\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)$$ > 0. As a result the work done by the gas during an isothermal expansion is positive.. Equation (4) is true for isothermal compression also. But in an isothermal compression $$\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}$$ < 1. So $$\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)$$ < 0.. Hence the work done on the gas in an isothermal compression is negative. In the PV diagram the work done during the isothermal expansion is equal to the area under the graph as shown in figure.. Similarly for an isothermal compression, the area under the PV graph is equal to the work done on the gas that turns out to be the area with a negative sign.. Question 15.. Explain in detail an adiabatic process.. It is a process in which no heat energy flows into or out of the system (Q = 0). But the gas can expand by spending its internal energy or gas can be compressed through some external work. Hence the pressure, volume and temperature of the system may change in an adiabatic process.. From the first law for an adiabatic process, ∆U = W. This implies that the work is done by the gas at the expense of internal energy or work is done on the system that increases its internal energy. The adiabatic process can be achieved by the following methods. (i) Thermally insulating the system from surroundings. So that no heat energy flows into or out of the system. For instance, when thermally insulated cylinder of gas is compressed (adiabatic compression) or expanded (adiabatic expansion) as shown in the Figure.. (ii) If the process occurs so quickly so that there is no time to exchange heat with surroundings even though there is no thermal insulation. The equation (1). Implies that if the gas goes from an equilibrium state (Pi, Vi) to another equilibrium state (Pf,Vf) adiabatically then it satisfies the relation,. PVγ = constant … (1). The equation (1) Here γ is called adiabatic exponent ( γ = $$\frac{C_{\mathrm{P}}}{C_{\mathrm{V}}}$$) which depends on the nature of the gas.. The equation (1). Implies that if the gas goes from an equilibrium state (Pi,Vi) to another equilibrium state (Pf, Vf) adiabatically then it satisfies the relation,. PiViγ = PfVfγ … (2). The PV diagram for an adiabatic process is also called adiabatic. It is noted that the PV diagram for isothermal and adiabatic processes look similar. But actually the adiabatic curve is steeper than isothermal curse.. Let us rewrite the equation (1) in terms of T and V. From ideal gas equation, the pressure P = $$\frac { μRT }{ V }$$. Substituting this equation in the equation (1), we have. $$\frac { μRT }{ V }$$Vγ = constant. (or) $$\frac { T }{ V }$$Vγ = $$\frac { constant }{ μR }$$. Note here that is another constant. So it can. be written as. TVγ-1 = constant … (3). From the equation (3) it is known that if the gas goes from an initial equilibrium state (Ti,Vi) to final equilibrium state (Tf, Vf) adiabatically then it satisfies the relation. $$\mathrm{T}_{i} \mathrm{~V}_{i}^{\gamma-1}=\mathrm{T}_{f} \mathrm{~V}_{f}^{\gamma-1}$$ … (4). The equation of state for adiabatic process can also be written in terms of T arid P as. $$\mathrm{T}^{\gamma} \mathrm{P}^{1-\gamma}$$ = constant. Question 16.. Derive the work done in an adiabatic process.. Let us consider p moles of an ideal gas enclosed in a cylinder having perfectly non conducting walls and base. A frictionless and insulating piston A is fitted in the cylinder as shown in Figure. It’s area of cross-section be A.. Let W be the work done when the system goes from the initial state (PiViTi) to the final state (PfVfTf) adiabatically.. W = $$\int_{V_{i}}^{\mathrm{V}_{f}} \mathrm{P} d \mathrm{~V}$$ … (1). It is assumed that the adiabatic process occurs quasi-statically, at every stage the ideal gas law is valid. Under this condition, the adiabatic equation of state is PVγ = constant (or) P = $$\frac{\text { constant }}{\mathrm{V}^{\gamma}}$$ can be substituted in the equation (1), we get. From ideal gas law,. In adiabatic expansion, work is done by the gas. i.e., Wadia is positive. As Ti > Tf the gas cools during adiabatic expansion.. In adiabatic compression, work is done on the gas. i.e., Wadia is negative. As Ti < Tf hence the temperature of the gas increases during adiabatic compression.. Question 17.. Explain the isobaric process and derive the work done in this process.. Isobaric process is a thermodynamic process that occurs at constant pressure. Inspite of the pressure consistency of this process, temperature, volume and internal energy are not constant. From the ideal gas equation, we have. V = ($$\frac { μR }{ P }$$)T … (1). Here, $$\frac { μR }{ P }$$ = constant. In an isobaric process the temperature is directly proportional to volume.. V ∝ T(Isobaric process) … (2). It is implied that for a isobaric process, the V-T graph is a straight line passing through the origin.. Work done by the gas. W= $$\int_{V_{i}}^{\mathrm{V}_{f}} \mathrm{P} d \mathrm{~V}$$. In an isobaric process, the pressure is constant, So P comes out of the integral,. ∆V – change in the volume. If AV is negative, W is also negative. It is implied that the work is done on the gas. If ∆V is positive, W is also positive, implying that work is done by the gas. From the ideal gas equation.. equation (4) can be rewritten as. PV = μRT and V = $$\frac { μRT }{ P }$$. Substituting this in equation (4) we get,. W = μRTf(1 – $$\frac{\mathrm{T}_{i}}{\mathrm{~T}_{f}}$$) … (5). In the PV diagram, area under the isobaric curve is equal to the work done in isobaric process. The shaded area in the following Figure is equal to the work done by the gas.. The first law of thermodynamics for isobaric process is given by,. ∆U = Q – P∆V. Question 18.. Explain in detail the isochoric process.. Isochoric process is a thermodynamic process in which the volume of the system is kept constant. But pressure, temperature and internal energy continue to be variables. The pressure-volume graph for an isochoric process is a vertical line parallel to pressure axis as shown in Figure.. The equation of state for an isochoric process is given by,. P = $$\frac { μR }{ V }$$ … (1). Here $$\frac { μR }{ V }$$ = constant. ∴ P ∝ T. It is inferred that the pressure is directly proportional to temperature. This implies that the P-T graph for an isochoric process is a straight line passing through origin. If a gas goes from state (Pi, Ti) to (Pf, Tf) at constant volume, then the system satisfies the following equation,. $$\frac{P_{i}}{T_{i}}=\frac{P_{f}}{T_{f}}$$. For an isochoric processes, AV=0 and W-0. Then the first law of thermodynamics becomes. ∆U = Q … (3). It is implied that the heat supplied is used to increase only the internal energy. As a result the temperature increases and pressure also increases.. Question 19.. What are the limitations of the first law of thermodynamics?. The first law of thermodynamics explains well the inter convertibility of heat and work. But is not indicated the direction of change.. Example:. (i) When a hot object is in contact with a cold object, heat always flows from the hot object to cold object but not in the reverse direction. According to first law of thermodynamics, it is possible for the energy to flow from hot object to cold object and viceversa. But in nature the direction of heat flow is always from higher temperature to lower temperature.. (ii) When brakes are applied, a car stops due to friction and the work done against friction is converted into heat. But this heat is not reconverted into the kinetic energy of the car. Hence the first law is not sufficient to explain many of natural phenomena.. Question 20.. Explain the heat engine and obtain its efficiency.. Heat engine is a device which takes heat as input and converts this heat into work by undergoing a cyclic process.. A heat engine has three parts:. (i) Hot reservoir. (ii) Working substance. (iii) Cold reservoir. A Schematic diagram for heat engine is given below in the figure.. • Hot reservoir (or) Source: It supplies heat to the engine. Which is always maintained at a high temperature TH.. • Working substance: It is a substance like gas or water, that converts the heat supplied into work.. • Cold reservoir (or) Sink: The heat engine ejects some amount of heat (QL) into cold reservoir after it doing work. It is always maintained at a low temperature TL.. The heat engine works in a cyclic process. After a cyclic process it returns to the same state. Since the heat engine returns to the same state after it ejects heat, the change in the internal energy of the heat engine is zero.. The efficiency of the heat engine is defined as the ratio of the work done (output) to the heat absorbed (input) in one cyclic process. Let the working substance absorb heat QH units from the source and reject QL units to the sink after doing work W units.. We can write,. Input heat = Work done + ejected heat. QH = W + QL. W = QH – QL. Then the efficiency of heat engine. Question 21.. Explain in detail Carnot heat engine.. A reversible heat engine operating in a cycle between two temperatures in a particular way is called a Carnot Engine. The carnot engine consists of four parts that are given below.. 1. Source: It is the source of heat maintained at constant high temperature TH. Without changing its temperature any amount of heat can be extracted from it.. 2. Sink: It is a cold body maintained at a constant low temperature TL. It can absorb any amount of heat.. 3. Insulating stand: It is made of perfectly non-conducting material. Heat is not conducted through this stand.. 4. Working substance: It is an ideal gas enclosed in a cylinder with perfectly non-conducting walls and perfectly conducting bottom. A non-conducting and frictionless piston is fitted in it.. The four parts are shown in the following Figure.. Carnot’s cycle: In camot’s cycle, the working substance is subjected to four successive reversible processes.. Let the initial pressure, volume of the working substance be P1, V1.. When the cylinder is placed on the source, the heat (QH) flows from source to the working substance (ideal gas) through the bottom of the cylinder.. The input heat increases the volume of the gas.So the piston is allowed to move out very’ slowly.. W1 is the work done by the gas it expands from volume V1 to volume V2 with a decrease of pressure from P1 to P2.. Then the work done by the gas (working substance) is given by. ∴ QH = QA→B $$\int_{V_{1}}^{\mathrm{V}_{2}} \mathrm{P} d \mathrm{~V}$$. When the cylinder is placed on the insulating stand and the piston is allowed to move out, the gas expands adiabatically from volume V2 to volume V3 the pressure falls from P2 to P3. The temperature falls to TLK.. The work done by the gas in an adiabatic expansion is given by,. Area under the curve BC.. Isothermal compression from (P3, V3, TL) to (P4, V4, TL).. Let the cylinder is placed on the sink and the gas is isothermally compressed until the pressure and volume become P4 and V4 respectively.. Let the cylinder is placed on the insulating stand again and the gas is compressed adiabatically till it attains the initial pressure P1, volume V2 and temperature TH, Which is shown by the curve DA in the P – V diagram.. = – Area under the curve DA. Let ‘W’ be the net work done by the working substance in one cycle. The net work done by the Carnot engine in one cycle is given by,. W = WA→B – WC→D. Net work done by the working substance in one cycle is equal to the area (enclosed by ABCD) of the P-V diagram.. Question 22.. Derive the expression for Carnot engine efficiency.. Efficiency is defined as the ratio of work done by the working substance in one cycle to the amount of heat extracted from the source.. Let us omit the negative sign. Since we are interested in only the amount of heat (QL) ejected into the sink, we have. By applying adiabatic conditions, we get,. TH V2γ-1 = TL V3γ-1. TH V1γ-1 = TL V4γ-1. By dividing the above two equations, we get. $$\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)$$γ-1 = $$\left(\frac{\mathrm{V}_{3}}{\mathrm{~V}_{4}}\right)$$γ-1. Which implies that. $$\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}=\frac{\mathrm{V}_{3}}{\mathrm{~V}_{4}}$$ … (5). Substituting equation (5) and (4), we get. $$\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{Q}_{\mathrm{H}}}=\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$$. ∴ The efficiency. η = 1 – $$\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$$. Question 23.. Explain the second law of thermodynamics in terms of entropy.. $$\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{Q}_{\mathrm{H}}}$$ = $$\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$$ … (1). According to efficiency of Carnot’s engine equation, quantity $$\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}$$ = $$\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{L}}}$$.. The quantity $$\frac { Q }{ T }$$ is called entropy. It is a very important thermodynamic property of a System, which is also a state variable.. $$\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}$$ is the entropy received by the Carnot engine from hot reservoir and $$\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{L}}}$$ is entropy given out by the Carnot engineLto the cold reservoir. For reversible engines (Carnot Engine) both entropies should be same, so that the change in entropy of the Carnot engine in one cycle is zero. It is proved in equation (1).. But for all practical engines like diesel and petrol engines which are not reversible engines, they satisfy the relation $$\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{L}}}$$ > $$\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}$$.. In fact the second law of thermodynamics can be reformulated as follows “For all the processes that occur in nature(irreversible process), the entropy always increases. For reversible process entropy will not change”. Entropy determines the direction in which natural process must occur.. Question 24.. Explain in detail the working of a refrigerator.. A refrigerator is a Carnot’s engine working in the reverse order which is shown in the figure.. 8.17 Schematic diagram of a refrigerator The working substance (gas) absorbs a quantity of heat QL from the cold body (sink) at a lower temperature TL. A certain amount of work W is done on the working substance by the compressor and a quantity of heat QH is rejected to the hot body (source) i.e., the atmosphere at TH. When we stand beneath of the refrigerator, we can feel warmth air.. From the first law of thermodynamics, we have QL + W = QH. Hence the cold reservoir (refrigerator) further cools down and the surroundings (kitchen or atmosphere) gets hotter.. IV. Numerical problems:. Question 1.. Calculate the number of moles of air is in the inflated balloon at room temperature as shown in the figure.. The radius of the balloon is 10 cm, and pressure inside the balloon is 180 kPa.. The pressure inside the balloon P = 1.8 x 105 P. Room temperature. T = 273 + 30 = 303K.. Let the radius of the balloon be R = 10 x 10-2m.. Volume of the balloon. Question 2.. In the planet Mars, the average temperature is around – 53°C and atmospheric pressure is 0.9 kPa. Calculate the number of moles of. the molecules in unit volume in the planet Mars? Is this greater than that in earth?. = 0.9 x 10³Pa. PV = NkT. Since volume is unity V = 1. Equation (1) becomes. P = NkT. Where k is Boltzmann constant. Question 3.. An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V1 and contains ideal gas at pressure P1 and temperature T1. The other chamber has volume V1 and contains ideal gas at pressure P2 and temperature T2. If the partition is removed without doing any work on the gases, calculate the final equilibrium temperature of the container.. In first chamber let the pressure be P1. In second chamber let the pressure be P2. In first chamber let the volume be V1. In second chamber let the volume be V2. In first chamber let the temperature be T1. In second chamber let the temperature be T2. According to conservation of energy,. Question 4.. The temperature of a uniform rod of length L having a coefficient of linear expansion αL is changed by ∆T. Calculate the new moment of inertia of the uniform rod about axis passing through its centre and perpendicular to an axis of the rod.. Moment of inertia of uniform rod of mass M, and length L about axis passing through its centre and perpendicular to an axis of rod is given by. I = $$\frac{\mathrm{ML}^{2}}{12}$$. The increase in length of the rod. ∆l = LαL∆T. αL – Coefficient of linear expansion.. ∆T – change in temperature. After the rod is heated,. Moment of inertia,. Question 5.. Draw the TP diagram (P-x axis, T-y axis), VT(T-x axis, V-y axis) diagram for. (a) Isochoric process. (b) Isothermal process. (c) isobaric process.. (a) For isochoric process:. V = V0 = constant. The points a and b are represented as a = (P1V0T0) b = (P2, V0, T2). (b) For isothermal process:. T = T0 = constant. PV = nRT0. P(T) = multivalue. P(T) = $$\frac{n \mathrm{RT}_{0}}{\mathrm{~V}}$$. The points a and b are represented as. a = (P1V1,T) b = (P2V0T0). (c) For isobaric process:. P = P0 = constant. T(V) = $$\frac{\mathrm{P}_{0} \mathrm{~V}}{n^{2}}$$. P0V = nRT. P(T) = P0. Question 6.. A man starts bicycling in the morning at a temperature around 25°C, he checked the pressure of tire which is equal to be 500 kPa. Afternoon he found that the absolute pressure in the tyre is increased to 520 kPa. By assuming the expansion of tyre is negligible, what is the temperature of tyre at afternoon?. Question 7.. The temperature of a normal human body is 98.6°F. During high fever if the temperature increases to 104°F, what is the change in peak wavelength that emitted by our body? (Assume human body is a black body). Normal temperature of human body = 98.6°F. Conversion of F° to C°. C = $$\frac { (F-32) }{ 1.8 }$$. = $$\frac { (98.6-32 }{ 1.8 }$$. = $$\frac { 66.6 }{ 1.8 }$$. = 37°C. Conversion of C° to Kelvin. T = 37 + 273 = 310. Peak wavelength at 98.6°F is. λm = $$\frac { b }{ T }$$. = $$\frac{2.898 \times 10^{-3}}{310}$$. = 0.009348 x 10-3 m. = 9348 x 10-9. = 9348 nm. Final temperature of the human body = 104°F. Conversion of F° to C°. C = $$\frac { F-32 }{ 1.8 }$$. = $$\frac { 104-32 }{ 1.8 }$$. = $$\frac { 72) }{ 1.8 }$$. Conversion of 40° to Kelvin. T = 40+ 273 = 313 K. Peak wavelength at 104°F is. λm = $$\frac { b }{ T }$$. = $$\frac{2.898 \times 10^{3}}{313}$$. = 0.009258 x 10-3 m. = 9258 x 10-9 = 9258 nm. λmax = 9348 nm at 98.6°F. λmax = 9258 nm at 104°F. Question 8.. In an adiabatic expansion of the air, the volume is increased by 4%, what is percentage change in pressure? (For air γ = 1.4). Percentage of increased volume = 4%. $$\frac { ∆V }{ V }$$ x 100 = 4%. γ = 1.4. PVγ = Constant. Question 9.. In a petrol engine, (internal combustion engine) air at atmospheric pressure and temperature of 20°C is compressed in the cylinder by the piston to 1/8 of its original volume. Calculate the temperature of the compressed air. (For air γ = 1.4). P1 = 1 atmospheric pressure, V1 = V, V2 = $$\frac { V }{ 8 }$$. T1 = 20 + 273 – 293K, T2 – ?, γ = 1.4.. ∴ The temperature of the compressed air = 400°C. Question 10.. Consider the following cyclic process consist of isotherm, isochoric and isobar which is given in the figure.. Draw the same cyclic process qualitatively in the V-T diagram where T is taken along x direction and V is taken along y-direction. Analyze the nature of heat exchange in each process.. Process 1 to 2: increase in volume. So heat must be added.. Process 2 to 3: Volume remains constant. Increase in temperature. The given heat is used to increase the internal energy.. Process 3 to 1: Pressure remains constant. Volume and Temperature are reduced. Heat flows out of the system. It is an isobaric compression where the work is done on the system.. Explanation: In the graph, during the process (1 to 2), the gas undergoes isothermal expansion. It receives certain amount of heat from the surroundings. It uses this heat in doing the work. Hence internal energy of the gas remains unchanged.. During the process represented by (2 to 3) the gas is heated at constant volume. Since no work is done and volume does not change, the process is isochoric process.. Since heat is transferred to the gas from the surroundings, the internal energy of the gas is increased. During the process represented by (3 to 1) the gas is compressed isobarically. Work is done on the gas. Since temperature drops internal energy is reached. Hence the gas gives up heat to the surroundings.. Question 11.. An ideal gas is taken in a cyclic process as shown in the figure. Calculate. (a) work done by the gas.. (b) work done on the gas.. (c) Net work done in the process. (a) Let the work done by the gas along AB be W. From the fig pressure,. P = 600 N/m². From the fig change in volume,. ∆V = 3. ∴Work done,. W = PAV. = 600 x 3 = 1800J. = 1.8 kJ. (b) Isobaric compression take place and work is done on the gas along BC.. W = – PAV. ∆V = (6 – 3) = 3. P = +400N/m². ∴ W= – 400 x 3. = – 1200. = – 1.2 kJ. (c). Question 12.. For a given ideal gas 6 x 105 J heat energy is supplied and the volume of gas is increased from 4 m³ to 6 m² at atmospheric pressure. Calculate. (a) the work done by the gas. (b) change in internal energy of the gas. (c) graph this process in PV and TV diagram.. Heat energy supplied to the gas. Q = 6 x 105J. (b) Change in internal energy. ∆U = ∆Q – AW. = ∆Q – P∆V. = 6 x 105 – 2.026 x 105. = 3,974 x 103. ∆U = 3.974 kJ. (c). Question 13.. Suppose a person wants to increase the efficiency of the reversible heat engine that is operating between 100°C and 300°C. He had two ways to increase the efficiency.. (a) By decreasing the cold reservoir temperature from 100°C to 50°C and keeping the hot reservoir temperature constant. (b) By increasing the temperature of the hot reservoir from 300°C tc 350°C by keeping the cold reservoir temperature constant. Which is the suitable method?. Temperature of cold reservoir. T1 = 100°C. = 100 + 273 – 373 K. Temperature of hot reservoir. TH = 300 + 273 = 573 K. η = 1 – $$\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$$. η = 1 – $$\frac { 373 }{ 573 }$$. η = 1 – $$\frac { 573 – 373 }{ 573 }$$. = $$\frac { 200 }{ 753 }$$. = 0.349 x 100. Initial efficiency = 34.9%. (a) By decreasing the temperature of the cold reservoir from 100°C to 50PC.. (b) By increasing the temperature of the hot reservoir from 300°C to 350°C,. T1 = 350 + 273 = 623K. T1 = 100 + 273 =373K. ∴ Efficiency,. η = 1 – $$\frac{\mathrm{T}_{\mathrm{2}}}{\mathrm{T}_{\mathrm{1}}}$$. η = 1 – $$\frac { 373 }{ 623 }$$. η = $$\frac { 623 – 373 }{ 623 }$$. η = $$\frac { 250 }{ 623 }$$. η = 0.4012. η = 0.4012 x 100 = 40.12%. ∴ Method (a) is more efficient than method (b).. Question 14.. A Carnot engine whose efficiency is 45% takes heat from a source maintained atatemperature of 327°C. To have an engine of efficiency 60% what must be the intake temperature for the same exhaust (sink) temperature?. Temperature of a source,. T1 = 327°C = 327 + 273 = 600 k. η = $$\frac { 45 }{ 100 }$$ = 0.45. Efficiency η = 1 – $$\frac{\mathrm{T}_{\mathrm{2}}}{\mathrm{T}_{\mathrm{1}}}$$. ∴ 0.45 = 1 – $$\frac{\mathrm{T}_{2}}{600}$$. $$\frac{\mathrm{T}_{2}}{600}$$ = 1 – 0.45 = 0.55. ∴ T2 = 0.55 x 600 = 330 K. η = 60% = $$\frac { 60 }{ 100 }$$ = 0.6. 0.6 = 1 – $$\frac{\mathrm{T}_{\mathrm{2}}}{\mathrm{T}_{\mathrm{1}}}$$ = 1 – $$\frac{330}{\mathrm{~T}_{1}}$$. $$\frac{330}{\mathrm{~T}_{1}}$$ = 1 – 0.6 = 0.4. In take temperature. T1 = $$\frac { 330 }{ 0.4 }$$ = 825K. = 825 – 273. = 552°C. The intake temperature = 552°C. Question 15.. An ideal refrigerator keeps its content at 0°C while the room temperature is 27°C. Calculate its coefficient of performance.
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Break-even analysis also can help companies determine the level of sales (in dollars or in units) that is needed to make a desired profit. The process for factoring a desired level of profit into a break-even analysis is to add the desired level of profit to the fixed costs and then calculate a new break-even point. We know that Hicks Manufacturing breaks even at 225 Blue Jay birdbaths, but what if they have a target profit for the month of July? By calculating a target profit, they will produce and (hopefully) sell enough bird baths to cover both fixed costs and the target profit.
• Reducing expenses lowers your break-even point and increases your opportunities for profits.
• Alternatively, you can find the break-even point in sales dollars and then find the number of units by dividing by the selling price per unit.
• Things like looking for an affordable office or warehouse to rent will decrease BPE.
• We will use this ratio (Figure 3.9) to calculate the break-even point in dollars.
The contribution margin may not always be the same from month to month. Note that in either scenario, the break-even point is the same in dollars and units, regardless of approach. Thus, you can always find the break-even point (or a desired profit) in units and then convert it to sales by multiplying by the selling price per unit. Alternatively, you can find the break-even point in sales dollars and then find the number of units by dividing by the selling price per unit.
## Logistics Calculators
Neil has a protein supplement company that wants to introduce a new flavour. Before launching this new flavour, he wants to determine how it will impact his company’s finances. That’s why he decided to calculate the break-even point to find out if it was worth the investment. The cosmetic company must generate \$379,746 in lipsticks sales dollars to break even. Doing a break even analysis can provide deep insight into financial performance, profitability, and how to grow the business further.
The concept of break-even analysis is concerned with the contribution margin of a product. The contribution margin is the excess between the selling price of the product and the total variable costs. This \$40 reflects the amount of revenue collected to cover the remaining fixed costs, which are excluded when figuring the contribution margin. The total fixed costs, variable costs, unit or service sales are calculated on a monthly basis in this calculator. Meaning that adding the total for all products and services monthly should account for all products and services.
Now, if you want to pay your \$20,000 commercial loan within 3 years, you must add \$5 per unit on your product. By knowing you’re within your competitor’s price range, you are also not overpricing your product. When you analyze the BEP, you might find that \$25 is too steep a price for a new product. Depending on your market, it might not be a competitive price that entices sales. Once you know this, you can adjust your loan repayment to a three-year period instead. This way, you can keep the selling price more reasonable while still paying off your commercial loan.
• For working capital, it can range for seven to ten years, while purchasing equipment can have a ten-year payment term.
• What happens when Hicks has a busy month and sells 300 Blue Jay birdbaths?
• One way to decrease BEP is to reduce the variable cost needed to produce a product.
• Barbara is the managerial accountant in charge of a large furniture factory’s production lines and supply chains.
It can also refer to the amount of money for which a product or service must be sold to cover the costs of manufacturing or providing it. Another good way to engage potential clients is by being active on social media. Depending on your business, you can stay connected with your consumers via Facebook, Instagram, or Twitter. These days, it’s definitely to worthwhile to invest in social media marketing. Social media can help promote events, post about promos and discounts, and create an image that is accessible for consumers. According to eMarketer.com, in 2021, around 91.9% of U.S. marketers in companies with over 100 employees were expected to use social media to market products and services.
There are two main business factors that impact BPE, these are fixed costs and variable costs. The fixed costs refer to necessary expenses such as rent or mortgage payments, utilities, marketing, research and development, etc. These are essential operational expenses that keep your business afloat even when you’re not producing goods. Meanwhile, variable costs are expenditures that increase when you raise your production.
Sales Price per Unit- This is how much a company is going to charge consumers for just one of the products that the calculation is being done for. A gross break-even point is often not entirely correct for figuring out exactly where you would break even on a trade, investment, or project. This is because taxes, fees, and other charges are often involved that must be taken into account.
## AccountingTools
You can then generate BEP reports and share them across your company to encourage different departments to implement actionable changes. However, Company V gives sales commissions based on total revenue, so it also needs to know the total dollar amount it’d need to sell this quarter to break even. Even the smallest expenses can add up over time, and if companies aren’t keeping tabs on these costs, it can lead to major surprises down the road.
## Break even analysis example
On the other hand, you can keep the same price (\$20), which allow you to make more profits per unit. On top of this, what if you have a startup loan you need to repay? Let’s say you have a \$20,000 commercial loan that you bad debt expense want to pay off in two years. To do this, you must put an additional \$10 per unit if you intend to sell 2,000 products in two years. Thus, your BEP selling point will be \$25 per product during the first two years.
Next, Barbara can translate the number of units into total sales dollars by multiplying the 2,500 units by the total sales price for each unit of \$500. Barbara is the managerial accountant in charge of a large furniture factory’s production lines and supply chains. Determine fixed costs
You’ll first need to identify fixed costs for your business – essentially, costs that don’t change even if the business output is high or low. Also, by understanding the contribution margin, businesses can make informed decisions about the pricing of their products and their levels of production. Businesses can even develop cost management strategies to improve efficiencies.
## How to Calculate Break Even Point in Units
BEP could be stated as the necessary number of units sold or hours of services rendered to equal the amount of revenue. This equation looks similar to the previous BEP analysis formula, but it has one key difference. Instead of dividing the fixed costs by the profit gained from each sale, it uses the percentage of how much value you’re getting from each unit. Companies have many fixed overhead expenses such as rent, salaries, taxes, and insurance. Add in the variable expenses of supplies, materials, research and development, labor costs, and marketing (among others), and you get total expenses.
## How to Connect Your Payoneer and eBay Accounts: 6 Simple Steps
This includes term loans, business lines of credit, and even equipment loans. Just remember that qualifications, rates, and terms vary per lender. Either way, you should have an effective means to track if your campaigns are generating awareness. If your campaign has been on for a few months with hardly any improvements to your sales, it’s likely better to cancel them.
## Contribution Margin
When the market price increases, that’s when investors earn profit. Arm your business with the tools you need to boost your income with our interactive profit margin calculator and guide. There are a few ways to calculate your BEP, but if you have a strong CRM like Zendesk Sell, it can calculate the values for you.
By reaching this number of unit sales, the company has not gained profits yet. Once Company A sells over 500,000 units, that’s when it will earn profits. One can determine the break-even point in sales dollars (instead of units) by dividing the company’s total fixed expenses by the contribution margin ratio. Simply enter your fixed and variable costs, the selling price per unit and the number of units expected to be sold. In the first calculation, divide the total fixed costs by the unit contribution margin. In the example above, assume the value of the entire fixed costs is \$20,000.
Equipment failures also mean higher operational costs and, therefore, a higher break-even. As you can see, the \$38,400 in revenue will not only cover the \$14,000 in fixed costs, but will supply Marshall & Hirito with the \$10,000 in profit (net income) they desire. By knowing at what level sales are sufficient to cover fixed expenses is critical, but companies want to be able to make a profit and can use this break-even analysis to help them. When a company breaks even, it’s reached a point where it does not have profits or loss. This is called the break even point (BPE), when a business’s revenue is equal to its expenses.
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Break-even analysis also can help companies determine the level of sales (in dollars or in units) that is needed to make a desired profit. The process for factoring a desired level of profit into a break-even analysis is to add the desired level of profit to the fixed costs and then calculate a new break-even point. We know that Hicks Manufacturing breaks even at 225 Blue Jay birdbaths, but what if they have a target profit for the month of July? By calculating a target profit, they will produce and (hopefully) sell enough bird baths to cover both fixed costs and the target profit.. • Reducing expenses lowers your break-even point and increases your opportunities for profits.. • Alternatively, you can find the break-even point in sales dollars and then find the number of units by dividing by the selling price per unit.. • Things like looking for an affordable office or warehouse to rent will decrease BPE.. • We will use this ratio (Figure 3.9) to calculate the break-even point in dollars.. The contribution margin may not always be the same from month to month. Note that in either scenario, the break-even point is the same in dollars and units, regardless of approach. Thus, you can always find the break-even point (or a desired profit) in units and then convert it to sales by multiplying by the selling price per unit. Alternatively, you can find the break-even point in sales dollars and then find the number of units by dividing by the selling price per unit.. ## Logistics Calculators. Neil has a protein supplement company that wants to introduce a new flavour. Before launching this new flavour, he wants to determine how it will impact his company’s finances. That’s why he decided to calculate the break-even point to find out if it was worth the investment. The cosmetic company must generate \$379,746 in lipsticks sales dollars to break even. Doing a break even analysis can provide deep insight into financial performance, profitability, and how to grow the business further.. The concept of break-even analysis is concerned with the contribution margin of a product. The contribution margin is the excess between the selling price of the product and the total variable costs. This \$40 reflects the amount of revenue collected to cover the remaining fixed costs, which are excluded when figuring the contribution margin. The total fixed costs, variable costs, unit or service sales are calculated on a monthly basis in this calculator. Meaning that adding the total for all products and services monthly should account for all products and services.. Now, if you want to pay your \$20,000 commercial loan within 3 years, you must add \$5 per unit on your product. By knowing you’re within your competitor’s price range, you are also not overpricing your product. When you analyze the BEP, you might find that \$25 is too steep a price for a new product. Depending on your market, it might not be a competitive price that entices sales. Once you know this, you can adjust your loan repayment to a three-year period instead. This way, you can keep the selling price more reasonable while still paying off your commercial loan.. • For working capital, it can range for seven to ten years, while purchasing equipment can have a ten-year payment term.. • What happens when Hicks has a busy month and sells 300 Blue Jay birdbaths?. • One way to decrease BEP is to reduce the variable cost needed to produce a product.. • Barbara is the managerial accountant in charge of a large furniture factory’s production lines and supply chains.. It can also refer to the amount of money for which a product or service must be sold to cover the costs of manufacturing or providing it. Another good way to engage potential clients is by being active on social media. Depending on your business, you can stay connected with your consumers via Facebook, Instagram, or Twitter. These days, it’s definitely to worthwhile to invest in social media marketing. Social media can help promote events, post about promos and discounts, and create an image that is accessible for consumers. According to eMarketer.com, in 2021, around 91.9% of U.S. marketers in companies with over 100 employees were expected to use social media to market products and services.. There are two main business factors that impact BPE, these are fixed costs and variable costs. The fixed costs refer to necessary expenses such as rent or mortgage payments, utilities, marketing, research and development, etc. These are essential operational expenses that keep your business afloat even when you’re not producing goods. Meanwhile, variable costs are expenditures that increase when you raise your production.. Sales Price per Unit- This is how much a company is going to charge consumers for just one of the products that the calculation is being done for.
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A gross break-even point is often not entirely correct for figuring out exactly where you would break even on a trade, investment, or project. This is because taxes, fees, and other charges are often involved that must be taken into account.. ## AccountingTools. You can then generate BEP reports and share them across your company to encourage different departments to implement actionable changes. However, Company V gives sales commissions based on total revenue, so it also needs to know the total dollar amount it’d need to sell this quarter to break even. Even the smallest expenses can add up over time, and if companies aren’t keeping tabs on these costs, it can lead to major surprises down the road.. ## Break even analysis example. On the other hand, you can keep the same price (\$20), which allow you to make more profits per unit. On top of this, what if you have a startup loan you need to repay? Let’s say you have a \$20,000 commercial loan that you bad debt expense want to pay off in two years. To do this, you must put an additional \$10 per unit if you intend to sell 2,000 products in two years. Thus, your BEP selling point will be \$25 per product during the first two years.. Next, Barbara can translate the number of units into total sales dollars by multiplying the 2,500 units by the total sales price for each unit of \$500. Barbara is the managerial accountant in charge of a large furniture factory’s production lines and supply chains. Determine fixed costs. You’ll first need to identify fixed costs for your business – essentially, costs that don’t change even if the business output is high or low. Also, by understanding the contribution margin, businesses can make informed decisions about the pricing of their products and their levels of production. Businesses can even develop cost management strategies to improve efficiencies.. ## How to Calculate Break Even Point in Units. BEP could be stated as the necessary number of units sold or hours of services rendered to equal the amount of revenue. This equation looks similar to the previous BEP analysis formula, but it has one key difference. Instead of dividing the fixed costs by the profit gained from each sale, it uses the percentage of how much value you’re getting from each unit. Companies have many fixed overhead expenses such as rent, salaries, taxes, and insurance. Add in the variable expenses of supplies, materials, research and development, labor costs, and marketing (among others), and you get total expenses.. ## How to Connect Your Payoneer and eBay Accounts: 6 Simple Steps. This includes term loans, business lines of credit, and even equipment loans. Just remember that qualifications, rates, and terms vary per lender. Either way, you should have an effective means to track if your campaigns are generating awareness. If your campaign has been on for a few months with hardly any improvements to your sales, it’s likely better to cancel them.. ## Contribution Margin. When the market price increases, that’s when investors earn profit. Arm your business with the tools you need to boost your income with our interactive profit margin calculator and guide. There are a few ways to calculate your BEP, but if you have a strong CRM like Zendesk Sell, it can calculate the values for you.. By reaching this number of unit sales, the company has not gained profits yet. Once Company A sells over 500,000 units, that’s when it will earn profits. One can determine the break-even point in sales dollars (instead of units) by dividing the company’s total fixed expenses by the contribution margin ratio. Simply enter your fixed and variable costs, the selling price per unit and the number of units expected to be sold. In the first calculation, divide the total fixed costs by the unit contribution margin. In the example above, assume the value of the entire fixed costs is \$20,000.. Equipment failures also mean higher operational costs and, therefore, a higher break-even. As you can see, the \$38,400 in revenue will not only cover the \$14,000 in fixed costs, but will supply Marshall & Hirito with the \$10,000 in profit (net income) they desire. By knowing at what level sales are sufficient to cover fixed expenses is critical, but companies want to be able to make a profit and can use this break-even analysis to help them. When a company breaks even, it’s reached a point where it does not have profits or loss. This is called the break even point (BPE), when a business’s revenue is equal to its expenses.
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# Detailed Lesson Plan In Mathematics VI
```Detailed Lesson Plan
In Mathematics VI
I.
Objectives:
At the end of the class 100% of the students should be able to learn 75% of the lesson and
be able to:
a. Visualize and describe the different solid figures
b. Identify the faces vertex and edges of a solid figure.
c. Sketches the different solid figures in a real world.
d. Display cooperativeness during group activity.
II.
Subject Matter:
a. Topic: Solid Figures
b. References: 21st Century Matheletics by Marjoseph H. Perez, Donnel P. Placer, Jaime
Pg.186-197
c. Materials: Pictures, real object, chart
d. Values: Active Participation
III.
Procedure:
Teacher Activity
A. Preliminary Activities
1. Prayer:
Let us stand and pray
2. Greeting:
Good Morning Children
You may now take your sit
3. Checking of Attendance:
Is there any absent today?
Very good!
Students Activity
In the name of Father
and the Son and the
Holy Spirit Amen!
Good Morning Ma’am
Princess
Thank you ma’am
Princess
None Ma’am Princess
4. Classroom Management:
Can you please sit properly so we can start our
lesson today!
B. Preparatory Activities:
1. Drill:
Choose the letter of the correct answer.
1.This polygon has 5 sides.
b. heptagon
c. pentagon
d. triangle
2.This figure is called ______
a. triangle
b. hexagon
c.octagon
d. nonagon
3.A decagon has ____ sides.
a. 8
b. 9
c. 10 d. 5
4. What do you call a polygon with 9 sides?
a. octagon
b.decagon
c.pentagon
d.nonagon
5.How many sides does an octagon has?
a. 8 b. 9
c. 10
d.11
1. C
2. B
3. C
4. D
5. A
2. Review
Yesterday we discuss the adding integers. Right?
So I guess all of you can answer this equation
Yes Ma’am!
(call a student who want to answer the equation)
1. (-8) + (-7) =
2. (+11) + (+23) =
3. (-43) + (+17) =
4. (+59) + (+73) =
5. (+102) + (-102) =
-15
+34 or 34
-26
+14 or 14
0
C. Developmental Activities:
1. Motivation
I have here a big box inside of this box has different
shape all you have to do is to pick one shape in this
box and tell to everyone what kind of shape did you
pick.
Is that clear?
So who wants to pick first in my big box? Anyone?
Cube
.
sphere
cone
sphere
cylinder
cube
cone
cylinder
2. Discussion
Can you guess what is our lesson for the day?
Very good!
So how we will describe the solid figure?
Solid figure Ma’am
I have here a box.
(Show them a big box.)
What is the shape of the bottom part of a box?
Is it the same with the shape of the top part?
Rectangle
yes!
What is the shape of the right side of the box?
Does it have with the same shape with the one on
the left side?
Square
Yes!
Do the back and front sides of the box have the
same shape?
Yes!
So how many faces does the box have?
6 faces ma’am
Very good!
How many edges that the box have?
12 edges Ma’am
Very good!
How about vertices? How many vertices that the
box have?
8 vertices
Very good!
Thus the box has three- dimensional object?
Yes!
Because it has length
width and height.
Very good!
So how we will describe a solid figure?
Very good!
Solid figure has threedimensional object if
they have length, width
and height.
So class this means that This box is a solid figure
called rectangular prism, because of its rectangular
base, has 6 faces, 8 vertices and 12 edges. That
you said a while ago.
So what is prism?
(one student will read the definition of a prism)
A Prism is a polyhedron that has two congruent
parallel faces called base.
When we say polyhedron this means a solid figure
whose side are polygon. Each side is called faces.
Two faces that intersect in a line segment is called
an edge. And two edges that intersect in a point is
called a vertex.
(The teacher pointed out where’s the faces edges
and vertex in a box)
A Prism is a polyhedron
that has two congruent
parallel faces called
base.
When we say two congruent parallel faces called
base, meaning to say two same bases was aligned
in bout side.
Can you follow?
Yes Ma’am
I have here some prism that was named according
to the shape of its base.
name
Shape of
the base
Number definition
example
of
lateral
faces
Triangular
triangle
3
Composed
prism
of two
triangular
bases and
three
rectangular
lateral
faces.
Rectangular rectangle
4
Composed
prism
of two
rectangular
bases and
four
rectangular
lateral
faces
cube
Square
4
A cube is a
prism with
a square
base. all its
faces are
square
Pentagonal Pentagonal 5
Has two
Prism
pentagonal
bases and
five
rectangular
lateral
faces.
The lateral faces are the faces that join the base of a solid
figure. Always remember that each faces is a polygon.
Is that clear?
I have here another solid figure can you tell me what is the
name of this solid figure?
Yes ma’am
This are the clues how will you observe a figure
1. The sides are all polygon
2. The figure is a prism
3. Its base are hexagons
The figure is a
hexagonal Prism
8 faces
18 edges
12 vertices
How many faces?
How many edges?
How many vertices?
Another prism is pyramid which is also named according to
the shape of its base.
(one student will read the definition of the pyramid)
A pyramid is a polyhedron whose base is a polygon and the
lateral faces are triangles.
Meaning to say if the prism has 2 equal bases the pyramid
has only one base but the lateral faces are triangle.
Is that clear?
Yes Ma’am
Here are some example of pyramids that was name according
to the shape of its base.
name
Triangular
pyramid
Rectangular
pyramid
Square
pyramid
Shape of the
base
triangle
Number of
lateral faces
3
rectangle
4
square
4
example
Is that clear?
None ma’am
Can you name the figure? Count the number of faces, edges
and vertices.
This are the clues
1. The lateral faces are all triangles
2. The figure is a pyramid
3. The base is a pentagon
Very good!
Other solid figures have curved surfaces
name
definition
Example
the figure is a
pentagonal pyramid
It has 6 faces, 10 edges
and 6 vertices.
cylinder
cone
sphere
Has two
circular bases
that are
congruent
and parallel
Has one
circular base
Is a curved
surface of
points that
are all the
same
distance from
the center
3 .Generalization:
What is solid figure?
Very good!
Can you give me some example of solid figure?
Solid figure has threedimensional object if
they have length, width
and height.
Hexagonal prism
Rectangular prism
Triangular prism
Square pyramid
Cone
Sphere
Cylinder
Etc.
Okay very good!
4. .Application:
(The class will group into 5 group)
What we should do when doing an activity?
Very good children.
1. Be quite
2. And participate in
a group activity.
Activity group 1
Identify the picture what kind of solid figure it is.
Cube
1.____________
Cylinder
2. ______________
Rectangular prism
3.____________
Sphere
4.______________
cone
5.______________
Activity group 2
Match the column A (2D shape) to column B (3D
shape)
Activity group 3
Identify how many edges, faces and vertex did the
solid figure below.
HEXAGONAL PRISM
EDGES:__________________
FACES:__________________
18 edges
8 faces
12 vertex
VERTEX:_________________
PENTAGONAL TRIANGLE
EDGES:__________________
FACES:__________________
10 edges
6 faces
6 vertex
VERTEX:_________________
Activity group 4
Write check ( / ) if the picture is solid figure and ( x ) if
is not.
1.__________
2. _____________
3. ____________
4. _____________
5._________
1.
2.
3.
4.
5.
X
/
X
/
/
IV.
Evaluation:
Complete the table below
Solid Figure
Illustration
Number of vertices Number of faces Number of edges
(draw the figure)
Rectangular prism
Square pyramid
Cube
Triangular prism
Rectangular pyramid
V.
Assignment
1. Draw and color a robot using the solid figure.
2. Study your mathematic book on page 200.
PREPARED BY:
GACAYAN, PRINCESS ALORA V.
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# Detailed Lesson Plan In Mathematics VI. ```Detailed Lesson Plan. In Mathematics VI. I.. Objectives:. At the end of the class 100% of the students should be able to learn 75% of the lesson and. be able to:. a. Visualize and describe the different solid figures. b. Identify the faces vertex and edges of a solid figure.. c. Sketches the different solid figures in a real world.. d. Display cooperativeness during group activity.. II.. Subject Matter:. a. Topic: Solid Figures. b. References: 21st Century Matheletics by Marjoseph H. Perez, Donnel P. Placer, Jaime. Pg.186-197. c. Materials: Pictures, real object, chart. d. Values: Active Participation. III.. Procedure:. Teacher Activity. A. Preliminary Activities. 1. Prayer:. Let us stand and pray. 2. Greeting:. Good Morning Children. You may now take your sit. 3. Checking of Attendance:. Is there any absent today?. Very good!. Students Activity. In the name of Father. and the Son and the. Holy Spirit Amen!. Good Morning Ma’am. Princess. Thank you ma’am. Princess. None Ma’am Princess. 4. Classroom Management:. Can you please sit properly so we can start our. lesson today!. B. Preparatory Activities:. 1. Drill:. Choose the letter of the correct answer.. 1.This polygon has 5 sides.. b. heptagon. c. pentagon. d. triangle. 2.This figure is called ______. a. triangle. b. hexagon. c.octagon. d. nonagon. 3.A decagon has ____ sides.. a. 8. b. 9. c. 10 d. 5. 4. What do you call a polygon with 9 sides?. a. octagon. b.decagon. c.pentagon. d.nonagon. 5.How many sides does an octagon has?. a. 8 b. 9. c. 10. d.11. 1. C. 2. B. 3. C. 4. D. 5. A. 2. Review. Yesterday we discuss the adding integers. Right?. So I guess all of you can answer this equation. Yes Ma’am!. (call a student who want to answer the equation). 1. (-8) + (-7) =. 2. (+11) + (+23) =. 3. (-43) + (+17) =. 4. (+59) + (+73) =. 5. (+102) + (-102) =. -15. +34 or 34. -26. +14 or 14. 0. C. Developmental Activities:. 1. Motivation. I have here a big box inside of this box has different. shape all you have to do is to pick one shape in this. box and tell to everyone what kind of shape did you. pick.. Is that clear?. So who wants to pick first in my big box? Anyone?. Cube. .. sphere. cone. sphere. cylinder. cube. cone. cylinder. 2. Discussion. Can you guess what is our lesson for the day?. Very good!. So how we will describe the solid figure?. Solid figure Ma’am. I have here a box.. (Show them a big box.). What is the shape of the bottom part of a box?. Is it the same with the shape of the top part?. Rectangle. yes!. What is the shape of the right side of the box?. Does it have with the same shape with the one on. the left side?. Square. Yes!. Do the back and front sides of the box have the. same shape?. Yes!. So how many faces does the box have?. 6 faces ma’am. Very good!. How many edges that the box have?. 12 edges Ma’am. Very good!. How about vertices? How many vertices that the. box have?. 8 vertices. Very good!. Thus the box has three- dimensional object?. Yes!. Because it has length. width and height.. Very good!. So how we will describe a solid figure?. Very good!. Solid figure has threedimensional object if. they have length, width. and height.. So class this means that This box is a solid figure. called rectangular prism, because of its rectangular. base, has 6 faces, 8 vertices and 12 edges. That. you said a while ago.. So what is prism?. (one student will read the definition of a prism). A Prism is a polyhedron that has two congruent. parallel faces called base.. When we say polyhedron this means a solid figure. whose side are polygon. Each side is called faces.. Two faces that intersect in a line segment is called. an edge. And two edges that intersect in a point is. called a vertex.. (The teacher pointed out where’s the faces edges. and vertex in a box). A Prism is a polyhedron. that has two congruent. parallel faces called. base.. When we say two congruent parallel faces called. base, meaning to say two same bases was aligned. in bout side.. Can you follow?. Yes Ma’am. I have here some prism that was named according. to the shape of its base.. name. Shape of. the base. Number definition. example. of. lateral. faces. Triangular. triangle. 3. Composed. prism. of two. triangular. bases and.
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three. rectangular. lateral. faces.. Rectangular rectangle. 4. Composed. prism. of two. rectangular. bases and. four. rectangular. lateral. faces. cube. Square. 4. A cube is a. prism with. a square. base. all its. faces are. square. Pentagonal Pentagonal 5. Has two. Prism. pentagonal. bases and. five. rectangular. lateral. faces.. The lateral faces are the faces that join the base of a solid. figure. Always remember that each faces is a polygon.. Is that clear?. I have here another solid figure can you tell me what is the. name of this solid figure?. Yes ma’am. This are the clues how will you observe a figure. 1. The sides are all polygon. 2. The figure is a prism. 3. Its base are hexagons. The figure is a. hexagonal Prism. 8 faces. 18 edges. 12 vertices. How many faces?. How many edges?. How many vertices?. Another prism is pyramid which is also named according to. the shape of its base.. (one student will read the definition of the pyramid). A pyramid is a polyhedron whose base is a polygon and the. lateral faces are triangles.. Meaning to say if the prism has 2 equal bases the pyramid. has only one base but the lateral faces are triangle.. Is that clear?. Yes Ma’am. Here are some example of pyramids that was name according. to the shape of its base.. name. Triangular. pyramid. Rectangular. pyramid. Square. pyramid. Shape of the. base. triangle. Number of. lateral faces. 3. rectangle. 4. square. 4. example. Is that clear?. None ma’am. Can you name the figure? Count the number of faces, edges. and vertices.. This are the clues. 1. The lateral faces are all triangles. 2. The figure is a pyramid. 3. The base is a pentagon. Very good!. Other solid figures have curved surfaces. name. definition. Example. the figure is a. pentagonal pyramid. It has 6 faces, 10 edges. and 6 vertices.. cylinder. cone. sphere. Has two. circular bases. that are. congruent. and parallel. Has one. circular base. Is a curved. surface of. points that. are all the. same. distance from. the center. 3 .Generalization:. What is solid figure?. Very good!. Can you give me some example of solid figure?. Solid figure has threedimensional object if. they have length, width. and height.. Hexagonal prism. Rectangular prism. Triangular prism. Square pyramid. Cone. Sphere. Cylinder. Etc.. Okay very good!. 4. .Application:. (The class will group into 5 group). What we should do when doing an activity?. Very good children.. 1. Be quite. 2. And participate in. a group activity.. Activity group 1. Identify the picture what kind of solid figure it is.. Cube. 1.____________. Cylinder. 2. ______________. Rectangular prism. 3.____________. Sphere. 4.______________. cone. 5.______________. Activity group 2. Match the column A (2D shape) to column B (3D. shape). . . . . . . . . . . Activity group 3. Identify how many edges, faces and vertex did the. solid figure below.. HEXAGONAL PRISM. EDGES:__________________. FACES:__________________. 18 edges. 8 faces. 12 vertex. VERTEX:_________________. PENTAGONAL TRIANGLE. EDGES:__________________. FACES:__________________. 10 edges. 6 faces. 6 vertex. VERTEX:_________________. Activity group 4. Write check ( / ) if the picture is solid figure and ( x ) if. is not.. 1.__________. 2. _____________. 3. ____________. 4. _____________. 5._________. 1.. 2.. 3.. 4.. 5.. X. /. X. /. /. IV.. Evaluation:. Complete the table below. Solid Figure. Illustration. Number of vertices Number of faces Number of edges. (draw the figure). Rectangular prism. Square pyramid. Cube. Triangular prism. Rectangular pyramid. V.. Assignment. 1. Draw and color a robot using the solid figure.. 2. Study your mathematic book on page 200.. PREPARED BY:. GACAYAN, PRINCESS ALORA V.. ```.
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1. ## conditional probability problem
A doctor assumes that a patient has one of three diseases d1, d2, or d3. Before any test, he assumes
an equal probability for each disease. He carries out a test that will be positive with probability 0.8 if the patient
has d1, 0.6 if he has disease d2, and 0.4 if he has disease d3. Given that the outcome of the test was positive,
what probabilities should the doctor now assign to the three possible diseases?
i know it involves the conditional probability formula but i can't figure out which numbers to use for each part. thanks!
2. This seems to be Bayes, but then you cannot have 2 diseases at once.
That should be a partition and I don't see it here.
Let P=psoitive and D1=Disease 1, D2=Disease 2 and D3=Disease 3.
P(D1)=P(D2)=P(D3)=1/3 our partition?
P(P)=P(P)P(P|D1)+P(D2)P(P|D2)+P(D3)P(P|D3)=[.8+.6+.4]/3=.6
Thus P(D1|P)=P(D1P)/P(P)=(.8)(1/3)/.6=4/9.
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1. ## conditional probability problem. A doctor assumes that a patient has one of three diseases d1, d2, or d3. Before any test, he assumes. an equal probability for each disease. He carries out a test that will be positive with probability 0.8 if the patient. has d1, 0.6 if he has disease d2, and 0.4 if he has disease d3. Given that the outcome of the test was positive,. what probabilities should the doctor now assign to the three possible diseases?. i know it involves the conditional probability formula but i can't figure out which numbers to use for each part.
|
thanks!. 2. This seems to be Bayes, but then you cannot have 2 diseases at once.. That should be a partition and I don't see it here.. Let P=psoitive and D1=Disease 1, D2=Disease 2 and D3=Disease 3.. P(D1)=P(D2)=P(D3)=1/3 our partition?. P(P)=P(P)P(P|D1)+P(D2)P(P|D2)+P(D3)P(P|D3)=[.8+.6+.4]/3=.6. Thus P(D1|P)=P(D1P)/P(P)=(.8)(1/3)/.6=4/9.
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# A model train with a mass of 3 kg is moving along a track at 8 (cm)/s. If the curvature of the track changes from a radius of 15 cm to 18 cm, by how much must the centripetal force applied by the tracks change?
Apr 26, 2017
The centripetal force changes by $= 0.021 N$
#### Explanation:
The centripetal force is
$F = \frac{m {v}^{2}}{r}$
mass, $m = 3 k g$
speed, $v = 0.08 m {s}^{-} 1$
radius, $= r$
The variation in centripetal force is
$\Delta F = {F}_{2} - {F}_{1}$
${F}_{1} = m {v}^{2} / {r}_{1} = 3 \cdot {0.08}^{2} / 0.15 = 0.128 N$
${F}_{2} = m {v}^{2} / {r}_{2} = 3 \cdot {0.08}^{2} / 0.18 = 0.107 N$
$\Delta F = 0.128 - 0.107 = 0.021 N$
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# A model train with a mass of 3 kg is moving along a track at 8 (cm)/s. If the curvature of the track changes from a radius of 15 cm to 18 cm, by how much must the centripetal force applied by the tracks change?. Apr 26, 2017. The centripetal force changes by $= 0.021 N$. #### Explanation:. The centripetal force is. $F = \frac{m {v}^{2}}{r}$. mass, $m = 3 k g$.
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speed, $v = 0.08 m {s}^{-} 1$. radius, $= r$. The variation in centripetal force is. $\Delta F = {F}_{2} - {F}_{1}$. ${F}_{1} = m {v}^{2} / {r}_{1} = 3 \cdot {0.08}^{2} / 0.15 = 0.128 N$. ${F}_{2} = m {v}^{2} / {r}_{2} = 3 \cdot {0.08}^{2} / 0.18 = 0.107 N$. $\Delta F = 0.128 - 0.107 = 0.021 N$.
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https://www.first-learn.com/word-problems-on-hcf-and-lcm.html
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# Word Problems on H.C.F. and L.C.M.
We have already learnt H.C.F, L.C.M and the relationship between the two and now we will learn on how to do word problems on H.C.F. and L.C.M.
Note: In word problems on H.C.F. and L.C.M we have to remember two points:
• When we are asked to find out maximum, highest or largest of anything we are required to do H.C.F as it is Highest Common factor.
• When we are asked to find out minimum, lowest or smallest of anything we are required to do L.C.M as it is Lowest Common Multiple
Here few examples are illustrated on word problem of H.C.F. and L.C.M
1. There are two pots each having a volume of 256 cu.cm and 264 cu.cm respectively. Find the maximum volume of same pot that should be made which will fit into both the pots.
Solution:
The maximum volume of pot that can be made to fit into both the pots are: H.C.F of 256 and 264
Factors of 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2^8
Factors of 264 = 2 × 2 × 2 × 3 = 2^3 × 3 × 11
H.C.F of 256 and 264 = 2^3 = 8
The greatest volume of the pot that can be fitted into both the pots = 8 cu. cm
Explanation:
The problem asks to find out the greatest size of pot that can fit into both the pots hence we need to find out the H.C.F of 256 and 264.
2. Find the greatest number that can divide 12 and 48 exactly without leaving any remainder.
Solution:
Greatest number = H.C.F of 12 and 48
Factors of 12 = 3× 2^2
Factors of 48 = 3 × 2^4
Therefore, H.C.F of 12 and 48 = 2^2 × 3 = 4 × 3 = 12
Greatest number = 12
Explanation:
This sum has also asked to find out the greatest number hence we need to do H.C.F of 12 and 48.
3. In a hostel room there are three girls. During winter vacation all the three girls left for their home and after coming back each of the girls decided to visit their home town after 2 months, 4 months and 6 months respectively. After how many months again three girls will together go to their home to spend their vacation?
Solution:
We have to find out the L.C.M of 2, 4 and 6
Factors of 2 = 2
Factors of 4 = 2 × 2
Factors of 6 = 2 × 3
L.C.M of 2, 4 and 6 = 2 × 2 × 3 = 12
The three girls will again go together after 12 months to spend their vacation in their hometown.
Explanation:
The sum has asked that after how months the three girls will again visit their hometown together if they go at an interval of 2, 4 and 6 months respectively. Hence, we need to carry out the L.C.M of 2, 4 and 6.
From Word Problems on H.C.F. and L.C.M. to HOME PAGE
Have your say about what you just read! Leave me a comment in the box below.
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# Word Problems on H.C.F. and L.C.M.. We have already learnt H.C.F, L.C.M and the relationship between the two and now we will learn on how to do word problems on H.C.F. and L.C.M.. Note: In word problems on H.C.F. and L.C.M we have to remember two points:. • When we are asked to find out maximum, highest or largest of anything we are required to do H.C.F as it is Highest Common factor.. • When we are asked to find out minimum, lowest or smallest of anything we are required to do L.C.M as it is Lowest Common Multiple. Here few examples are illustrated on word problem of H.C.F. and L.C.M. 1. There are two pots each having a volume of 256 cu.cm and 264 cu.cm respectively. Find the maximum volume of same pot that should be made which will fit into both the pots.. Solution:. The maximum volume of pot that can be made to fit into both the pots are: H.C.F of 256 and 264. Factors of 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2^8. Factors of 264 = 2 × 2 × 2 × 3 = 2^3 × 3 × 11. H.C.F of 256 and 264 = 2^3 = 8. The greatest volume of the pot that can be fitted into both the pots = 8 cu. cm. Explanation:. The problem asks to find out the greatest size of pot that can fit into both the pots hence we need to find out the H.C.F of 256 and 264.. 2. Find the greatest number that can divide 12 and 48 exactly without leaving any remainder.. Solution:. Greatest number = H.C.F of 12 and 48. Factors of 12 = 3× 2^2. Factors of 48 = 3 × 2^4. Therefore, H.C.F of 12 and 48 = 2^2 × 3 = 4 × 3 = 12. Greatest number = 12. Explanation:. This sum has also asked to find out the greatest number hence we need to do H.C.F of 12 and 48.. 3. In a hostel room there are three girls. During winter vacation all the three girls left for their home and after coming back each of the girls decided to visit their home town after 2 months, 4 months and 6 months respectively.
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After how many months again three girls will together go to their home to spend their vacation?. Solution:. We have to find out the L.C.M of 2, 4 and 6. Factors of 2 = 2. Factors of 4 = 2 × 2. Factors of 6 = 2 × 3. L.C.M of 2, 4 and 6 = 2 × 2 × 3 = 12. The three girls will again go together after 12 months to spend their vacation in their hometown.. Explanation:. The sum has asked that after how months the three girls will again visit their hometown together if they go at an interval of 2, 4 and 6 months respectively. Hence, we need to carry out the L.C.M of 2, 4 and 6.. From Word Problems on H.C.F. and L.C.M. to HOME PAGE. Have your say about what you just read! Leave me a comment in the box below.. ## Recent Articles. 1. ### Respiratory Balance Sheet | TCA Cycle | ATP Consumption Process. Feb 18, 24 01:56 PM. The major component that produced during the photosynthesis is Glucose which is further metabolised by the different metabolic pathways like glycolysis, Krebs cycle, TCA cycle and produces energy whic…. 2. ### Electron Transport System and Oxidative Phosphorylation | ETC |Diagram. Feb 04, 24 01:57 PM. It is also called ETC. Electron transfer means the process where one electron relocates from one atom to the other atom. Definition of electron transport chain - The biological process where a chains…. 3. ### Tricarboxylic Acid Cycle | Krebs Cycle | Steps | End Products |Diagram. Jan 28, 24 12:39 PM. This is a type of process which execute in a cyclical form and final common pathway for oxidation of Carbohydrates fat protein through which acetyl coenzyme a or acetyl CoA is completely oxidised to c…. 4. ### Aerobic Respiration | Definition of Aerobic Respiration | Glycolysis. Dec 15, 23 08:42 AM. This is a type of respiration where molecular free oxygen is used as the final acceptor and it is observed in cell. Site of Aerobic Respiration - Aerobic respiration is observed in most of the eukaryo….
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http://www.ck12.org/book/CK-12-Middle-School-Math-Concepts---Grade-7/r4/section/9.11/
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# 9.11: Estimation of Parallelogram Area in Scale Drawings
Difficulty Level: At Grade Created by: CK-12
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Practice Estimation of Parallelogram Area in Scale Drawings
MEMORY METER
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Progress
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Remember how Miguel's sister was working with tile? Well, look at the dilemma his Mother has.
Mrs. Vasquez, Miguel's Mother is buying new carpet to redecorate her parallelogram-shaped store. To determine how much carpet she needs, she measured the length of the room and found it was 36 meters. Then she measured across the room with a line perpendicular to the first and found it was 16 meters across.
How many square meters of carpet does Mrs. Vasquez need to buy?
Use what you have learned to solve this problem. We will return to it at the end of the Concept.
### Guidance
Sometimes, you will see a scale drawing. A scale drawing is a drawing that uses a small measurement to represent a real-world measurement. We can work with a scale drawing and estimate actual areas of a parallelogram given the scale.
Estimate the area of this garden.
To work on solving this problem, we need to look at what information is given to us in the drawing.
We know that the scale says that 1” is equal to 3 feet.
The base of the garden in the drawing is 8” and the height is 3”.
We want to estimate the area.
Let’s start by estimating the base. 8×3\begin{align*}8 \times 3\end{align*}. We know that 8×3\begin{align*}8 \times 3\end{align*} is 24. This gives us a total of 24 feet for the base of the garden.
Let’s estimate the height. 3×3\begin{align*}3 \times 3\end{align*}. We know that 3×3=9\begin{align*}3 \times 3 = 9\end{align*}. The height of the garden is about 9 feet.
Next, we estimate the area of the garden.
24 rounds down to 20 and 9 rounds up to 10
20×10=200 square feet\begin{align*}20 \times 10 = 200 \ square \ feet\end{align*}
How close it our estimate?
Let’s do the actual multiplying to figure this out.
24×9=216 square feet\begin{align*}24 \times 9 = 216 \ square \ feet\end{align*}
We have an estimate that is reasonable.
Now it's time for you to try a few on your own. Use a scale of 1" = 2 feet.
#### Example A
Base of 6 inches and a height of 4 inches
Solution: 48 square feet
#### Example B
Base of 9 inches and height of 5 inches
Solution:90 square feet
#### Example C
Base of 12 inches and height of 9 inches
Solution: 216 square feet
Here is the original problem once again.
Mrs. Vasquez, Miguel's Mother is buying new carpet to redecorate her parallelogram-shaped store. To determine how much carpet she needs, she measured the length of the room and found it was 36 meters. Then she measured across the room with a line perpendicular to the first and found it was 16 meters across.
How many square meters of carpet does Mrs. Vasquez need to buy?
Let’s begin by figuring out what the problem is asking us to find. We need to find how much carpet Mrs. Vasquez needs to cover the floor of her store, so we have to find the area of the store. That means we will use the area formula to solve for A. In order to use the formula, we need to know the base and height of the store.
We know that one side of the store will be 36 meters. Let’s call this the base. We also know that Mrs. Vasquez made a perpendicular line in order to measure the height, or the distance across the room. The height given in the problem is 16 meters. Let’s put this information into the formula and solve for area.
AAA=bh=36(16)=576 m2\begin{align*}A & = bh\\ A & = 36(16) \\ A & = 576 \ m^2\end{align*}
Mrs. Vasquez will need to buy 576 square meters of carpet.
### Vocabulary
Here are the vocabulary words in this Concept.
Parallelogram
a quadrilateral with opposite sides parallel.
Perimeter
the distance around a figure.
Area
the amount of space contained inside a two-dimensional figure.
Scale Drawing
a drawing of a life size image where the drawing is made smaller than the actual image using a scale.
### Guided Practice
Here is one for you to try on your own.
If you use the scale 1" = 3 feet, what is the area of a parallelogram with a height of 3 inches and a base of 6 inches?
To figure this out, we can use the formula for finding the area of a parallelogram. We will work in inches first and then convert that into square feet.
A=bh\begin{align*}A = bh\end{align*}
A=(3)(6)\begin{align*}A = (3)(6)\end{align*}
A=18 sq.in\begin{align*}A = 18 \ sq. in\end{align*}
Now if 1" = 3 feet, then we need to take the area of the room in inches and multiply it by three.
A=18×3=54 sq.feet\begin{align*}A = 18 \times 3 = 54 \ sq. feet\end{align*}
### Video Review
Here are videos for review.
### Practice
Directions: For each parallelogram, find each new area using the scale 1" = 2 feet.
1. Base of 6 inches, height of 4 inches
2. Base of 8 inches, height of 6 inches
3. Base of 4 inches, height of 4 inches
4. Base of 5 inches, height of 4 inches
5. Base of 6 inches, height of 6 inches
6. Base of 10 inches, height of 8 inches
7. Base of 11 inches, height of 12 inches
8. Base of 15 inches, height of 9 inches
9. Base of 15 inches, height of 12 inches
Directions: Solve each problem.
10. A parallelogram has an area of 390 square centimeters. If its height is 15 cm, what is its base?
11. What is the height of a parallelogram whose base is 28 inches and area is 1,176 square inches?
12. Donna wants to cover her parallelogram-shaped crafts box in fabric. The base of the lid is 32.7 cm and the height is 12.2 cm. What is the area of the lid?
13. John is planting grass in a patch of lawn that is shaped like a parallelogram. The height of the parallelogram is 34 feet. The other border is 65 feet. How many square feet of grass will John plant?
14. Kara and Sharice are in a quilting competition. Both are stitching parallelogram-shaped quilts. So far Kara’s has an area of 2,278 square inches and a height of 44 inches. Sharice’s quilt has an area of 2,276 square inches and a height of 47 inches. Whose quilt is longer? By how many inches is it longer?
15. Denise bought a picture frame in the shape of a parallelogram. The area of the picture frame is 36,795 square centimeters. If its height is 165 centimeters, what is its base?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Show Hide Details
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Difficulty Level:
Authors:
Tags:
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# 9.11: Estimation of Parallelogram Area in Scale Drawings. Difficulty Level: At Grade Created by: CK-12. Estimated4 minsto complete. %. Progress. Practice Estimation of Parallelogram Area in Scale Drawings. MEMORY METER. This indicates how strong in your memory this concept is. Progress. Estimated4 minsto complete. %. Estimated4 minsto complete. %. MEMORY METER. This indicates how strong in your memory this concept is. Remember how Miguel's sister was working with tile? Well, look at the dilemma his Mother has.. Mrs. Vasquez, Miguel's Mother is buying new carpet to redecorate her parallelogram-shaped store. To determine how much carpet she needs, she measured the length of the room and found it was 36 meters. Then she measured across the room with a line perpendicular to the first and found it was 16 meters across.. How many square meters of carpet does Mrs. Vasquez need to buy?. Use what you have learned to solve this problem. We will return to it at the end of the Concept.. ### Guidance. Sometimes, you will see a scale drawing. A scale drawing is a drawing that uses a small measurement to represent a real-world measurement. We can work with a scale drawing and estimate actual areas of a parallelogram given the scale.. Estimate the area of this garden.. To work on solving this problem, we need to look at what information is given to us in the drawing.. We know that the scale says that 1” is equal to 3 feet.. The base of the garden in the drawing is 8” and the height is 3”.. We want to estimate the area.. Let’s start by estimating the base. 8×3\begin{align*}8 \times 3\end{align*}. We know that 8×3\begin{align*}8 \times 3\end{align*} is 24. This gives us a total of 24 feet for the base of the garden.. Let’s estimate the height. 3×3\begin{align*}3 \times 3\end{align*}. We know that 3×3=9\begin{align*}3 \times 3 = 9\end{align*}. The height of the garden is about 9 feet.. Next, we estimate the area of the garden.. 24 rounds down to 20 and 9 rounds up to 10. 20×10=200 square feet\begin{align*}20 \times 10 = 200 \ square \ feet\end{align*}. How close it our estimate?. Let’s do the actual multiplying to figure this out.. 24×9=216 square feet\begin{align*}24 \times 9 = 216 \ square \ feet\end{align*}. We have an estimate that is reasonable.. Now it's time for you to try a few on your own. Use a scale of 1" = 2 feet.. #### Example A. Base of 6 inches and a height of 4 inches. Solution: 48 square feet. #### Example B. Base of 9 inches and height of 5 inches. Solution:90 square feet. #### Example C. Base of 12 inches and height of 9 inches. Solution: 216 square feet. Here is the original problem once again.. Mrs. Vasquez, Miguel's Mother is buying new carpet to redecorate her parallelogram-shaped store. To determine how much carpet she needs, she measured the length of the room and found it was 36 meters. Then she measured across the room with a line perpendicular to the first and found it was 16 meters across.. How many square meters of carpet does Mrs. Vasquez need to buy?. Let’s begin by figuring out what the problem is asking us to find. We need to find how much carpet Mrs. Vasquez needs to cover the floor of her store, so we have to find the area of the store. That means we will use the area formula to solve for A. In order to use the formula, we need to know the base and height of the store.. We know that one side of the store will be 36 meters. Let’s call this the base. We also know that Mrs. Vasquez made a perpendicular line in order to measure the height, or the distance across the room. The height given in the problem is 16 meters. Let’s put this information into the formula and solve for area.. AAA=bh=36(16)=576 m2\begin{align*}A & = bh\\ A & = 36(16) \\ A & = 576 \ m^2\end{align*}. Mrs. Vasquez will need to buy 576 square meters of carpet.. ### Vocabulary.
|
Here are the vocabulary words in this Concept.. Parallelogram. a quadrilateral with opposite sides parallel.. Perimeter. the distance around a figure.. Area. the amount of space contained inside a two-dimensional figure.. Scale Drawing. a drawing of a life size image where the drawing is made smaller than the actual image using a scale.. ### Guided Practice. Here is one for you to try on your own.. If you use the scale 1" = 3 feet, what is the area of a parallelogram with a height of 3 inches and a base of 6 inches?. To figure this out, we can use the formula for finding the area of a parallelogram. We will work in inches first and then convert that into square feet.. A=bh\begin{align*}A = bh\end{align*}. A=(3)(6)\begin{align*}A = (3)(6)\end{align*}. A=18 sq.in\begin{align*}A = 18 \ sq. in\end{align*}. Now if 1" = 3 feet, then we need to take the area of the room in inches and multiply it by three.. A=18×3=54 sq.feet\begin{align*}A = 18 \times 3 = 54 \ sq. feet\end{align*}. ### Video Review. Here are videos for review.. ### Practice. Directions: For each parallelogram, find each new area using the scale 1" = 2 feet.. 1. Base of 6 inches, height of 4 inches. 2. Base of 8 inches, height of 6 inches. 3. Base of 4 inches, height of 4 inches. 4. Base of 5 inches, height of 4 inches. 5. Base of 6 inches, height of 6 inches. 6. Base of 10 inches, height of 8 inches. 7. Base of 11 inches, height of 12 inches. 8. Base of 15 inches, height of 9 inches. 9. Base of 15 inches, height of 12 inches. Directions: Solve each problem.. 10. A parallelogram has an area of 390 square centimeters. If its height is 15 cm, what is its base?. 11. What is the height of a parallelogram whose base is 28 inches and area is 1,176 square inches?. 12. Donna wants to cover her parallelogram-shaped crafts box in fabric. The base of the lid is 32.7 cm and the height is 12.2 cm. What is the area of the lid?. 13. John is planting grass in a patch of lawn that is shaped like a parallelogram. The height of the parallelogram is 34 feet. The other border is 65 feet. How many square feet of grass will John plant?. 14. Kara and Sharice are in a quilting competition. Both are stitching parallelogram-shaped quilts. So far Kara’s has an area of 2,278 square inches and a height of 44 inches. Sharice’s quilt has an area of 2,276 square inches and a height of 47 inches. Whose quilt is longer? By how many inches is it longer?. 15. Denise bought a picture frame in the shape of a parallelogram. The area of the picture frame is 36,795 square centimeters. If its height is 165 centimeters, what is its base?. ### Notes/Highlights Having trouble? Report an issue.. Color Highlighted Text Notes. Show Hide Details. Description. Difficulty Level:. Authors:. Tags:. Subjects:.
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https://nrich.maths.org/public/topic.php?code=148&cl=3&cldcmpid=7500
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# Resources tagged with: Length/distance
Filter by: Content type:
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Challenge level:
### There are 21 results
Broad Topics > Measuring and calculating with units > Length/distance
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Can you rank these sets of quantities in order, from smallest to largest? Can you provide convincing evidence for your rankings?
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##### Age 11 to 14 Challenge Level:
My measurements have got all jumbled up! Swap them around and see if you can find a combination where every measurement is valid.
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##### Age 7 to 14
Mathematics has allowed us now to measure lots of things about eclipses and so calculate exactly when they will happen, where they can be seen from, and what they will look like.
### Swimmers
##### Age 14 to 16 Challenge Level:
Swimmers in opposite directions cross at 20m and at 30m from each end of a swimming pool. How long is the pool ?
### Uniform Units
##### Age 14 to 16 Challenge Level:
Can you choose your units so that a cube has the same numerical value for it volume, surface area and total edge length?
### Walk and Ride
##### Age 7 to 14 Challenge Level:
How far have these students walked by the time the teacher's car reaches them after their bus broke down?
### Speed-time Problems at the Olympics
##### Age 14 to 16 Challenge Level:
Have you ever wondered what it would be like to race against Usain Bolt?
### N Is a Number
##### Age 11 to 14 Challenge Level:
N people visit their friends staying N kilometres along the coast. Some walk along the cliff path at N km an hour, the rest go by car. How long is the road?
##### Age 14 to 16 Challenge Level:
Four vehicles travel along a road one afternoon. Can you make sense of the graphs showing their motion?
### Where Am I?
##### Age 11 to 16 Challenge Level:
From the information you are asked to work out where the picture was taken. Is there too much information? How accurate can your answer be?
##### Age 14 to 16 Challenge Level:
Four vehicles travelled on a road. What can you deduce from the times that they met?
### Olympic Measures
##### Age 11 to 14 Challenge Level:
These Olympic quantities have been jumbled up! Can you put them back together again?
##### Age 7 to 14
A paradox is a statement that seems to be both untrue and true at the same time. This article looks at a few examples and challenges you to investigate them for yourself.
### A Question of Scale
##### Age 14 to 16 Challenge Level:
Use your skill and knowledge to place various scientific lengths in order of size. Can you judge the length of objects with sizes ranging from 1 Angstrom to 1 million km with no wrong attempts?
### How Far Does it Move?
##### Age 11 to 14 Challenge Level:
Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects the distance it travels at each stage.
### Up and Across
##### Age 11 to 14 Challenge Level:
Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the dot affects its vertical and horizontal movement at each stage.
### 2010: A Year of Investigations
##### Age 5 to 14
This article for teachers suggests ideas for activities built around 10 and 2010.
### A Scale for the Solar System
##### Age 14 to 16 Challenge Level:
The Earth is further from the Sun than Venus, but how much further? Twice as far? Ten times?
### Alternative Record Book
##### Age 14 to 18 Challenge Level:
In which Olympic event does a human travel fastest? Decide which events to include in your Alternative Record Book.
### Pentagonal
##### Age 14 to 16 Challenge Level:
Can you prove that the sum of the distances of any point inside a square from its sides is always equal (half the perimeter)? Can you prove it to be true for a rectangle or a hexagon?
### Rolling Around
##### Age 11 to 14 Challenge Level:
A circle rolls around the outside edge of a square so that its circumference always touches the edge of the square. Can you describe the locus of the centre of the circle?
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# Resources tagged with: Length/distance. Filter by: Content type:. Age range:. Challenge level:. ### There are 21 results. Broad Topics > Measuring and calculating with units > Length/distance. ##### Age 11 to 14 Challenge Level:. Can you rank these sets of quantities in order, from smallest to largest? Can you provide convincing evidence for your rankings?. ### All in a Jumble. ##### Age 11 to 14 Challenge Level:. My measurements have got all jumbled up! Swap them around and see if you can find a combination where every measurement is valid.. ### Eclipses of the Sun. ##### Age 7 to 14. Mathematics has allowed us now to measure lots of things about eclipses and so calculate exactly when they will happen, where they can be seen from, and what they will look like.. ### Swimmers. ##### Age 14 to 16 Challenge Level:. Swimmers in opposite directions cross at 20m and at 30m from each end of a swimming pool. How long is the pool ?. ### Uniform Units. ##### Age 14 to 16 Challenge Level:. Can you choose your units so that a cube has the same numerical value for it volume, surface area and total edge length?. ### Walk and Ride. ##### Age 7 to 14 Challenge Level:. How far have these students walked by the time the teacher's car reaches them after their bus broke down?. ### Speed-time Problems at the Olympics. ##### Age 14 to 16 Challenge Level:. Have you ever wondered what it would be like to race against Usain Bolt?. ### N Is a Number. ##### Age 11 to 14 Challenge Level:. N people visit their friends staying N kilometres along the coast. Some walk along the cliff path at N km an hour, the rest go by car. How long is the road?. ##### Age 14 to 16 Challenge Level:. Four vehicles travel along a road one afternoon. Can you make sense of the graphs showing their motion?. ### Where Am I?. ##### Age 11 to 16 Challenge Level:. From the information you are asked to work out where the picture was taken.
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Is there too much information? How accurate can your answer be?. ##### Age 14 to 16 Challenge Level:. Four vehicles travelled on a road. What can you deduce from the times that they met?. ### Olympic Measures. ##### Age 11 to 14 Challenge Level:. These Olympic quantities have been jumbled up! Can you put them back together again?. ##### Age 7 to 14. A paradox is a statement that seems to be both untrue and true at the same time. This article looks at a few examples and challenges you to investigate them for yourself.. ### A Question of Scale. ##### Age 14 to 16 Challenge Level:. Use your skill and knowledge to place various scientific lengths in order of size. Can you judge the length of objects with sizes ranging from 1 Angstrom to 1 million km with no wrong attempts?. ### How Far Does it Move?. ##### Age 11 to 14 Challenge Level:. Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects the distance it travels at each stage.. ### Up and Across. ##### Age 11 to 14 Challenge Level:. Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the dot affects its vertical and horizontal movement at each stage.. ### 2010: A Year of Investigations. ##### Age 5 to 14. This article for teachers suggests ideas for activities built around 10 and 2010.. ### A Scale for the Solar System. ##### Age 14 to 16 Challenge Level:. The Earth is further from the Sun than Venus, but how much further? Twice as far? Ten times?. ### Alternative Record Book. ##### Age 14 to 18 Challenge Level:. In which Olympic event does a human travel fastest? Decide which events to include in your Alternative Record Book.. ### Pentagonal. ##### Age 14 to 16 Challenge Level:. Can you prove that the sum of the distances of any point inside a square from its sides is always equal (half the perimeter)? Can you prove it to be true for a rectangle or a hexagon?. ### Rolling Around. ##### Age 11 to 14 Challenge Level:. A circle rolls around the outside edge of a square so that its circumference always touches the edge of the square. Can you describe the locus of the centre of the circle?.
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# I Know What This Hole Is For Vol 18
I Know What This Hole is For Vol 18 I Know What This Hole is For Vol Like. About Share. views. 0%. 0 0. From: admin. Category: ้เด็ก
Use these formulas to calculate volume or, for round holes, use our cylinder cubic yardage calculator. Rectangle: volume = length x width x depth Cylinder: volume = radius 2 x ПЂ x depth (ПЂ = ) Calculate Concrete Volume. With the volume of the submerged post and the volume of the hole figured out, subtract the post volume from the hole volume, and you will have the volume of concrete needed for each hole.
volume=1/3 Г— ПЂ Г— 4( 2 + Г— + 2) = in 3. Ellipsoid. An ellipsoid is the three-dimensional counterpart of an ellipse, and is a surface that can be described as the deformation of a sphere through scaling of directional elements. The center of an ellipsoid is the point at which three pairwise perpendicular axes of symmetry.
Apr 14, В В· The formula behind the volume of a hollow cylinder is: cylinder_volume = ПЂ * (RВІ - rВІ) * cylinder_height. where R - external radius, and r - internal radius. To calculate the volume of a cylindrical shell, let's take some real-life example, maybe a roll of toilet paper, because why not?:) Enter the external radius of the cylinder. The standard is equal to approximately cm.
This is the size of the woofer. Available sizes: 8, 10, 12, 15, 18 inches. Cutout Diameter (unit of measure - inch) This is the hole cutout needed to "drop-in" a woofer and secure it to the front panel. Most manufacturers provide this spec in the woofer's datasheet. Mounting Depth (unit of measure - inch).
Use this post hole volume calculator to accurately figure the amount of concrete required for any sized post hole with or without any common sized post. Returns both cubic yards and how many bags. This free tool is designed as an aid for building fences, barns, post frame buildings, or just anything that has a post in the ground.
it through a small hole at the wall of the duct; or a series of holes positioned at right angles to the flow in a surface lying parallel to the lines of flow. The pitot static tube is an example of this. Figure 4 shows the principle of the pitot static tube. It will be seen that connecting tubes to a manometer makes the measurement of pressure.
If you know 2 of the 3 variables the third can be calculated. As usual, here at wwwcom, we have a calculator that will do all the work for you. This ultra calculator is special by allowing you to choose among a great variety of units. Unlike other calculators, you are NOT confined to inputting volume in liters, time in minutes, etc.
Dec 26, В В· Breaking taps is a pain! (by 'a little', I mean or oversize on 1/, or on 1/2 NC, etc) from what I can see, the ideal is, whiich is close to a F letter size drill for a 77% thread A G size will give you an easier time tapping the hole, and reduce the strength by something around 10% overall. Last edited: Dec 26,
Jun 05,  · Some guitar wiring circuit designs use k or even 1 Meg pots. Here’s what you need to know about choosing the right [pots for your guitar. Generally speaking, to control volume, humbuckers should be paired to k pots, while single-coil pickups should be .
Search the world's information, including webpages, images, videos and more. Google has many special features to help you find exactly what you're looking [HOST]g: Hole.
Google's free service instantly translates words, phrases, and web pages between English and over other [HOST]g: Hole.
Calculate cubic yards, cubic feet or cubic meters for landscape material, mulch, land fill, gravel, cement, sand, containers, etc. Enter measurements in US or metric units and get volume conversions to other units. How to calculate cubic yards for rectangular, circular, annular and triangular areas. Calculate project cost based on price per cubic foot, cubic yard or cubic meter.
50 x Volume 1 (V1) = x ; V1 = 3 gallons The final mixture (Volume 2 or V2) is the amount of the concentrate (V1) plus the required amount to make up to V2. If V1 = 3 gallons and the required amount is gallons, add 97 gallons of water to 3 gallons of concentrate. Formula 11 C1 x V1 = C2 x V2 C1= % of a.i. in concentrateMissing: Hole.
Shop [HOST] for Every Day Low Prices. Free Shipping on Orders \$35+ or Pickup In-Store and get a Pickup Discount. Open a Walmart Credit Card to Save Even More!
November 27, John Michell anticipates black holes. We think of black holes as a 20th century invention, dating back to , when Albert Einstein first published his theory of general relativity and fellow physicist Karl Schwarzschild used those equations to envision a spherical section of spacetime so badly warped around a concentrated mass that it is invisible to the .
I want to know its volume in cubic inches and therefore its liquid capacity (how much water I can fit in the tank) in litres. I would enter the values in the calculator as shown in the figure above, selecting the correct units for each measurement from the drop down options.
The larger your subwoofer is, the more volume of space is needed around it. According to the JL Audio website, the recommended volume space for a inch subwoofer is cubic feet. For a inch woofer, the volume recommendation is cubic feet, and the recommendation for an eight-inch speaker is cubic feet.
Mandingo, Actor: Mandingo Rocks That Ass. Mandingo was born on February 25, in Mississippi, USA as Frederick Lamont. He is an actor and director.
Nov 16,  · Regardless of its origins, the 18″ AR barrel is very popular in the competition world. It offers a very smooth recoil impulse due to its full-length rifle gas system while shaving a few ounces off of the end of the rifle. The length also maintains good velocity for reaching out with a flatter trajectory.
Jul 25, В В· And she thinks I look good. And she sees my chest pubes all the way down to my ball fro, and she says iv'e had the old bull, now I want the old calve. Then she grabs me by the wein. Dale Doback.
Jun 03, В В· Input the container's length into the first field of the box volume calculator. It's 12 inches in our case. Enter the width of the box. Put 10 inches into the proper field. Finally, input the height of your container, 8 inches. And there it is: the volume of a rectangular prism calculator did the job. Now we know that our cat's volume is
approximately 53 times more units of volume through a hole than water if the water is able to flow at all. • The characteristics of the hole (or leak) that determine whether it will leak are the smallest diameter in the hole path, the path length of the hole through the part material, surface tension, and the surface finish of hole (leak).
Saturday Nite Southern Soul Juke Joint Jamz!Compiled,Mixed & Arranged by [HOST]pyright Disclaimer Under Section of the Copyright Act , allowance is Missing: Hole.
Unlock the power of video and join over M professionals, teams, and organizations who use Vimeo to create, collaborate and [HOST]g: Hole.
Related Surface Area Calculator | Volume Calculator. Area is a quantity that describes the size or extent of a two-dimensional figure or shape in a plane. It can be visualized as the amount of paint that would be necessary to cover a surface, and is the two-dimensional counterpart of the one-dimensional length of a curve, and three-dimensional volume of a solid.
Faye Reagan, Actress: Pin-Up Girls 5. Buxom, freckled, and slender redhead stunner Faye Reagan was born Faye Jillian Henning on September 19, , in Nashville, Tennessee. Faye moved with her family to Las Vegas, Nevada when she was only eight months old. She first began performing in explicit X-rated fare at age 19 in Reagan did five sex scenes in a single .
equal to the concentration of holes in the valence band. We may denote, n i: intrinsic electron concentration p i: intrinsic hole concentration However, n i = p i Simply, n i:intrinsic carrier concentration, which refers to either the intrinsic electron or hole concentration Commonly accepted values of n i at T = В°K Silicon x cm
Mar 24, В В· Costa & Hanna Finsen - I Will Wait For You (Radio Edit) Somna & Sarah Russell - Story Untold (Radio Edit) Hazem Beltagui & Julie Scott - When Nothing's True (Radio Edit) 28 Mino Safy & Maria Nayler - A Second Breath (Radio Edit) Store N Forward & Neev Kennedy - Brought Me Back To Life (Radio Edit) Missing: Hole.
plenum, there is an available air volume of 1, cfm and an available velocity of ft/min. After the third branch run, a total of cfm has been distributed to the branches and the velocity in the plenum has been reduced to ft/[HOST] conditions indicate that the proper location for the reduction in the plenum is.
– 18″ ( cm) Square (1) – Several flower boxes (like ish) The answer is maybe. You want to discover the total volume of the containers that you have. We have some larger diameter containers, but they are not very tall, so they do not work well for some of our larger plants.
ISABELLE RONIN (@ISABELLERONIN) Chasing Red was one of ’s most-read stories on Wattpad -- and that was just the beginning for this Winnipeg-Manitoba-based writer. In a single year, her explosive hit has racked up over million reads on Wattpad. Newly edited and expanded, the book was split into two and hit bookstore shelves in Missing: Hole.
Use this form to estimate the brake horsepower required. Brake horsepower is the power out of the drive motor, and the power into the water pump and is how most pumps and drive motors are specified. Also calculate the total power requirements (energy use rate). The pump efficiency depends on the pump and the pressure and flow that the pump is Missing: Hole.
Nov 14,  · But hold on, we need to subtract the inner cylinder, otherwise, it’s like we’re filling up the entire bracelet without a hole for your wrist. Inner cylinder volume equals: x (/2) 2 x which equals cubic inches. That means the volume of the bangle is cubic inches minus cubic inches which equals cubic inches.
the hole is drilled, it may be perforated by down-the-hole tools, forming a screen opposite the water-producing formations. With most methods of down-the-hole perforating, a small aperture cannot be formed nor can the aperture size be precisely controlled. Consequently, finer-grained aquifers must be avoided.
Short Stories: The Red-Headed League by Arthur Conan Doyle. I had called upon my friend, Mr. Sherlock Holmes, one day in the autumn of last year, and found him in deep conversation with a very stout, florid-faced elderly gentleman, with fiery red hair. With an apology for my intrusion, I was about to withdraw, when Holmes pulled me abruptly.
I Know What This Hole is For Vol 21 - Kumpulan Video Film Indo, Sikwap Streaming Cewek Jilbab Ngentot, Sedarah Ibu Dan Anak, XXX Indo Porno Memek Perawan Sd, Sex - I Know What This Hole is For Vol 21п»ї. 12 min. Anal The Forbidden Fruit is Sweet Vol 17 min. I Know What This Hole is For Vol 17 min. I Know What This Hole is For Vol 6. 17 min.
Groot (/ ЙЎ r uЛђ t /) is a fictional character appearing in American comic books published by Marvel [HOST]d by Stan Lee, Larry Lieber and Jack Kirby, the character first appeared in Tales to Astonish #13 (November ). An extraterrestrial, sentient tree-like creature, the original Groot first appeared as an invader that intended to capture humans for [HOST]g: Hole.
HESI RN EXIT EXAM V1 HESI EXIT EXAM V1 1. Which information is a priority for the RN to reinforce to an older client after intravenous pylegraphy? A) Eat a light diet for the rest of the day B) Rest for the next 24 hours since the preparation and the test is tiring. C) During waking hours drink at least 1 8-ounce glass of fluid every hour for the next 2 days D) Measure the urine .
HESI RN COMPREHENSIVE EXAMHESI RN Comprehensive Exam 1 A client with asthma receives a prescription for high blood pressure during a clinic visit. Which prescription should the nurse anticipate the client to receive that is at least likely to exacerbate asthma? Metoprolol Tartrate (Lopressor) The best antihypertensive agent for clients with asthma is metoprolol .
41 minutes agoВ В· She continued her roll, ending the day with seven more birdies, including one on her final hole (No. 9), and 24 putts compared to 33 in the first round. When asked how she would grade her day, Ko.I Know What This Hole is For Vol 18Trans magrinha gostosa Mixed girl TiffanyTease from onlyfans drips cum First ever ginger babe Felenk Roka on a massage table oiled Wife joyfully exposes herself in the city while she masturbates Una rica verga en mi boca College Girls Hardcore Group Party Sex Nuestro viciosillo jugando Shy but horny mature nurse use her body to help patient to collect semen！害羞但淫蕩熟女и·еЈ«е€©з”Ёиє«й«”幫男人收集精液！ Horny teens suck and ride cock Vanessa P C gets leche tribute por el padrino
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# I Know What This Hole Is For Vol 18. I Know What This Hole is For Vol 18 I Know What This Hole is For Vol Like. About Share. views. 0%. 0 0. From: admin. Category: ้เด็ก. Use these formulas to calculate volume or, for round holes, use our cylinder cubic yardage calculator. Rectangle: volume = length x width x depth Cylinder: volume = radius 2 x π x depth (π = ) Calculate Concrete Volume. With the volume of the submerged post and the volume of the hole figured out, subtract the post volume from the hole volume, and you will have the volume of concrete needed for each hole.. volume=1/3 × π × 4( 2 + × + 2) = in 3. Ellipsoid. An ellipsoid is the three-dimensional counterpart of an ellipse, and is a surface that can be described as the deformation of a sphere through scaling of directional elements. The center of an ellipsoid is the point at which three pairwise perpendicular axes of symmetry.. Apr 14,  · The formula behind the volume of a hollow cylinder is: cylinder_volume = π * (R² - r²) * cylinder_height. where R - external radius, and r - internal radius. To calculate the volume of a cylindrical shell, let's take some real-life example, maybe a roll of toilet paper, because why not?:) Enter the external radius of the cylinder. The standard is equal to approximately cm.. This is the size of the woofer. Available sizes: 8, 10, 12, 15, 18 inches. Cutout Diameter (unit of measure - inch) This is the hole cutout needed to "drop-in" a woofer and secure it to the front panel. Most manufacturers provide this spec in the woofer's datasheet. Mounting Depth (unit of measure - inch).. Use this post hole volume calculator to accurately figure the amount of concrete required for any sized post hole with or without any common sized post. Returns both cubic yards and how many bags. This free tool is designed as an aid for building fences, barns, post frame buildings, or just anything that has a post in the ground.. it through a small hole at the wall of the duct; or a series of holes positioned at right angles to the flow in a surface lying parallel to the lines of flow. The pitot static tube is an example of this. Figure 4 shows the principle of the pitot static tube. It will be seen that connecting tubes to a manometer makes the measurement of pressure.. If you know 2 of the 3 variables the third can be calculated. As usual, here at wwwcom, we have a calculator that will do all the work for you. This ultra calculator is special by allowing you to choose among a great variety of units. Unlike other calculators, you are NOT confined to inputting volume in liters, time in minutes, etc.. Dec 26,  · Breaking taps is a pain! (by 'a little', I mean or oversize on 1/, or on 1/2 NC, etc) from what I can see, the ideal is, whiich is close to a F letter size drill for a 77% thread A G size will give you an easier time tapping the hole, and reduce the strength by something around 10% overall. Last edited: Dec 26,. Jun 05,  · Some guitar wiring circuit designs use k or even 1 Meg pots. Here’s what you need to know about choosing the right [pots for your guitar. Generally speaking, to control volume, humbuckers should be paired to k pots, while single-coil pickups should be .. Search the world's information, including webpages, images, videos and more. Google has many special features to help you find exactly what you're looking [HOST]g: Hole.. Google's free service instantly translates words, phrases, and web pages between English and over other [HOST]g: Hole.. Calculate cubic yards, cubic feet or cubic meters for landscape material, mulch, land fill, gravel, cement, sand, containers, etc. Enter measurements in US or metric units and get volume conversions to other units. How to calculate cubic yards for rectangular, circular, annular and triangular areas. Calculate project cost based on price per cubic foot, cubic yard or cubic meter.. 50 x Volume 1 (V1) = x ; V1 = 3 gallons The final mixture (Volume 2 or V2) is the amount of the concentrate (V1) plus the required amount to make up to V2. If V1 = 3 gallons and the required amount is gallons, add 97 gallons of water to 3 gallons of concentrate. Formula 11 C1 x V1 = C2 x V2 C1= % of a.i. in concentrateMissing: Hole.. Shop [HOST] for Every Day Low Prices. Free Shipping on Orders \$35+ or Pickup In-Store and get a Pickup Discount. Open a Walmart Credit Card to Save Even More!. November 27, John Michell anticipates black holes. We think of black holes as a 20th century invention, dating back to , when Albert Einstein first published his theory of general relativity and fellow physicist Karl Schwarzschild used those equations to envision a spherical section of spacetime so badly warped around a concentrated mass that it is invisible to the .. I want to know its volume in cubic inches and therefore its liquid capacity (how much water I can fit in the tank) in litres. I would enter the values in the calculator as shown in the figure above, selecting the correct units for each measurement from the drop down options.. The larger your subwoofer is, the more volume of space is needed around it. According to the JL Audio website, the recommended volume space for a inch subwoofer is cubic feet. For a inch woofer, the volume recommendation is cubic feet, and the recommendation for an eight-inch speaker is cubic feet.. Mandingo, Actor: Mandingo Rocks That Ass. Mandingo was born on February 25, in Mississippi, USA as Frederick Lamont. He is an actor and director.. Nov 16,  · Regardless of its origins, the 18″ AR barrel is very popular in the competition world. It offers a very smooth recoil impulse due to its full-length rifle gas system while shaving a few ounces off of the end of the rifle. The length also maintains good velocity for reaching out with a flatter trajectory.. Jul 25,  · And she thinks I look good.
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And she sees my chest pubes all the way down to my ball fro, and she says iv'e had the old bull, now I want the old calve. Then she grabs me by the wein. Dale Doback.. Jun 03, В В· Input the container's length into the first field of the box volume calculator. It's 12 inches in our case. Enter the width of the box. Put 10 inches into the proper field. Finally, input the height of your container, 8 inches. And there it is: the volume of a rectangular prism calculator did the job. Now we know that our cat's volume is. approximately 53 times more units of volume through a hole than water if the water is able to flow at all. • The characteristics of the hole (or leak) that determine whether it will leak are the smallest diameter in the hole path, the path length of the hole through the part material, surface tension, and the surface finish of hole (leak).. Saturday Nite Southern Soul Juke Joint Jamz!Compiled,Mixed & Arranged by [HOST]pyright Disclaimer Under Section of the Copyright Act , allowance is Missing: Hole.. Unlock the power of video and join over M professionals, teams, and organizations who use Vimeo to create, collaborate and [HOST]g: Hole.. Related Surface Area Calculator | Volume Calculator. Area is a quantity that describes the size or extent of a two-dimensional figure or shape in a plane. It can be visualized as the amount of paint that would be necessary to cover a surface, and is the two-dimensional counterpart of the one-dimensional length of a curve, and three-dimensional volume of a solid.. Faye Reagan, Actress: Pin-Up Girls 5. Buxom, freckled, and slender redhead stunner Faye Reagan was born Faye Jillian Henning on September 19, , in Nashville, Tennessee. Faye moved with her family to Las Vegas, Nevada when she was only eight months old. She first began performing in explicit X-rated fare at age 19 in Reagan did five sex scenes in a single .. equal to the concentration of holes in the valence band. We may denote, n i: intrinsic electron concentration p i: intrinsic hole concentration However, n i = p i Simply, n i:intrinsic carrier concentration, which refers to either the intrinsic electron or hole concentration Commonly accepted values of n i at T = В°K Silicon x cm. Mar 24, В В· Costa & Hanna Finsen - I Will Wait For You (Radio Edit) Somna & Sarah Russell - Story Untold (Radio Edit) Hazem Beltagui & Julie Scott - When Nothing's True (Radio Edit) 28 Mino Safy & Maria Nayler - A Second Breath (Radio Edit) Store N Forward & Neev Kennedy - Brought Me Back To Life (Radio Edit) Missing: Hole.. plenum, there is an available air volume of 1, cfm and an available velocity of ft/min. After the third branch run, a total of cfm has been distributed to the branches and the velocity in the plenum has been reduced to ft/[HOST] conditions indicate that the proper location for the reduction in the plenum is.. – 18″ ( cm) Square (1) – Several flower boxes (like ish) The answer is maybe. You want to discover the total volume of the containers that you have. We have some larger diameter containers, but they are not very tall, so they do not work well for some of our larger plants.. ISABELLE RONIN (@ISABELLERONIN) Chasing Red was one of ’s most-read stories on Wattpad -- and that was just the beginning for this Winnipeg-Manitoba-based writer. In a single year, her explosive hit has racked up over million reads on Wattpad. Newly edited and expanded, the book was split into two and hit bookstore shelves in Missing: Hole.. Use this form to estimate the brake horsepower required. Brake horsepower is the power out of the drive motor, and the power into the water pump and is how most pumps and drive motors are specified. Also calculate the total power requirements (energy use rate). The pump efficiency depends on the pump and the pressure and flow that the pump is Missing: Hole.. Nov 14, В В· But hold on, we need to subtract the inner cylinder, otherwise, it’s like we’re filling up the entire bracelet without a hole for your wrist. Inner cylinder volume equals: x (/2) 2 x which equals cubic inches. That means the volume of the bangle is cubic inches minus cubic inches which equals cubic inches.. the hole is drilled, it may be perforated by down-the-hole tools, forming a screen opposite the water-producing formations. With most methods of down-the-hole perforating, a small aperture cannot be formed nor can the aperture size be precisely controlled. Consequently, finer-grained aquifers must be avoided.. Short Stories: The Red-Headed League by Arthur Conan Doyle. I had called upon my friend, Mr. Sherlock Holmes, one day in the autumn of last year, and found him in deep conversation with a very stout, florid-faced elderly gentleman, with fiery red hair. With an apology for my intrusion, I was about to withdraw, when Holmes pulled me abruptly.. I Know What This Hole is For Vol 21 - Kumpulan Video Film Indo, Sikwap Streaming Cewek Jilbab Ngentot, Sedarah Ibu Dan Anak, XXX Indo Porno Memek Perawan Sd, Sex - I Know What This Hole is For Vol 21п»ї. 12 min. Anal The Forbidden Fruit is Sweet Vol 17 min. I Know What This Hole is For Vol 17 min. I Know What This Hole is For Vol 6. 17 min.. Groot (/ ЙЎ r uЛђ t /) is a fictional character appearing in American comic books published by Marvel [HOST]d by Stan Lee, Larry Lieber and Jack Kirby, the character first appeared in Tales to Astonish #13 (November ). An extraterrestrial, sentient tree-like creature, the original Groot first appeared as an invader that intended to capture humans for [HOST]g: Hole.. HESI RN EXIT EXAM V1 HESI EXIT EXAM V1 1. Which information is a priority for the RN to reinforce to an older client after intravenous pylegraphy? A) Eat a light diet for the rest of the day B) Rest for the next 24 hours since the preparation and the test is tiring. C) During waking hours drink at least 1 8-ounce glass of fluid every hour for the next 2 days D) Measure the urine .. HESI RN COMPREHENSIVE EXAMHESI RN Comprehensive Exam 1 A client with asthma receives a prescription for high blood pressure during a clinic visit. Which prescription should the nurse anticipate the client to receive that is at least likely to exacerbate asthma? Metoprolol Tartrate (Lopressor) The best antihypertensive agent for clients with asthma is metoprolol .. 41 minutes agoВ В· She continued her roll, ending the day with seven more birdies, including one on her final hole (No. 9), and 24 putts compared to 33 in the first round. 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DIAMMIBENVERMk1
2022-07-08
Using Logarithms
$\begin{array}{rl}-{2}^{n-1}\mathrm{ln}2& =-100\mathrm{ln}10\\ & \\ -100\mathrm{ln}10& =-230\\ & \\ \frac{-230}{\mathrm{ln}\left(2\right)}& =-333\\ & \\ -{2}^{n-1}& >-333\\ & \\ \left(n-1\right)\mathrm{ln}\left(-2\right)& >\mathrm{ln}\left(-333\right)\end{array}$
Here is where I am stuck.
I am not sure if this part is correct: $-n-1=8$. Then solving we would get $-9$
Camron Herrera
Expert
Log is an increasing function. So you can take log in both side of an inequality and that is fine. But you can't take log of negative values like -2 The mistake is at
$-{2}^{n-1}=-333.$
You missed negative sign at 333. Then if you correct the next steps it's fine. Also when you multiply both sides of an inequality by negative the sign changes.
$-{2}^{n-1}ln2<-100\ast ln10$
$-{2}^{n-1}<\frac{-100\ast ln10}{ln2}$
${2}^{n-1}>\frac{100\ast ln10}{ln2}$
$n-1\ast ln\left(2\right)>ln\left(\frac{100\ast ln10}{ln2}\right).$
And this gives $n>\frac{ln\left(\frac{100\ast ln10}{ln2}\right)}{ln2}+1$
Do you have a similar question?
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DIAMMIBENVERMk1. 2022-07-08. Using Logarithms. $\begin{array}{rl}-{2}^{n-1}\mathrm{ln}2& =-100\mathrm{ln}10\\ & \\ -100\mathrm{ln}10& =-230\\ & \\ \frac{-230}{\mathrm{ln}\left(2\right)}& =-333\\ & \\ -{2}^{n-1}& >-333\\ & \\ \left(n-1\right)\mathrm{ln}\left(-2\right)& >\mathrm{ln}\left(-333\right)\end{array}$. Here is where I am stuck.. I am not sure if this part is correct: $-n-1=8$. Then solving we would get $-9$. Camron Herrera. Expert. Log is an increasing function. So you can take log in both side of an inequality and that is fine. But you can't take log of negative values like -2 The mistake is at.
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$-{2}^{n-1}=-333.$. You missed negative sign at 333. Then if you correct the next steps it's fine. Also when you multiply both sides of an inequality by negative the sign changes.. $-{2}^{n-1}ln2<-100\ast ln10$. $-{2}^{n-1}<\frac{-100\ast ln10}{ln2}$. ${2}^{n-1}>\frac{100\ast ln10}{ln2}$. $n-1\ast ln\left(2\right)>ln\left(\frac{100\ast ln10}{ln2}\right).$. And this gives $n>\frac{ln\left(\frac{100\ast ln10}{ln2}\right)}{ln2}+1$. Do you have a similar question?.
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https://www.byronjohn.co.za/best-wdng/wave-equation-triangle-bd9af7
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We conclude that the most general solution to the wave equation, , is a ... For instance, suppose that we have a triangular wave pulse of the form (749) (See Figure 52.) If it isn't possible with this equation, is there any general equation representing a Triangular wave with … A Triangular Waveform . As such, the wave speed can also be calculated using the equation: The functions and are shown in Figure 52. Triangle Waves. The analytical fuzzy triangular solutions for both one-dimensional homogeneous and non-homogeneous wave equations with emphasis on the type of [gH-p]-differentiability of solutions are obtained by using the fuzzy D’Alembert’s formulas. This is pretty tedious and not very fun, but here we go: The Fourier Transform of the triangle function is the sinc function squared. Journal of Computational and Applied Mathematics 206 :1, 420-431. collapse all. I can shift the wave vertically, but not horizontally. Now, you can go through and do that math yourself if you want. (Dated: November 20, 2010) It is well known that symmetry considerations can often be a powerful tool for simplifying physical systems. The wave equation is one of the most important equations in mechanics. Few types of waves like square wave, sawtooth wave, triangular wave, etc. We could calculate the RMS value by splitting the signal in 3, from 0 to t1, then from t1 to t3, and then from t3 to T. However, we already know the RMS value of the waveform from 0 to t2. T = 10*(1/50); fs = 1000; t = 0:1/fs:T-1/fs; x = sawtooth(2*pi*50*t); plot(t,x) grid on. For a bipolar triangle, the waveform looks like the one in Figure 7. To get a variation of the triangular wave that has curves rather than straight lines, you just need to introduce an exponent into the equation to make it quadratic. [091] Core Loss Modeling - Part III Sinewave Versus Triangle Wave Losses. Plot the power spectrum of the wave. All forum topics; Previous Topic; Next Topic; 4 REPLIES 4. ptc-1368288. In terms of its harmonic content, the triangle wave has a strong fundamental, much weaker and rapidly decreasing odd harmonics (much more so than the square wave), and no even harmonics. (2007) Numerical solution of the acoustic wave equation using Raviart–Thomas elements. All waves, including sound waves and electromagnetic waves, follow this equation. For triangle wave like for sine wave the half of the pulses increase and other half decrease. Table of Contents. So here's the first harmonic over a fundamental f0, the third harmonic, the fifth harmonic and so on, 7, 9, 11. Applets related to section 4.3. It is given in equation (15). The numerical approximation of this equation is … Standing wave as a result of reflection; w_n Building block; v_n Building block; A great MIT applet ; Wave Pulse Reflection (Free & Fixed Ends) Other useful applets. Powerpoint for the Wave Equation and using the equation triangle Generate 10 periods of a sawtooth wave with a fundamental frequency of 50 Hz. To make use of that assumption you also need to use a vector identity. The sample rate is 1 kHz. So you need to use that. It describes not only the movement of strings and wires, but also the movement of fluid surfaces, e.g., water waves. Square-to-Triangle Wave Converter Circuit. x^2 shape): y = pow(abs((x++ % 6) - 3), 2.0); Concave curves (i.e. – Robert Rouhani Sep 8 '12 at 16:31. The first would be pylab.arange(0,0.5,25) – 8765674 Sep 8 '12 at 16:36. The first two parts of this article showed how the core losses for real waveforms could be modeled better. I'm pretty sure that the claim is only true under irrotational assumptions; at least, all other previous times I've seen a wave equation derived from Navier-Stokes/Euler the irrotational assumption is enforced. The most important feature of a triangular wave is that it has equal rise and fall times while a Triangle waveform in mind: Stack Overflow. Although called a triangular waveform, the triangular wave is actually more of a symmetrical linear ramp waveform because it is simply a slow rising and falling voltage signal at a constant frequency or rate. Calculating the difference between sinewave and triangle wave excitation shows that sinewave measurements are sufficient for loss calculations. Author: Menny. The triangle wave is the second common waveform examined in Electronic Music Interactive, and it has the following characteristics: Frequency Components: Odd Numbered Harmonics: Relative Amplitudes of Harmonics: 1/Harmonic Number Squared: Phase: Every Other Harmonic is 180 degrees Out of Phase: The ratio 1/harmonic number squared means that the first harmonic has an … And the difference here though, is that the coefficients in front of each of the harmonics has a 1 over n squared. Triangular waves are a periodic, non-sinusoidal waveform with a triangular shape. Actually, you've basically gotten there already. A formula triangle for the wave speed equation All waves, including sound waves and electromagnetic waves , follow this equation. The square waveform can be generated from a number of sources, including a function generator or a circuit that can produce square waves such as this 555 timer clock circuit. The Triangle Wave Function is a periodic function used in signal processing. The string is plucked into oscillation. Again, the recipe calls for all of the odd harmonics. This equation starts at $(0,0)$. We can simply substitute equation [1] into the formula for the definition of the Fourier Transform, then crank through all the math, and then get the result. The triangle wave can also be expressed as the integral of the square wave: A simple equation with a period of 4, with . (2007) A diagonal-mass-matrix triangular-spectral-element method based on cubature points. Figure 7. Its left and right hand ends are held fixed at height zero and we are told its initial configuration and speed. are often needed to test various signal processing techniques. The wave equation is surprisingly simple to derive and not very complicated to solve although it is a second-order PDE. For waves that travel at light speed the subscript v is dropped and the wave equation is written as ☐ψ=0. The wave Equation . I think it might be simpler to build the array using three parts, as y=(-)x. I.e. The sine Fourier transform of this pulse shape is zero by symmetry. Is there any way to do that? However, the cosine Fourier transform is (750) (See Exercise 7.) Some particular waves have their own specific speeds. A waveform which appears (on an oscilliscope or in a waveform editor) as a series of rising and falling straight lines of equal slope, resembling a row of roof peaks or triangles without the base segments. To better understand the code, be ready with your Matlab to test it as we go. The square-to-triangle wave generator circuit that we will build with only resistors and capacitors is shown below. As this only uses the modulo operation and absolute value, this can be used to simply implement a triangle wave on hardware electronics with less CPU power: ::(1) About; Products For Teams; Stack Overflow Public questions & answers; Stack ... but implementing one of those equations should be a nice starting point. However, I would like the equation to start from an arbitrary point $(x,y)$. Spaces of arbitrary dimension can be built without any reference to distances or coordinates based simply on points which each have exactly 3 neighbors. Introduction . Solution of the Wave Equation by Separation of Variables The Problem Let u(x,t) denote the vertical displacement of a string from the x axis at position x and time t. The string has length ℓ. General Triangular Waveform So to build a triangle wave, from sine waves, it's similar to a square wave. We investigate the internal observability of the wave equation with Dirichlet boundary conditions in a triangular domain. The triangular potential is not symmetric in x, thus the wave functions lack the even or odd symmetry that one obtains for the infinitely deep square well. Play overtones; Vibrating String; A small collection of resources; Wave Interference and Beat Frequency; Next. A triangular function (also known as a triangle function, hat function, or tent function) is a function whose graph takes the shape of a triangle.Often this is an isosceles triangle of height 1 and base 2 in which case it is referred to as the triangular function. It is an even function, which means it is symmetrical around the y-axis.. The rate at which the voltage changes between each ramp direction is equal during both halves of the cycle as shown below. For such phenomena, the wave equation serves as a model problem. We will derive the wave equation using the model of the suspended string (see Fig. Examples. So I've been wondering why can't simple harmonic motion be represented in form of triangular waves.Although the equations above involve angular momentum so I may be contradicting myself but fundamentally the velocity time is sine function : $$-\sin(x)$$ and the gradient represents the acceleration is non-uniformly increasing and decreasing. Set xmax to 0.5 to generate a standard triangle wave. People often get confused between the triangle and sawtooth waves. A formula triangle for the wave speed equation. Concave curves (i.e. Please anyone help!! Solving the wave equation in the time domain by numerical methods is a delicate but fundamental problem for modeling numerous physical phe-nomena such as acoustic, elastic, or electromagnetic waves. This function is sometimes also called the continuous sawtooth function, however, the actual “sawtooth” has a slightly different shape: Search completed in 0.019 … A triangle wave at 100Hz. Open Live Script. 50 Hz Sawtooth Wave. Thank you, Labels: Algebra_Geometry; 0 Kudos Reply. More precisely, the domain taken into exam is the half of the equilateral triangle. Standing Waves: the Equilateral Triangle Mark Semon⁄ Physics Department, Bates College Nathaniel Stambaughy Mathematics Department, Brandeis University. Use the formula triangle to help you rearrange the equation Wave speed is the distance moved by a wave front every second. The 1-D Wave Equation 18.303 Linear Partial Differential Equations Matthew J. Hancock Fall 2006 1 1-D Wave Equation : Physical derivation Reference: Guenther & Lee §1.2, Myint-U & Debnath §2.1-2.4 [Oct. 3, 2006] We consider a string of length l with ends fixed, and rest state coinciding with x-axis. Wave Equation--Triangle: Eric Weisstein's World of Mathematics [home, info] Words similar to wave equation triangle Usage examples for wave equation triangle Words that often appear near wave equation triangle Rhymes of wave equation triangle Invented words related to wave equation triangle: Search for wave equation triangle on Google or Wikipedia. In this article, I will provide a working Matlab code for generating triangular wave. Problem, do not know the equation for a triangle wave.
wave equation triangle
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We conclude that the most general solution to the wave equation, , is a ... For instance, suppose that we have a triangular wave pulse of the form (749) (See Figure 52.) If it isn't possible with this equation, is there any general equation representing a Triangular wave with … A Triangular Waveform . As such, the wave speed can also be calculated using the equation: The functions and are shown in Figure 52. Triangle Waves. The analytical fuzzy triangular solutions for both one-dimensional homogeneous and non-homogeneous wave equations with emphasis on the type of [gH-p]-differentiability of solutions are obtained by using the fuzzy D’Alembert’s formulas. This is pretty tedious and not very fun, but here we go: The Fourier Transform of the triangle function is the sinc function squared. Journal of Computational and Applied Mathematics 206 :1, 420-431. collapse all. I can shift the wave vertically, but not horizontally. Now, you can go through and do that math yourself if you want. (Dated: November 20, 2010) It is well known that symmetry considerations can often be a powerful tool for simplifying physical systems. The wave equation is one of the most important equations in mechanics. Few types of waves like square wave, sawtooth wave, triangular wave, etc. We could calculate the RMS value by splitting the signal in 3, from 0 to t1, then from t1 to t3, and then from t3 to T. However, we already know the RMS value of the waveform from 0 to t2. T = 10*(1/50); fs = 1000; t = 0:1/fs:T-1/fs; x = sawtooth(2*pi*50*t); plot(t,x) grid on. For a bipolar triangle, the waveform looks like the one in Figure 7. To get a variation of the triangular wave that has curves rather than straight lines, you just need to introduce an exponent into the equation to make it quadratic. [091] Core Loss Modeling - Part III Sinewave Versus Triangle Wave Losses. Plot the power spectrum of the wave. All forum topics; Previous Topic; Next Topic; 4 REPLIES 4. ptc-1368288. In terms of its harmonic content, the triangle wave has a strong fundamental, much weaker and rapidly decreasing odd harmonics (much more so than the square wave), and no even harmonics. (2007) Numerical solution of the acoustic wave equation using Raviart–Thomas elements. All waves, including sound waves and electromagnetic waves, follow this equation. For triangle wave like for sine wave the half of the pulses increase and other half decrease. Table of Contents. So here's the first harmonic over a fundamental f0, the third harmonic, the fifth harmonic and so on, 7, 9, 11. Applets related to section 4.3. It is given in equation (15). The numerical approximation of this equation is … Standing wave as a result of reflection; w_n Building block; v_n Building block; A great MIT applet ; Wave Pulse Reflection (Free & Fixed Ends) Other useful applets. Powerpoint for the Wave Equation and using the equation triangle Generate 10 periods of a sawtooth wave with a fundamental frequency of 50 Hz. To make use of that assumption you also need to use a vector identity. The sample rate is 1 kHz. So you need to use that. It describes not only the movement of strings and wires, but also the movement of fluid surfaces, e.g., water waves. Square-to-Triangle Wave Converter Circuit. x^2 shape): y = pow(abs((x++ % 6) - 3), 2.0); Concave curves (i.e. – Robert Rouhani Sep 8 '12 at 16:31. The first would be pylab.arange(0,0.5,25) – 8765674 Sep 8 '12 at 16:36. The first two parts of this article showed how the core losses for real waveforms could be modeled better. I'm pretty sure that the claim is only true under irrotational assumptions; at least, all other previous times I've seen a wave equation derived from Navier-Stokes/Euler the irrotational assumption is enforced. The most important feature of a triangular wave is that it has equal rise and fall times while a Triangle waveform in mind: Stack Overflow. Although called a triangular waveform, the triangular wave is actually more of a symmetrical linear ramp waveform because it is simply a slow rising and falling voltage signal at a constant frequency or rate. Calculating the difference between sinewave and triangle wave excitation shows that sinewave measurements are sufficient for loss calculations. Author: Menny. The triangle wave is the second common waveform examined in Electronic Music Interactive, and it has the following characteristics: Frequency Components: Odd Numbered Harmonics: Relative Amplitudes of Harmonics: 1/Harmonic Number Squared: Phase: Every Other Harmonic is 180 degrees Out of Phase: The ratio 1/harmonic number squared means that the first harmonic has an … And the difference here though, is that the coefficients in front of each of the harmonics has a 1 over n squared. Triangular waves are a periodic, non-sinusoidal waveform with a triangular shape. Actually, you've basically gotten there already. A formula triangle for the wave speed equation All waves, including sound waves and electromagnetic waves , follow this equation. The square waveform can be generated from a number of sources, including a function generator or a circuit that can produce square waves such as this 555 timer clock circuit. The Triangle Wave Function is a periodic function used in signal processing. The string is plucked into oscillation.
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Again, the recipe calls for all of the odd harmonics. This equation starts at $(0,0)$. We can simply substitute equation [1] into the formula for the definition of the Fourier Transform, then crank through all the math, and then get the result. The triangle wave can also be expressed as the integral of the square wave: A simple equation with a period of 4, with . (2007) A diagonal-mass-matrix triangular-spectral-element method based on cubature points. Figure 7. Its left and right hand ends are held fixed at height zero and we are told its initial configuration and speed. are often needed to test various signal processing techniques. The wave equation is surprisingly simple to derive and not very complicated to solve although it is a second-order PDE. For waves that travel at light speed the subscript v is dropped and the wave equation is written as ☐ψ=0. The wave Equation . I think it might be simpler to build the array using three parts, as y=(-)x. I.e. The sine Fourier transform of this pulse shape is zero by symmetry. Is there any way to do that? However, the cosine Fourier transform is (750) (See Exercise 7.) Some particular waves have their own specific speeds. A waveform which appears (on an oscilliscope or in a waveform editor) as a series of rising and falling straight lines of equal slope, resembling a row of roof peaks or triangles without the base segments. To better understand the code, be ready with your Matlab to test it as we go. The square-to-triangle wave generator circuit that we will build with only resistors and capacitors is shown below. As this only uses the modulo operation and absolute value, this can be used to simply implement a triangle wave on hardware electronics with less CPU power: ::(1) About; Products For Teams; Stack Overflow Public questions & answers; Stack ... but implementing one of those equations should be a nice starting point. However, I would like the equation to start from an arbitrary point $(x,y)$. Spaces of arbitrary dimension can be built without any reference to distances or coordinates based simply on points which each have exactly 3 neighbors. Introduction . Solution of the Wave Equation by Separation of Variables The Problem Let u(x,t) denote the vertical displacement of a string from the x axis at position x and time t. The string has length ℓ. General Triangular Waveform So to build a triangle wave, from sine waves, it's similar to a square wave. We investigate the internal observability of the wave equation with Dirichlet boundary conditions in a triangular domain. The triangular potential is not symmetric in x, thus the wave functions lack the even or odd symmetry that one obtains for the infinitely deep square well. Play overtones; Vibrating String; A small collection of resources; Wave Interference and Beat Frequency; Next. A triangular function (also known as a triangle function, hat function, or tent function) is a function whose graph takes the shape of a triangle.Often this is an isosceles triangle of height 1 and base 2 in which case it is referred to as the triangular function. It is an even function, which means it is symmetrical around the y-axis.. The rate at which the voltage changes between each ramp direction is equal during both halves of the cycle as shown below. For such phenomena, the wave equation serves as a model problem. We will derive the wave equation using the model of the suspended string (see Fig. Examples. So I've been wondering why can't simple harmonic motion be represented in form of triangular waves.Although the equations above involve angular momentum so I may be contradicting myself but fundamentally the velocity time is sine function : $$-\sin(x)$$ and the gradient represents the acceleration is non-uniformly increasing and decreasing. Set xmax to 0.5 to generate a standard triangle wave. People often get confused between the triangle and sawtooth waves. A formula triangle for the wave speed equation. Concave curves (i.e. Please anyone help!! Solving the wave equation in the time domain by numerical methods is a delicate but fundamental problem for modeling numerous physical phe-nomena such as acoustic, elastic, or electromagnetic waves. This function is sometimes also called the continuous sawtooth function, however, the actual “sawtooth” has a slightly different shape: Search completed in 0.019 … A triangle wave at 100Hz. Open Live Script. 50 Hz Sawtooth Wave. Thank you, Labels: Algebra_Geometry; 0 Kudos Reply. More precisely, the domain taken into exam is the half of the equilateral triangle. Standing Waves: the Equilateral Triangle Mark Semon⁄ Physics Department, Bates College Nathaniel Stambaughy Mathematics Department, Brandeis University. Use the formula triangle to help you rearrange the equation Wave speed is the distance moved by a wave front every second. The 1-D Wave Equation 18.303 Linear Partial Differential Equations Matthew J. Hancock Fall 2006 1 1-D Wave Equation : Physical derivation Reference: Guenther & Lee §1.2, Myint-U & Debnath §2.1-2.4 [Oct. 3, 2006] We consider a string of length l with ends fixed, and rest state coinciding with x-axis. Wave Equation--Triangle: Eric Weisstein's World of Mathematics [home, info] Words similar to wave equation triangle Usage examples for wave equation triangle Words that often appear near wave equation triangle Rhymes of wave equation triangle Invented words related to wave equation triangle: Search for wave equation triangle on Google or Wikipedia. In this article, I will provide a working Matlab code for generating triangular wave. Problem, do not know the equation for a triangle wave.. wave equation triangle.
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Pat has two sheets of large, square stamps. One sheet consists of 9 stamps arranged in a 3-by-3 square. (That's 3 stamps by 3 stamps, not 3 inches by 3 inches.) The other sheet consists of 100 stamps arranged in a 10-by-10 square. The stamps on each sheet are separated by perforated creases. All the stamps on both sheets are the same shape and size.
The total length of the creases on the 3-by-3 sheet is 18 inches.
How many inches of creases does the 10-by-10 sheet have?
Oct 11, 2020
#1
0
First, let's take a look at the 3 by 3 sheet of stamps.
There are two vertical creases between the three columns of stamps.
These two creases comprise a total of six vertical stamp-lengths.
Similarly for the two horizontal creases, six horizontal stamp-lengths.
That makes a total of 12 stamp-lengths of creases.
It's given that there are 18 inches of crease in the 3 x 3 sheet.
Therefore these square stamps are 1.5 inches on a side.
Figure the 10 by 10 sheet the same way.
Nine vertical creases, creating 90 stamp-lengths of vertical crease.
Nine horizontal creases, creating 90 stamp-lengths of horizontal crease.
That totals 180 stamp-lengths of crease in the 10 x 10 sheet.
Since each stamp is 1.5 inches on a side, the total length of crease is 1.5 • 180 = 270 inches
.
Oct 11, 2020
edited by Guest Oct 11, 2020
#2
+28025
+1
There are 4 creases in the 3x3 totaling 18 inches each crease is 4.5 in
10 x 10 will have 18 creases that are 10/3 longer
18 * 10/3 * 4.5 = 270 inches
Oct 11, 2020
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+0. # plz help. 0. 77. 2. Pat has two sheets of large, square stamps. One sheet consists of 9 stamps arranged in a 3-by-3 square. (That's 3 stamps by 3 stamps, not 3 inches by 3 inches.) The other sheet consists of 100 stamps arranged in a 10-by-10 square. The stamps on each sheet are separated by perforated creases. All the stamps on both sheets are the same shape and size.. The total length of the creases on the 3-by-3 sheet is 18 inches.. How many inches of creases does the 10-by-10 sheet have?. Oct 11, 2020. #1. 0. First, let's take a look at the 3 by 3 sheet of stamps.. There are two vertical creases between the three columns of stamps.. These two creases comprise a total of six vertical stamp-lengths.. Similarly for the two horizontal creases, six horizontal stamp-lengths.
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That makes a total of 12 stamp-lengths of creases.. It's given that there are 18 inches of crease in the 3 x 3 sheet.. Therefore these square stamps are 1.5 inches on a side.. Figure the 10 by 10 sheet the same way.. Nine vertical creases, creating 90 stamp-lengths of vertical crease.. Nine horizontal creases, creating 90 stamp-lengths of horizontal crease.. That totals 180 stamp-lengths of crease in the 10 x 10 sheet.. Since each stamp is 1.5 inches on a side, the total length of crease is 1.5 • 180 = 270 inches. .. Oct 11, 2020. edited by Guest Oct 11, 2020. #2. +28025. +1. There are 4 creases in the 3x3 totaling 18 inches each crease is 4.5 in. 10 x 10 will have 18 creases that are 10/3 longer. 18 * 10/3 * 4.5 = 270 inches. Oct 11, 2020.
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https://www.princeton.edu/~achaney/tmve/wiki100k/docs/Bresenham_s_line_algorithm.html
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# Bresenham's line algorithm
related topics {math, number, function} {system, computer, user} {line, north, south} {math, energy, light} {@card@, make, design} {work, book, publish} {day, year, event} {government, party, election}
The Bresenham line algorithm is an algorithm which determines which points in an n-dimensional raster should be plotted in order to form a close approximation to a straight line between two given points. It is commonly used to draw lines on a computer screen, as it uses only integer addition, subtraction and bit shifting, all of which are very cheap operations in standard computer architectures. It is one of the earliest algorithms developed in the field of computer graphics. A minor extension to the original algorithm also deals with drawing circles.
While algorithms such as Wu's algorithm are also frequently used in modern computer graphics because they can support antialiasing, the speed and simplicity of Bresenham's line algorithm mean that it is still important. The algorithm is used in hardware such as plotters and in the graphics chips of modern graphics cards. It can also be found in many software graphics libraries. Because the algorithm is very simple, it is often implemented in either the firmware or the hardware of modern graphics cards.
The label "Bresenham" is used today for a whole family of algorithms extending or modifying Bresenham's original algorithm. See further references below.
## Contents
### The algorithm
The common conventions that pixel coordinates increase in the down and right directions (e.g. that the pixel at (1,1) is directly above the pixel at (1,2)) and that the pixel centers that have integer coordinates will be used. The endpoints of the line are the pixels at (x0, y0) and (x1, y1), where the first coordinate of the pair is the column and the second is the row.
The algorithm will be initially presented only for the octant in which the segment goes down and to the right (x0x1 and y0y1), and its horizontal projection x1x0 is longer than the vertical projection y1y0 (the line has a slope whose absolute value is less than 1 and greater than 0.) In this octant, for each column x between x0 and x1, there is exactly one row y (computed by the algorithm) containing a pixel of the line, while each row between y0 and y1 may contain multiple rasterized pixels.
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# Bresenham's line algorithm. related topics {math, number, function} {system, computer, user} {line, north, south} {math, energy, light} {@card@, make, design} {work, book, publish} {day, year, event} {government, party, election}. The Bresenham line algorithm is an algorithm which determines which points in an n-dimensional raster should be plotted in order to form a close approximation to a straight line between two given points. It is commonly used to draw lines on a computer screen, as it uses only integer addition, subtraction and bit shifting, all of which are very cheap operations in standard computer architectures. It is one of the earliest algorithms developed in the field of computer graphics. A minor extension to the original algorithm also deals with drawing circles.. While algorithms such as Wu's algorithm are also frequently used in modern computer graphics because they can support antialiasing, the speed and simplicity of Bresenham's line algorithm mean that it is still important. The algorithm is used in hardware such as plotters and in the graphics chips of modern graphics cards. It can also be found in many software graphics libraries. Because the algorithm is very simple, it is often implemented in either the firmware or the hardware of modern graphics cards.
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The label "Bresenham" is used today for a whole family of algorithms extending or modifying Bresenham's original algorithm. See further references below.. ## Contents. ### The algorithm. The common conventions that pixel coordinates increase in the down and right directions (e.g. that the pixel at (1,1) is directly above the pixel at (1,2)) and that the pixel centers that have integer coordinates will be used. The endpoints of the line are the pixels at (x0, y0) and (x1, y1), where the first coordinate of the pair is the column and the second is the row.. The algorithm will be initially presented only for the octant in which the segment goes down and to the right (x0x1 and y0y1), and its horizontal projection x1x0 is longer than the vertical projection y1y0 (the line has a slope whose absolute value is less than 1 and greater than 0.) In this octant, for each column x between x0 and x1, there is exactly one row y (computed by the algorithm) containing a pixel of the line, while each row between y0 and y1 may contain multiple rasterized pixels.
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https://de.scribd.com/presentation/416090711/SD-04-Descriptive-Statistics-and-Central-Tendency
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Sie sind auf Seite 1von 24
# Descriptive
Statistics:
Central Tendency
Lesson 4
Psychology & Statistics
Goals of Psychology
Describe, predict, influence behavior
## & cognitive processes
Role of statistics
Descriptive statistics
## Describe, organize & summarize data
Efficient communication
Inferential statistics
Aid decision making ~
Organizing Data
Describing distribution of variables
enumeration: list raw data
Frequency distributions
organize tables or graphs
## range, most frequent value ~
Distributions as Tables
f = Frequency
# of times a value of variable occurs
Sf = n
## Frequency distribution tables
ordered list of all values of variable &
their frequencies
logical order (usually descending) ~
Enumeration Frequency
Distribution
# of presentations to be able to Table
recall 100%
8 12 18 15 10 X f
9 13 14 11 14 19 1
18 2
7 12 14 7 16
16 3
8 13 12 9 6 15 3
16 12 8 5 11 14 5
7 14 11 6 15 13 2
10 8 11 8 9 12 6
11 9 9 10 19 11 7
16 15 9 11 12 10 3
14 12 18 11 5 9 6
8 5
7 3
Sf = n 6 2
5 2
calculate proportions &
percentages 50
Grouped Frequency Distribution
Group by class intervals
report frequency for each interval
## Lose information: no exact values
General rules
each interval same width
## consecutive & do not overlap ~
Enumeration Frequency
Distribution
# of presentations to be able to Table
recall 100%
8 12 18 15 10 X f
9 13 14 11 14 19 1
18 2
7 12 14 7 16
16 3
8 13 12 9 6 15 3
16 12 8 5 11 14 5
7 14 11 6 15 13 2
10 8 11 8 9 12 6
11 9 9 10 19 11 7
16 15 9 11 12 10 3
14 12 18 11 5 9 6
8 5
7 3
Sf = n 6 2
5 2
calculate proportions &
percentages 50
Distributions as graphs
Summarizes data
focus on clear communication
Bar Graphs
nominal or ordinal data
discrete variables
## Histograms & Frequency Polygons
Interval/ratio data
## continuous & discrete variables
Relative frequency distributions
Y axis = proportions
Bar Graphs
Nominal Ordinal
18 18
14 14
10 10
f f
6 6
2 2
## Rep Dem Ind A B C D F
Histograms
X-axis
Class intervals
18
of variables
14
Y-axis
10
f
Frequencies 6
vertical bars ~ 2
5-6 7-8 9-10 11-12 13-14 15-16 17-18 19-20
5 7 9 11 13 15 17 19 21
# of presentations
Frequency polygons
Frequency represented as points
Contains same info as histogram ~
18 Relative Frequency
14
10
f f
5 7 9 11 13 15 17 19 21
# of presentations
# of presentations
Distributions: 3 useful features
Summarizes important characteristics of
data
## 3. How wide is distribution?
Shapes of distributions
Unimodal distribution
single value is most f
frequent
X
## Bimodal (or multimodal )
2 most frequently
f
occurring values
May indicate relevant
X
subgroups ~
Symmetry of distributions
Symmetric
if right side mirror-
f
image of left
Skewed - asymmetric
a few extreme values
Positively skewed:
f
right tail longer -4 -2 0 +2 +4
Negatively skewed:
X
left tail longer ~
f
-4 -2 0 +2 +4
X
The Normal Distribution
Bell-shaped
3 characteristics
Unimodal f
symmetric
asymptotic
## Many naturally-occurring variables
approximately normally distributed
Makes statistics useful ~
Central Tendency
Describes most typical values
Depends on level of measurement
## Mode (all levels)
Most frequently occurring value
## Median (only ordinal & interval/ratio)
value where ½ observations above
& ½ below
Mean (only interval/ratio)
Arithmetic average ~
Mode
Most frequently occurring value ~
18 18
14 14
10 f 10
f
6
6
2
2
A B C D F
Rep Dem Ind
Political affiliation
18
18
14
14
f 10
f 10
6
6
2
2
5 7 9 11 13 15 17 19 21
5 7 9 11 13 15 17 19 21
# of presentations
# of presentations
Median
Midpoint of a data set
values ½ smaller, ½ larger ~
10 20 30 40 50 60 70 80 90 10 20 30 40 50 60 70 80 90
Finding the Median
1. List all values from largest smallest
if f=3, then list 3 times
## 3. Even # entries = half way b/n middle 2
values ~
Mean
Summarizes quantitative data
May not be actual value in data set
Introduces error
## Computing the mean
Sum of all observations
Mean =
Number of observations
Statistical Notation
Formula for mean: X
N
Σ: summate
X: observation
value of an observation
N: number of observations
Or data points ~
Populations & Samples: Notation
Different symbols
Often different formulas for
calculation
Population: Greek letters
Population mean = μ
## Sample: Roman letters
Sample mean = X
APA style: M ~
Populations & Samples
Population
Parameter
Exact value
Population mean = μ
Sample
Statistic
estimate of parameter
introduces error
Sample mean = X ~
Formulas for Mean
Population mean
Parameter X
N
Sample mean
Statistic
X
X
Estimate / error
Sometimes n used for
N
sample ~
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Sie sind auf Seite 1von 24. # Descriptive. Statistics:. Central Tendency. Lesson 4. Psychology & Statistics. Goals of Psychology. Describe, predict, influence behavior. ## & cognitive processes. Role of statistics. Descriptive statistics. ## Describe, organize & summarize data. Efficient communication. Inferential statistics. Aid decision making ~. Organizing Data. Describing distribution of variables. enumeration: list raw data. Frequency distributions. organize tables or graphs. ## range, most frequent value ~. Distributions as Tables. f = Frequency. # of times a value of variable occurs. Sf = n. ## Frequency distribution tables. ordered list of all values of variable &. their frequencies. logical order (usually descending) ~. Enumeration Frequency. Distribution. # of presentations to be able to Table. recall 100%. 8 12 18 15 10 X f. 9 13 14 11 14 19 1. 18 2. 7 12 14 7 16. 16 3. 8 13 12 9 6 15 3. 16 12 8 5 11 14 5. 7 14 11 6 15 13 2. 10 8 11 8 9 12 6. 11 9 9 10 19 11 7. 16 15 9 11 12 10 3. 14 12 18 11 5 9 6. 8 5. 7 3. Sf = n 6 2. 5 2. calculate proportions &. percentages 50. Grouped Frequency Distribution. Group by class intervals. report frequency for each interval. ## Lose information: no exact values. General rules. each interval same width. ## consecutive & do not overlap ~. Enumeration Frequency. Distribution. # of presentations to be able to Table. recall 100%. 8 12 18 15 10 X f. 9 13 14 11 14 19 1. 18 2. 7 12 14 7 16. 16 3. 8 13 12 9 6 15 3. 16 12 8 5 11 14 5. 7 14 11 6 15 13 2. 10 8 11 8 9 12 6. 11 9 9 10 19 11 7. 16 15 9 11 12 10 3. 14 12 18 11 5 9 6. 8 5. 7 3. Sf = n 6 2. 5 2. calculate proportions &. percentages 50. Distributions as graphs. Summarizes data. focus on clear communication. Bar Graphs. nominal or ordinal data. discrete variables. ## Histograms & Frequency Polygons. Interval/ratio data. ## continuous & discrete variables. Relative frequency distributions. Y axis = proportions. Bar Graphs. Nominal Ordinal. 18 18. 14 14. 10 10. f f. 6 6. 2 2. ## Rep Dem Ind A B C D F. Histograms. X-axis. Class intervals. 18. of variables. 14. Y-axis. 10. f. Frequencies 6. vertical bars ~ 2. 5-6 7-8 9-10 11-12 13-14 15-16 17-18 19-20. 5 7 9 11 13 15 17 19 21. # of presentations. Frequency polygons. Frequency represented as points. Contains same info as histogram ~. 18 Relative Frequency. 14. 10. f f. 5 7 9 11 13 15 17 19 21. # of presentations. # of presentations. Distributions: 3 useful features. Summarizes important characteristics of. data. ## 3. How wide is distribution?. Shapes of distributions. Unimodal distribution.
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single value is most f. frequent. X. ## Bimodal (or multimodal ). 2 most frequently. f. occurring values. May indicate relevant. X. subgroups ~. Symmetry of distributions. Symmetric. if right side mirror-. f. image of left. Skewed - asymmetric. a few extreme values. Positively skewed:. f. right tail longer -4 -2 0 +2 +4. Negatively skewed:. X. left tail longer ~. f. -4 -2 0 +2 +4. X. The Normal Distribution. Bell-shaped. 3 characteristics. Unimodal f. symmetric. asymptotic. ## Many naturally-occurring variables. approximately normally distributed. Makes statistics useful ~. Central Tendency. Describes most typical values. Depends on level of measurement. ## Mode (all levels). Most frequently occurring value. ## Median (only ordinal & interval/ratio). value where ½ observations above. & ½ below. Mean (only interval/ratio). Arithmetic average ~. Mode. Most frequently occurring value ~. 18 18. 14 14. 10 f 10. f. 6. 6. 2. 2. A B C D F. Rep Dem Ind. Political affiliation. 18. 18. 14. 14. f 10. f 10. 6. 6. 2. 2. 5 7 9 11 13 15 17 19 21. 5 7 9 11 13 15 17 19 21. # of presentations. # of presentations. Median. Midpoint of a data set. values ½ smaller, ½ larger ~. 10 20 30 40 50 60 70 80 90 10 20 30 40 50 60 70 80 90. Finding the Median. 1. List all values from largest smallest. if f=3, then list 3 times. ## 3. Even # entries = half way b/n middle 2. values ~. Mean. Summarizes quantitative data. May not be actual value in data set. Introduces error. ## Computing the mean. Sum of all observations. Mean =. Number of observations. Statistical Notation. Formula for mean: X. N. Σ: summate. X: observation. value of an observation. N: number of observations. Or data points ~. Populations & Samples: Notation. Different symbols. Often different formulas for. calculation. Population: Greek letters. Population mean = μ. ## Sample: Roman letters. Sample mean = X. APA style: M ~. Populations & Samples. Population. Parameter. Exact value. Population mean = μ. Sample. Statistic. estimate of parameter. introduces error. Sample mean = X ~. Formulas for Mean. Population mean. Parameter X. N. Sample mean. Statistic. X . X. Estimate / error. Sometimes n used for. N. sample ~.
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https://www.tinkutara.com/2023/06/03/given-sin-a-sin-b-0-7-cos-a-cos-b-0-8-find-the-value-of-a-and-b/
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# Given-sin-A-sin-B-0-7-cos-A-cos-B-0-8-find-the-value-of-A-and-B-
Question Number 134581 by EDWIN88 last updated on 05/Mar/21
$$\mathrm{Given}\:\begin{cases}{\mathrm{sin}\:\mathrm{A}+\mathrm{sin}\:\mathrm{B}=\mathrm{0}.\mathrm{7}}\\{\mathrm{cos}\:\mathrm{A}+\mathrm{cos}\:\mathrm{B}=\mathrm{0}.\mathrm{8}}\end{cases} \\$$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}. \\$$
Answered by liberty last updated on 05/Mar/21
$$\Leftrightarrow\:\mathrm{2}+\mathrm{2cos}\:\left(\mathrm{A}−\mathrm{B}\right)=\:\frac{\mathrm{49}+\mathrm{64}}{\mathrm{100}} \\$$$$\Leftrightarrow\mathrm{2cos}\:\left(\mathrm{A}−\mathrm{B}\right)\:=\:−\frac{\mathrm{87}}{\mathrm{100}} \\$$$$\Leftrightarrow\:\mathrm{cos}\:\left(\mathrm{A}−\mathrm{B}\right)\:=−\frac{\mathrm{87}}{\mathrm{200}} \\$$$$\Leftrightarrow\:\mathrm{A}−\mathrm{B}\:\approx\:\mathrm{115}.\mathrm{8}°+\mathrm{k}.\mathrm{360}° \\$$$$\Leftrightarrow\:\mathrm{A}\:\approx\:\mathrm{B}+\mathrm{115}.\mathrm{8}°+\mathrm{k}.\mathrm{360}°\:,\mathrm{k}\epsilon\mathbb{Z} \\$$$$\\$$
Answered by mr W last updated on 05/Mar/21
$$\mathrm{sin}\:{A}+\mathrm{sin}\:{B}={p}\:\:\:\left(\mathrm{0}.\mathrm{7}\right) \\$$$$\mathrm{cos}\:{A}+\mathrm{cos}\:{B}={q}\:\:\:\left(\mathrm{0}.\mathrm{8}\right) \\$$$${A}=\frac{{A}+{B}}{\mathrm{2}}+\frac{{A}−{B}}{\mathrm{2}}={u}+{v} \\$$$${B}=\frac{{A}+{B}}{\mathrm{2}}−\frac{{A}−{B}}{\mathrm{2}}={u}−{v} \\$$$$\mathrm{sin}\:\left({u}+{v}\right)+\mathrm{sin}\:\left({u}−{v}\right)={p} \\$$$$\Rightarrow\mathrm{sin}\:{u}\mathrm{cos}\:{v}=\frac{{p}}{\mathrm{2}}\:\:\:…\left({I}\right) \\$$$$\mathrm{cos}\:\left({u}+{v}\right)+\mathrm{cos}\:\left({u}−{v}\right)={q} \\$$$$\Rightarrow\mathrm{cos}\:{u}\mathrm{cos}\:{v}=\frac{{q}}{\mathrm{2}}\:\:\:…\left({II}\right) \\$$$$\left({I}\right)\boldsymbol{\div}\left({II}\right): \\$$$$\mathrm{tan}\:{u}=\frac{{p}}{{q}} \\$$$$\Rightarrow{u}={m}\pi+\mathrm{tan}^{−\mathrm{1}} \frac{{p}}{{q}} \\$$$$\left({I}\right)^{\mathrm{2}} +\left({II}\right)^{\mathrm{2}} : \\$$$$\mathrm{cos}^{\mathrm{2}} \:{v}=\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }{\mathrm{4}} \\$$$$\mathrm{cos}\:{v}=\pm\frac{\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}{\mathrm{2}} \\$$$$\Rightarrow{v}={n}\pi\pm\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}{\mathrm{2}} \\$$$$\Rightarrow{A}={u}+{v}=\left({m}+{n}\right)\pi+\mathrm{tan}^{−\mathrm{1}} \frac{{p}}{{q}}\pm\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}{\mathrm{2}} \\$$$$\Rightarrow{B}={u}−{v}=\left({m}−{n}\right)\pi+\mathrm{tan}^{−\mathrm{1}} \frac{{p}}{{q}}\mp\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}{\mathrm{2}} \\$$
Answered by bobhans last updated on 05/Mar/21
$$\Rightarrow\mathrm{2sin}\:\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\right)=\frac{\mathrm{7}}{\mathrm{10}} \\$$$$\Rightarrow\mathrm{2cos}\:\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\right)=\frac{\mathrm{8}}{\mathrm{10}} \\$$$$\left(\mathrm{i}\right)\:\mathrm{tan}\:\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right)=\:\frac{\mathrm{7}}{\mathrm{8}};\:\mathrm{A}+\mathrm{B}=\mathrm{82}.\mathrm{24}°+\mathrm{2k}\pi \\$$$$\left(\mathrm{ii}\right)\mathrm{2}+\mathrm{2cos}\:\left(\mathrm{A}−\mathrm{B}\right)=\:\frac{\mathrm{113}}{\mathrm{100}} \\$$$$\Rightarrow\mathrm{cos}\:\left(\mathrm{A}−\mathrm{B}\right)=\:−\frac{\mathrm{87}}{\mathrm{200}}\:;\:\mathrm{A}−\mathrm{B}\:=\:\mathrm{115}.\mathrm{78}°+\mathrm{2k}\pi \\$$$$\mathrm{Thus}\:\begin{cases}{\mathrm{A}=\mathrm{99}.\mathrm{01}°+\mathrm{2k}\pi}\\{\mathrm{B}=−\mathrm{16}.\mathrm{77}°+\mathrm{2k}\pi}\end{cases} \\$$$$\\$$
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# Given-sin-A-sin-B-0-7-cos-A-cos-B-0-8-find-the-value-of-A-and-B-. Question Number 134581 by EDWIN88 last updated on 05/Mar/21. $$\mathrm{Given}\:\begin{cases}{\mathrm{sin}\:\mathrm{A}+\mathrm{sin}\:\mathrm{B}=\mathrm{0}.\mathrm{7}}\\{\mathrm{cos}\:\mathrm{A}+\mathrm{cos}\:\mathrm{B}=\mathrm{0}.\mathrm{8}}\end{cases} \\$$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}. \\$$. Answered by liberty last updated on 05/Mar/21. $$\Leftrightarrow\:\mathrm{2}+\mathrm{2cos}\:\left(\mathrm{A}−\mathrm{B}\right)=\:\frac{\mathrm{49}+\mathrm{64}}{\mathrm{100}} \\$$$$\Leftrightarrow\mathrm{2cos}\:\left(\mathrm{A}−\mathrm{B}\right)\:=\:−\frac{\mathrm{87}}{\mathrm{100}} \\$$$$\Leftrightarrow\:\mathrm{cos}\:\left(\mathrm{A}−\mathrm{B}\right)\:=−\frac{\mathrm{87}}{\mathrm{200}} \\$$$$\Leftrightarrow\:\mathrm{A}−\mathrm{B}\:\approx\:\mathrm{115}.\mathrm{8}°+\mathrm{k}.\mathrm{360}° \\$$$$\Leftrightarrow\:\mathrm{A}\:\approx\:\mathrm{B}+\mathrm{115}.\mathrm{8}°+\mathrm{k}.\mathrm{360}°\:,\mathrm{k}\epsilon\mathbb{Z} \\$$$$\\$$.
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Answered by mr W last updated on 05/Mar/21. $$\mathrm{sin}\:{A}+\mathrm{sin}\:{B}={p}\:\:\:\left(\mathrm{0}.\mathrm{7}\right) \\$$$$\mathrm{cos}\:{A}+\mathrm{cos}\:{B}={q}\:\:\:\left(\mathrm{0}.\mathrm{8}\right) \\$$$${A}=\frac{{A}+{B}}{\mathrm{2}}+\frac{{A}−{B}}{\mathrm{2}}={u}+{v} \\$$$${B}=\frac{{A}+{B}}{\mathrm{2}}−\frac{{A}−{B}}{\mathrm{2}}={u}−{v} \\$$$$\mathrm{sin}\:\left({u}+{v}\right)+\mathrm{sin}\:\left({u}−{v}\right)={p} \\$$$$\Rightarrow\mathrm{sin}\:{u}\mathrm{cos}\:{v}=\frac{{p}}{\mathrm{2}}\:\:\:…\left({I}\right) \\$$$$\mathrm{cos}\:\left({u}+{v}\right)+\mathrm{cos}\:\left({u}−{v}\right)={q} \\$$$$\Rightarrow\mathrm{cos}\:{u}\mathrm{cos}\:{v}=\frac{{q}}{\mathrm{2}}\:\:\:…\left({II}\right) \\$$$$\left({I}\right)\boldsymbol{\div}\left({II}\right): \\$$$$\mathrm{tan}\:{u}=\frac{{p}}{{q}} \\$$$$\Rightarrow{u}={m}\pi+\mathrm{tan}^{−\mathrm{1}} \frac{{p}}{{q}} \\$$$$\left({I}\right)^{\mathrm{2}} +\left({II}\right)^{\mathrm{2}} : \\$$$$\mathrm{cos}^{\mathrm{2}} \:{v}=\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }{\mathrm{4}} \\$$$$\mathrm{cos}\:{v}=\pm\frac{\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}{\mathrm{2}} \\$$$$\Rightarrow{v}={n}\pi\pm\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}{\mathrm{2}} \\$$$$\Rightarrow{A}={u}+{v}=\left({m}+{n}\right)\pi+\mathrm{tan}^{−\mathrm{1}} \frac{{p}}{{q}}\pm\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}{\mathrm{2}} \\$$$$\Rightarrow{B}={u}−{v}=\left({m}−{n}\right)\pi+\mathrm{tan}^{−\mathrm{1}} \frac{{p}}{{q}}\mp\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}{\mathrm{2}} \\$$. Answered by bobhans last updated on 05/Mar/21. $$\Rightarrow\mathrm{2sin}\:\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\right)=\frac{\mathrm{7}}{\mathrm{10}} \\$$$$\Rightarrow\mathrm{2cos}\:\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\right)=\frac{\mathrm{8}}{\mathrm{10}} \\$$$$\left(\mathrm{i}\right)\:\mathrm{tan}\:\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right)=\:\frac{\mathrm{7}}{\mathrm{8}};\:\mathrm{A}+\mathrm{B}=\mathrm{82}.\mathrm{24}°+\mathrm{2k}\pi \\$$$$\left(\mathrm{ii}\right)\mathrm{2}+\mathrm{2cos}\:\left(\mathrm{A}−\mathrm{B}\right)=\:\frac{\mathrm{113}}{\mathrm{100}} \\$$$$\Rightarrow\mathrm{cos}\:\left(\mathrm{A}−\mathrm{B}\right)=\:−\frac{\mathrm{87}}{\mathrm{200}}\:;\:\mathrm{A}−\mathrm{B}\:=\:\mathrm{115}.\mathrm{78}°+\mathrm{2k}\pi \\$$$$\mathrm{Thus}\:\begin{cases}{\mathrm{A}=\mathrm{99}.\mathrm{01}°+\mathrm{2k}\pi}\\{\mathrm{B}=−\mathrm{16}.\mathrm{77}°+\mathrm{2k}\pi}\end{cases} \\$$$$\\$$.
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https://physics.stackexchange.com/questions/469783/calculating-magnetic-strength-required-to-hold-an-objects-weight-wind-speed
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# Calculating magnetic strength required to hold an objects weight + wind speed?
Note: I'm a bit new to this subject, so I'm looking for some general answers, ballparks, and direction to continue my research.
Scenario: I have three electromagnets capable of X amount of holding force attached to a surface. There's a small metal object that weighs 10lbs that will be attached to the magnets. The object is rectangular 36"x5"x5". Wind will be blowing and/or the surface will be moving at a particular speed.
Question: How do I go about solving the strength my magnets need to be in order to safely/securely hold an object of a certain weight with windy conditions?
I started researching about MPH to PSF conversions where it reads that 100mph translates to 25.60psf.
• Is that accurate, and if so, is that additional force applied to my object?
• Since the object is 36" (3 feet), would that be 25.60psf mean: 10lbs over all / 3 feet = 3.334lbs per foot + 25.60 ... So 28.934lbs of force on each magnet? Is that in the ballpark or am I making this up?
I'm assuming that lift may play a part in this as well, or is that in some ways being accommodated by this MPH->PSF conversion?
I imagine the shape of the object would play a role with regards to lift, so would a rectangular simple object as stated have a minimal, near insignificant, amount of lift for this discussion? Again, I'm not exactly sure what I'm talking about here, so please correct me.
Assuming I can figure out the holding force required per magnet per object per conditions here... is there a safe amount/rule of thumb to add to account for unexpected shocks or gusts? Would a steady wind of 70mph be different than 20mph with a 70mph gust?
Sorry for all the questions. It's something I'm trying to read more about and understand better. Again, it's a bit of an open ended question because I'm trying to learn about it/understand it and don't actually have these materials yet.
There are two questions being asked:
1.) How do I calculate how much force my electromagnets can exert
and
2.) How do I calculate the aerodynamic forces acting on an object that will be blowing in the wind, so that I can know how strong my electromagnet needs to be.
Just for future reference, the force exerted by an electromagnet is: $$F = \frac{\mu^2 N^2 I^2 A}{2\mu_0 L^2}$$ where $$N$$ is the number of times your wire wraps around your electromagnet, $$I$$ is the electric current going through it, $$A$$ is the cross sectional area of your electromagnet, $$L$$ is the length of the electromagnet. $$\mu$$ is the magnetic permeability of whatever your electromagnet core is made of, and $$\mu_0$$ is the magnetic permeability of free space. This equation appears about half way down the Wikipedia article titled Electromagnet.
A couple of problems with this are that most electromagnets sold commercially do not state the exact number of turns $$N$$. Any number that they place on the packaging (if one is stated at all) is likely a rough estimate. Furthermore, most electromagnets sold commercially may state something like "iron core" on the packaging, but that doesn't mean it's made of pure iron. Hence, it is difficult to know for a fact exactly what the core is made of. This makes is more-or-less impossible to know $$\mu$$, the magnetic permeability of the material from which the core is made.
The answer to question 2 goes like this. In principle, the aerodynamic forces acting on your "object" might could be calculated. But I guarantee you that this is beyond the scope of your project. Such a feat would require years of work by a team of engineers working full time. This isn't the direction you want to go.
In all seriousness, the best approach I see is to just trial and error your way through whatever it is you are designing.
Lastly, while doing so, bear in mind that many electromagnets sold in science kits are not very powerful at all. They are wimpy little things that barely exert any force on anything - it require a delicate touch to determine that they are exerting any force on anything at all. Be sure that you get something stronger than this. The strongest electromagnet I've ever encountered outside of a laboratory was the electromagnet that held the back door security gate closed at a restaurant I worked at many years ago. These electromagnets must be professionally installed and cost many thousands of dollars. Not to mention the fact that they can be quite dangerous in a wide variety of bizarre ways.
Best of luck with your project.
• Thanks for the comment. Can't agree with the "most electromagnets are not very powerful at all" statement though. There are videos on YouTube showing an electromagnet made from a microwave transformer very easily lifting +200lbs. I consider that to be powerful, and it costs about \$30 on eBay to make. Apr 1, 2019 at 14:55
• Thanks Matt - I edited my post just a bit. I really should say that if you need a really strong one, you need to make your own or buy an industrial use one. Apr 4, 2019 at 0:40
• The formula you quote is, I believe, for the force exerted on a detachable piece of a closed-loop core by the rest of the core (the part that carries the coil). Jun 4, 2023 at 13:52
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# Calculating magnetic strength required to hold an objects weight + wind speed?. Note: I'm a bit new to this subject, so I'm looking for some general answers, ballparks, and direction to continue my research.. Scenario: I have three electromagnets capable of X amount of holding force attached to a surface. There's a small metal object that weighs 10lbs that will be attached to the magnets. The object is rectangular 36"x5"x5". Wind will be blowing and/or the surface will be moving at a particular speed.. Question: How do I go about solving the strength my magnets need to be in order to safely/securely hold an object of a certain weight with windy conditions?. I started researching about MPH to PSF conversions where it reads that 100mph translates to 25.60psf.. • Is that accurate, and if so, is that additional force applied to my object?. • Since the object is 36" (3 feet), would that be 25.60psf mean: 10lbs over all / 3 feet = 3.334lbs per foot + 25.60 ... So 28.934lbs of force on each magnet? Is that in the ballpark or am I making this up?. I'm assuming that lift may play a part in this as well, or is that in some ways being accommodated by this MPH->PSF conversion?. I imagine the shape of the object would play a role with regards to lift, so would a rectangular simple object as stated have a minimal, near insignificant, amount of lift for this discussion? Again, I'm not exactly sure what I'm talking about here, so please correct me.. Assuming I can figure out the holding force required per magnet per object per conditions here... is there a safe amount/rule of thumb to add to account for unexpected shocks or gusts? Would a steady wind of 70mph be different than 20mph with a 70mph gust?. Sorry for all the questions. It's something I'm trying to read more about and understand better. Again, it's a bit of an open ended question because I'm trying to learn about it/understand it and don't actually have these materials yet.. There are two questions being asked:. 1.) How do I calculate how much force my electromagnets can exert. and. 2.) How do I calculate the aerodynamic forces acting on an object that will be blowing in the wind, so that I can know how strong my electromagnet needs to be.. Just for future reference, the force exerted by an electromagnet is: $$F = \frac{\mu^2 N^2 I^2 A}{2\mu_0 L^2}$$ where $$N$$ is the number of times your wire wraps around your electromagnet, $$I$$ is the electric current going through it, $$A$$ is the cross sectional area of your electromagnet, $$L$$ is the length of the electromagnet. $$\mu$$ is the magnetic permeability of whatever your electromagnet core is made of, and $$\mu_0$$ is the magnetic permeability of free space. This equation appears about half way down the Wikipedia article titled Electromagnet.. A couple of problems with this are that most electromagnets sold commercially do not state the exact number of turns $$N$$. Any number that they place on the packaging (if one is stated at all) is likely a rough estimate. Furthermore, most electromagnets sold commercially may state something like "iron core" on the packaging, but that doesn't mean it's made of pure iron. Hence, it is difficult to know for a fact exactly what the core is made of.
|
This makes is more-or-less impossible to know $$\mu$$, the magnetic permeability of the material from which the core is made.. The answer to question 2 goes like this. In principle, the aerodynamic forces acting on your "object" might could be calculated. But I guarantee you that this is beyond the scope of your project. Such a feat would require years of work by a team of engineers working full time. This isn't the direction you want to go.. In all seriousness, the best approach I see is to just trial and error your way through whatever it is you are designing.. Lastly, while doing so, bear in mind that many electromagnets sold in science kits are not very powerful at all. They are wimpy little things that barely exert any force on anything - it require a delicate touch to determine that they are exerting any force on anything at all. Be sure that you get something stronger than this. The strongest electromagnet I've ever encountered outside of a laboratory was the electromagnet that held the back door security gate closed at a restaurant I worked at many years ago. These electromagnets must be professionally installed and cost many thousands of dollars. Not to mention the fact that they can be quite dangerous in a wide variety of bizarre ways.. Best of luck with your project.. • Thanks for the comment. Can't agree with the "most electromagnets are not very powerful at all" statement though. There are videos on YouTube showing an electromagnet made from a microwave transformer very easily lifting +200lbs. I consider that to be powerful, and it costs about \$30 on eBay to make. Apr 1, 2019 at 14:55. • Thanks Matt - I edited my post just a bit. I really should say that if you need a really strong one, you need to make your own or buy an industrial use one. Apr 4, 2019 at 0:40. • The formula you quote is, I believe, for the force exerted on a detachable piece of a closed-loop core by the rest of the core (the part that carries the coil). Jun 4, 2023 at 13:52.
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https://www.physicsforums.com/threads/how-does-an-explosion-affect-the-center-of-mass-in-projectile-motion.82307/
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How Does an Explosion Affect the Center of Mass in Projectile Motion?
• pez
In summary: Therefore, in summary, the distance x_cm from the mortar to the center of mass of the exploded pieces is equal to sqrt[(4/5)r^2 + d^2]. This is true regardless of the value of r, as long as the conditions given in the problem hold (i.e. negligible air resistance and explosive charge). I hope this helps with your homework problem. Best of luck!
pez
Hi, I was hoping to gather some feedback for one of my homework problems. Any help would be much appreciated. The problem is as follows:
A mortar fires a shell of mass m at speed v_0. The shell explodes at the top of its trajectory as designed. However, rather than creating a shower of colored flares, it breaks into just two pieces, a smaller piece of mass (1/5)m and a larger piece of mass (4/5)m. Both pieces land at exactly the same time. The smaller piece lands perilously close to the mortar (at a distance of zero from the mortar). The larger piece lands a distance d from the mortar. If there had been no explosion, the shell would have landed a distance r from the mortar. Assume that air resistance and the mass of the shell's explosive charge are negligible.
"The two exploded pieces of the shell land at the same time. At the moment of landing, what is the distance x_cm from the mortar to the center of mass of the exploded pieces?"
We're supposed to express it in terms of r. I had no problem w/ the second part of the problem, which wants us to express x_cm in terms of D. That was merely (4/5*m*d)/m, since the other component of the center of mass had a position of 0. I'm stumped on how I can approach the problem expressing it in terms of r, however. The hint given was that "The explosion only exerts internal forces on the particles. The only external force acting on the two-piece system is gravity, so the center of mass will continue along the original trajectory of the shell," which I didn't really find helpful. The position of the first piece would still be 0, wouldn't it? So wouldn't that just leave us with (4/5)*r/m ? This, however, is incorrect.
Edit: After thinking about it some more, would x_cm merely be r?
Edit2: Sweet, it is. Thanks for looking.
Last edited:
Hi there,
I can provide some insight into this problem. First, let's define some variables for clarity:
- m = mass of the shell before explosion
- v_0 = initial speed of the shell
- (1/5)m = mass of the smaller piece after explosion
- (4/5)m = mass of the larger piece after explosion
- r = distance from the mortar to where the shell would have landed without explosion
- d = distance from the mortar to where the larger piece lands after explosion
- x_cm = distance from the mortar to the center of mass of the exploded pieces
Now, let's consider the problem in terms of conservation of momentum and energy. Before the explosion, the shell has a certain momentum and energy, given by:
- p = mv_0 (momentum)
- E = (1/2)mv_0^2 (energy)
After the explosion, the two pieces will have a combined momentum and energy, given by:
- p' = (1/5)m(0) + (4/5)m(v_0) = (4/5)mv_0 (momentum)
- E' = (1/2)(1/5)m(0)^2 + (1/2)(4/5)m(v_0)^2 = (2/5)mv_0^2 (energy)
We can see that the momentum is conserved, but the energy is not. This is because some of the energy is lost in the explosion. However, we can still use the conservation of momentum to find the velocity of the center of mass of the two pieces after the explosion:
- p' = (4/5)mv_0 = (m')(v_cm) (where m' is the total mass of the two pieces and v_cm is the center of mass velocity)
- Therefore, v_cm = (4/5)v_0
Now, let's consider the motion of the center of mass. We know that it will continue along the original trajectory of the shell, so its horizontal displacement will be equal to r. However, we also know that it will have a vertical displacement equal to d, as the larger piece lands at that distance from the mortar. Using the Pythagorean theorem, we can find the distance x_cm from the mortar to the center of mass:
- x_cm = sqrt(r^2 + d^2)
- Sub
Hi there,
It seems like you have already arrived at the correct answer, which is great! The key concept to understand in this problem is the conservation of momentum. When the shell explodes, the total momentum of the system (shell + explosive charge) remains the same, but it is now distributed between the two pieces.
The hint given about the internal and external forces is trying to convey that the explosion does not change the total momentum of the system, but only redistributes it within the system. This means that the center of mass will still follow the same trajectory as the original shell, which is why the position of the first piece is still 0.
Using this understanding, we can see that the center of mass will be located at a distance r from the mortar, since the total mass of the system is still m and the position of the first piece is 0. Therefore, x_cm = r.
I hope this helps clarify any confusion and good job on solving the problem! Keep up the good work.
What is the center of mass?
The center of mass is the point in a system where the mass is evenly distributed in all directions. It is the average position of all the mass in a given object or system.
Why is the center of mass important?
The center of mass is important because it helps us understand the overall motion and stability of a system. It is also a useful concept in physics and engineering for analyzing the forces acting on an object.
How is the center of mass calculated?
The center of mass can be calculated by finding the weighted average of the positions of all the individual masses in a system. This can be done using a mathematical formula or by physically balancing the system on a pivot point.
Can the center of mass be outside of an object?
Yes, the center of mass can be outside of an object. This can happen when there is a non-uniform distribution of mass in the object, or when there are external forces acting on the system.
What is the difference between center of mass and center of gravity?
The center of mass and center of gravity are often used interchangeably, but there is a subtle difference. The center of mass is the average position of mass in a system, while the center of gravity is the point where the force of gravity can be considered to act on an object. In most cases, the two points will be in the same location, but they may differ in situations where there are non-gravitational forces at play.
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How Does an Explosion Affect the Center of Mass in Projectile Motion?. • pez. In summary: Therefore, in summary, the distance x_cm from the mortar to the center of mass of the exploded pieces is equal to sqrt[(4/5)r^2 + d^2]. This is true regardless of the value of r, as long as the conditions given in the problem hold (i.e. negligible air resistance and explosive charge). I hope this helps with your homework problem. Best of luck!. pez. Hi, I was hoping to gather some feedback for one of my homework problems. Any help would be much appreciated. The problem is as follows:. A mortar fires a shell of mass m at speed v_0. The shell explodes at the top of its trajectory as designed. However, rather than creating a shower of colored flares, it breaks into just two pieces, a smaller piece of mass (1/5)m and a larger piece of mass (4/5)m. Both pieces land at exactly the same time. The smaller piece lands perilously close to the mortar (at a distance of zero from the mortar). The larger piece lands a distance d from the mortar. If there had been no explosion, the shell would have landed a distance r from the mortar. Assume that air resistance and the mass of the shell's explosive charge are negligible.. "The two exploded pieces of the shell land at the same time. At the moment of landing, what is the distance x_cm from the mortar to the center of mass of the exploded pieces?". We're supposed to express it in terms of r. I had no problem w/ the second part of the problem, which wants us to express x_cm in terms of D. That was merely (4/5*m*d)/m, since the other component of the center of mass had a position of 0. I'm stumped on how I can approach the problem expressing it in terms of r, however. The hint given was that "The explosion only exerts internal forces on the particles. The only external force acting on the two-piece system is gravity, so the center of mass will continue along the original trajectory of the shell," which I didn't really find helpful. The position of the first piece would still be 0, wouldn't it? So wouldn't that just leave us with (4/5)*r/m ? This, however, is incorrect.. Edit: After thinking about it some more, would x_cm merely be r?. Edit2: Sweet, it is. Thanks for looking.. Last edited:. Hi there,. I can provide some insight into this problem. First, let's define some variables for clarity:. - m = mass of the shell before explosion. - v_0 = initial speed of the shell. - (1/5)m = mass of the smaller piece after explosion. - (4/5)m = mass of the larger piece after explosion. - r = distance from the mortar to where the shell would have landed without explosion. - d = distance from the mortar to where the larger piece lands after explosion. - x_cm = distance from the mortar to the center of mass of the exploded pieces. Now, let's consider the problem in terms of conservation of momentum and energy. Before the explosion, the shell has a certain momentum and energy, given by:. - p = mv_0 (momentum). - E = (1/2)mv_0^2 (energy). After the explosion, the two pieces will have a combined momentum and energy, given by:. - p' = (1/5)m(0) + (4/5)m(v_0) = (4/5)mv_0 (momentum). - E' = (1/2)(1/5)m(0)^2 + (1/2)(4/5)m(v_0)^2 = (2/5)mv_0^2 (energy). We can see that the momentum is conserved, but the energy is not. This is because some of the energy is lost in the explosion. However, we can still use the conservation of momentum to find the velocity of the center of mass of the two pieces after the explosion:. - p' = (4/5)mv_0 = (m')(v_cm) (where m' is the total mass of the two pieces and v_cm is the center of mass velocity). - Therefore, v_cm = (4/5)v_0. Now, let's consider the motion of the center of mass. We know that it will continue along the original trajectory of the shell, so its horizontal displacement will be equal to r. However, we also know that it will have a vertical displacement equal to d, as the larger piece lands at that distance from the mortar. Using the Pythagorean theorem, we can find the distance x_cm from the mortar to the center of mass:. - x_cm = sqrt(r^2 + d^2). - Sub. Hi there,. It seems like you have already arrived at the correct answer, which is great! The key concept to understand in this problem is the conservation of momentum. When the shell explodes, the total momentum of the system (shell + explosive charge) remains the same, but it is now distributed between the two pieces.. The hint given about the internal and external forces is trying to convey that the explosion does not change the total momentum of the system, but only redistributes it within the system. This means that the center of mass will still follow the same trajectory as the original shell, which is why the position of the first piece is still 0.. Using this understanding, we can see that the center of mass will be located at a distance r from the mortar, since the total mass of the system is still m and the position of the first piece is 0. Therefore, x_cm = r.. I hope this helps clarify any confusion and good job on solving the problem! Keep up the good work.. What is the center of mass?.
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The center of mass is the point in a system where the mass is evenly distributed in all directions. It is the average position of all the mass in a given object or system.. Why is the center of mass important?. The center of mass is important because it helps us understand the overall motion and stability of a system. It is also a useful concept in physics and engineering for analyzing the forces acting on an object.. How is the center of mass calculated?. The center of mass can be calculated by finding the weighted average of the positions of all the individual masses in a system. This can be done using a mathematical formula or by physically balancing the system on a pivot point.. Can the center of mass be outside of an object?. Yes, the center of mass can be outside of an object. This can happen when there is a non-uniform distribution of mass in the object, or when there are external forces acting on the system.. What is the difference between center of mass and center of gravity?. The center of mass and center of gravity are often used interchangeably, but there is a subtle difference. The center of mass is the average position of mass in a system, while the center of gravity is the point where the force of gravity can be considered to act on an object. In most cases, the two points will be in the same location, but they may differ in situations where there are non-gravitational forces at play.. • Introductory Physics Homework Help. Replies. 19. Views. 3K. • Introductory Physics Homework Help. Replies. 1. Views. 3K. • Introductory Physics Homework Help. Replies. 4. Views. 694. • Introductory Physics Homework Help. Replies. 335. Views. 9K. • Introductory Physics Homework Help. Replies. 2. Views. 1K. • Introductory Physics Homework Help. Replies. 13. Views. 1K. • Introductory Physics Homework Help. Replies. 27. Views. 2K. • Introductory Physics Homework Help. Replies. 8. Views. 1K. • Introductory Physics Homework Help. Replies. 2. Views. 854. • Introductory Physics Homework Help. Replies. 23. Views. 473.
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# Homework Help: Oscillations Car carrying four people find how much body rises
1. Nov 23, 2004
A 1175 kg car carrying four 80 kg people travels over a rough "washboard" dirt road with corrugations 4.0 m apart which causes the car to bounce on its spring suspension. The car bounces with maximum amplitude when its speed is 17 km/h. The car now stops, and the four people get out. By how much does the car body rise on its suspension owing to this decrease in weight?
I first tried figuring the natural frequency of the car:
$$\omega_o=(2\pi\upsilon)/(\Delta(x))=(2\pi(17km/h)(1hr/60s)/(.004km)$$ This gives 445.059 rad/s.
I then tried to figure out k:
$$\kappa=(m1+m2)\omega_o^2=((1175 kg+(4*180kg))(445.059rad/s)^2$$ This gives 3.75e8.
Finally, I tried finding $$\Delta(x)=\Delta(F)/\kappa=(m_2*g)/(\kappa)=((180 kg*4)(9.8m/s^2))/(3.75*10^8)$$This gives 1.8e-5 m, or .00188 cm, this is wrong, where did I go wrong? Help!
PS This is the first time I use Latex so if it looks odd I'm sorry!
2. Nov 23, 2004
Help...Anyone lol...
3. Nov 27, 2004
### vtech
your conversion for the first part seems to be off... recheck it... and make sure you convert to meter seconds... but you're on the right track
4. Nov 30, 2011
### mmayorga
i just solved my homework problem with your equations and it worked. the only problem with your method is your unit conversions: to go from km/h to m/s you multiply by 1000/(60^2)
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# Homework Help: Oscillations Car carrying four people find how much body rises. 1. Nov 23, 2004. A 1175 kg car carrying four 80 kg people travels over a rough "washboard" dirt road with corrugations 4.0 m apart which causes the car to bounce on its spring suspension. The car bounces with maximum amplitude when its speed is 17 km/h. The car now stops, and the four people get out. By how much does the car body rise on its suspension owing to this decrease in weight?. I first tried figuring the natural frequency of the car:. $$\omega_o=(2\pi\upsilon)/(\Delta(x))=(2\pi(17km/h)(1hr/60s)/(.004km)$$ This gives 445.059 rad/s.. I then tried to figure out k:. $$\kappa=(m1+m2)\omega_o^2=((1175 kg+(4*180kg))(445.059rad/s)^2$$ This gives 3.75e8.. Finally, I tried finding $$\Delta(x)=\Delta(F)/\kappa=(m_2*g)/(\kappa)=((180 kg*4)(9.8m/s^2))/(3.75*10^8)$$This gives 1.8e-5 m, or .00188 cm, this is wrong, where did I go wrong? Help!. PS This is the first time I use Latex so if it looks odd I'm sorry!.
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2. Nov 23, 2004. Help...Anyone lol.... 3. Nov 27, 2004. ### vtech. your conversion for the first part seems to be off... recheck it... and make sure you convert to meter seconds... but you're on the right track. 4. Nov 30, 2011. ### mmayorga. i just solved my homework problem with your equations and it worked. the only problem with your method is your unit conversions: to go from km/h to m/s you multiply by 1000/(60^2).
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+0
# 1 over 9 to the power of 3 over 2
+1
344
1
1 over 9 to the power of 3 over 2
Guest Apr 7, 2017
#1
+7324
+2
$$(\frac19)^{\frac32}=\sqrt{\frac19^3}=\sqrt{\frac19\cdot\frac19\cdot\frac19}=\frac19\sqrt{\frac19}=\frac19\cdot\frac{\sqrt1}{\sqrt9}=\frac{1}{9}\cdot\frac13\mathbf{=\frac{1}{27}}$$
hectictar Apr 7, 2017
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+0. # 1 over 9 to the power of 3 over 2. +1. 344. 1. 1 over 9 to the power of 3 over 2. Guest Apr 7, 2017.
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#1. +7324. +2. $$(\frac19)^{\frac32}=\sqrt{\frac19^3}=\sqrt{\frac19\cdot\frac19\cdot\frac19}=\frac19\sqrt{\frac19}=\frac19\cdot\frac{\sqrt1}{\sqrt9}=\frac{1}{9}\cdot\frac13\mathbf{=\frac{1}{27}}$$. hectictar Apr 7, 2017.
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# Integration in a Polynomial for a given value
Given string str which represents a polynomial and a number N, the task is to integrate this polynomial with respect to X at the given value N
Examples:
Input: str = “90x4 + 24x3 + 18x2 + 18x”, N = 1.
Output: 39
Explanation:
Given, dy/dx = 90*(X4) + 24* (X3) + 18* (X2) + 18*(X). On integrating this equation, we get 18*(X5) + 6*(X4) + 6*(X3) + 9*(X2) and substituting the value X = 1, we get:
18 + 6 + 6 + 9 = 39.
Input: str = “4x3 + 2x2 + 3x”, N = 2
Output: 27
Approach: The idea is to use the identity of the integration. For some given function X with the power of N, the integration of this term is given by:
Therefore, the following steps are followed to compute the answer:
• Get the string.
• Split the string and perform the integration based on the above formula.
• Substitute the value of N in the obtained expression.
• Add all the individual values to get the final integral value.
Below is the implementation of the above approach:
## C++
// C++ program to find the integration// of the given polynomial for the// value N #include using namespace std;typedef long long ll; // Function to return the integral// of the given termdouble inteTerm(string pTerm, ll val){ // Get the coefficient string coeffStr = ""; int i; // Loop to iterate through the string // and get the coefficient for (i = 0; pTerm[i] != 'x'; i++) coeffStr.push_back(pTerm[i]); ll coeff = atol(coeffStr.c_str()); // Get the Power string powStr = ""; // Loop to skip 2 characters for x and ^ for (i = i + 2; i != pTerm.size(); i++) powStr.push_back(pTerm[i]); ll power = atol(powStr.c_str()); // Return the computed integral return (coeff * pow(val, power + 1)) / (power + 1);} // Functionto find the integration// of the given polynomial for the// value Ndouble integrationVal(string poly, int val){ ll ans = 0; // Using string stream to get the // input in tokens istringstream is(poly); string pTerm; while (is >> pTerm) { // If the token is equal to '+' then // continue with the string if (pTerm == "+") continue; // Otherwise find the integration // of that particular term else ans = (ans + inteTerm(pTerm, val)); } return ans;} // Driver codeint main(){ string str = "4x^3 + 3x^1 + 2x^2"; int val = 2; cout << integrationVal(str, val); return 0;}
## Java
// Java program for the above approachpublic class GFG{ // Function to return the integral // of the given term static int inteTerm(String pTerm, int val) { // Get the coefficient String coeffStr = ""; // Loop to iterate through // the string and get the // coefficient int i = 0; while (i < pTerm.length() && pTerm.charAt(i) != 'x') { coeffStr += pTerm.charAt(i); i += 1; } int coeff = Integer.parseInt(coeffStr); // Get the Power String powStr = ""; // Loop to skip 2 characters // for x and ^ int j = i + 2; while(j< pTerm.length()) { powStr += (pTerm.charAt(j)); j += 1; } int power = Integer.parseInt(powStr); // Return the computed integral return ((coeff * (int)Math.pow(val, power + 1)) / (power + 1)); } // Functionto find the integration // of the given polynomial for the // value N static int integrationVal(String poly, int val) { int ans = 0; // Using string stream to // get the input in tokens String[] stSplit = poly.split(" \\+ "); int i = 0; while(i < stSplit.length) { ans = (ans + inteTerm(stSplit[i], val)); i += 1; } return ans; } // Driver code public static void main(String[] args) { String st = "4x^3 + 3x^1 + 2x^2"; int val = 2; System.out.println(integrationVal(st, val)); }} // This code is contributed by divyesh072019.
## Python3
# Python3 program to find # the integration of the # given polynomial for the# value N # Function to return the integral# of the given termdef inteTerm(pTerm, val): # Get the coefficient coeffStr = "" # Loop to iterate through # the string and get the # coefficient i = 0 while (i < len(pTerm) and pTerm[i] != 'x'): coeffStr += pTerm[i] i += 1 coeff = int(coeffStr) # Get the Power powStr = "" # Loop to skip 2 characters # for x and ^ j = i + 2 while j< len(pTerm): powStr += (pTerm[j]) j += 1 power = int(powStr) # Return the computed integral return ((coeff * pow(val, power + 1)) // (power + 1)) # Functionto find the integration# of the given polynomial for the# value Ndef integrationVal(poly, val): ans = 0 # Using string stream to # get the input in tokens stSplit = poly.split("+") i = 0 while i < len(stSplit): ans = (ans + inteTerm(stSplit[i], val)) i += 1 return ans # Driver codeif __name__ == "__main__": st = "4x^3 + 3x^1 + 2x^2" val = 2 print(integrationVal(st, val)) # This code is contributed by Chitranayal
## C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG { // Function to return the integral // of the given term static int inteTerm(string pTerm, int val) { // Get the coefficient string coeffStr = ""; // Loop to iterate through // the string and get the // coefficient int i = 0; while (i < pTerm.Length && pTerm[i] != 'x') { coeffStr += pTerm[i]; i += 1; } int coeff = Convert.ToInt32(coeffStr); // Get the Power string powStr = ""; // Loop to skip 2 characters // for x and ^ int j = i + 2; while(j< pTerm.Length) { powStr += (pTerm[j]); j += 1; } int power = Convert.ToInt32(powStr); // Return the computed integral return ((coeff * (int)Math.Pow(val, power + 1)) / (power + 1)); } // Functionto find the integration // of the given polynomial for the // value N static int integrationVal(string poly, int val) { int ans = 0; // Using string stream to // get the input in tokens string[] stSplit = poly.Split('+'); int i = 0; while(i < stSplit.Length) { ans = (ans + inteTerm(stSplit[i], val)); i += 1; } return ans; } // Driver code static void Main() { string st = "4x^3 + 3x^1 + 2x^2"; int val = 2; Console.WriteLine(integrationVal(st, val)); }} // This code is contributed by divyeshrabadiya07.
## Javascript
Output:
27
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# Integration in a Polynomial for a given value. Given string str which represents a polynomial and a number N, the task is to integrate this polynomial with respect to X at the given value N. Examples:. Input: str = “90x4 + 24x3 + 18x2 + 18x”, N = 1.. Output: 39. Explanation:. Given, dy/dx = 90*(X4) + 24* (X3) + 18* (X2) + 18*(X). On integrating this equation, we get 18*(X5) + 6*(X4) + 6*(X3) + 9*(X2) and substituting the value X = 1, we get:. 18 + 6 + 6 + 9 = 39.. Input: str = “4x3 + 2x2 + 3x”, N = 2. Output: 27. Approach: The idea is to use the identity of the integration. For some given function X with the power of N, the integration of this term is given by:. Therefore, the following steps are followed to compute the answer:. • Get the string.. • Split the string and perform the integration based on the above formula.. • Substitute the value of N in the obtained expression.
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• Add all the individual values to get the final integral value.. Below is the implementation of the above approach:. ## C++. // C++ program to find the integration// of the given polynomial for the// value N #include using namespace std;typedef long long ll; // Function to return the integral// of the given termdouble inteTerm(string pTerm, ll val){ // Get the coefficient string coeffStr = ""; int i; // Loop to iterate through the string // and get the coefficient for (i = 0; pTerm[i] != 'x'; i++) coeffStr.push_back(pTerm[i]); ll coeff = atol(coeffStr.c_str()); // Get the Power string powStr = ""; // Loop to skip 2 characters for x and ^ for (i = i + 2; i != pTerm.size(); i++) powStr.push_back(pTerm[i]); ll power = atol(powStr.c_str()); // Return the computed integral return (coeff * pow(val, power + 1)) / (power + 1);} // Functionto find the integration// of the given polynomial for the// value Ndouble integrationVal(string poly, int val){ ll ans = 0; // Using string stream to get the // input in tokens istringstream is(poly); string pTerm; while (is >> pTerm) { // If the token is equal to '+' then // continue with the string if (pTerm == "+") continue; // Otherwise find the integration // of that particular term else ans = (ans + inteTerm(pTerm, val)); } return ans;} // Driver codeint main(){ string str = "4x^3 + 3x^1 + 2x^2"; int val = 2; cout << integrationVal(str, val); return 0;}. ## Java. // Java program for the above approachpublic class GFG{ // Function to return the integral // of the given term static int inteTerm(String pTerm, int val) { // Get the coefficient String coeffStr = ""; // Loop to iterate through // the string and get the // coefficient int i = 0; while (i < pTerm.length() && pTerm.charAt(i) != 'x') { coeffStr += pTerm.charAt(i); i += 1; } int coeff = Integer.parseInt(coeffStr); // Get the Power String powStr = ""; // Loop to skip 2 characters // for x and ^ int j = i + 2; while(j< pTerm.length()) { powStr += (pTerm.charAt(j)); j += 1; } int power = Integer.parseInt(powStr); // Return the computed integral return ((coeff * (int)Math.pow(val, power + 1)) / (power + 1)); } // Functionto find the integration // of the given polynomial for the // value N static int integrationVal(String poly, int val) { int ans = 0; // Using string stream to // get the input in tokens String[] stSplit = poly.split(" \\+ "); int i = 0; while(i < stSplit.length) { ans = (ans + inteTerm(stSplit[i], val)); i += 1; } return ans; } // Driver code public static void main(String[] args) { String st = "4x^3 + 3x^1 + 2x^2"; int val = 2; System.out.println(integrationVal(st, val)); }} // This code is contributed by divyesh072019.. ## Python3. # Python3 program to find # the integration of the # given polynomial for the# value N # Function to return the integral# of the given termdef inteTerm(pTerm, val): # Get the coefficient coeffStr = "" # Loop to iterate through # the string and get the # coefficient i = 0 while (i < len(pTerm) and pTerm[i] != 'x'): coeffStr += pTerm[i] i += 1 coeff = int(coeffStr) # Get the Power powStr = "" # Loop to skip 2 characters # for x and ^ j = i + 2 while j< len(pTerm): powStr += (pTerm[j]) j += 1 power = int(powStr) # Return the computed integral return ((coeff * pow(val, power + 1)) // (power + 1)) # Functionto find the integration# of the given polynomial for the# value Ndef integrationVal(poly, val): ans = 0 # Using string stream to # get the input in tokens stSplit = poly.split("+") i = 0 while i < len(stSplit): ans = (ans + inteTerm(stSplit[i], val)) i += 1 return ans # Driver codeif __name__ == "__main__": st = "4x^3 + 3x^1 + 2x^2" val = 2 print(integrationVal(st, val)) # This code is contributed by Chitranayal. ## C#. // C# program for the above approachusing System;using System.Collections.Generic;class GFG { // Function to return the integral // of the given term static int inteTerm(string pTerm, int val) { // Get the coefficient string coeffStr = ""; // Loop to iterate through // the string and get the // coefficient int i = 0; while (i < pTerm.Length && pTerm[i] != 'x') { coeffStr += pTerm[i]; i += 1; } int coeff = Convert.ToInt32(coeffStr); // Get the Power string powStr = ""; // Loop to skip 2 characters // for x and ^ int j = i + 2; while(j< pTerm.Length) { powStr += (pTerm[j]); j += 1; } int power = Convert.ToInt32(powStr); // Return the computed integral return ((coeff * (int)Math.Pow(val, power + 1)) / (power + 1)); } // Functionto find the integration // of the given polynomial for the // value N static int integrationVal(string poly, int val) { int ans = 0; // Using string stream to // get the input in tokens string[] stSplit = poly.Split('+'); int i = 0; while(i < stSplit.Length) { ans = (ans + inteTerm(stSplit[i], val)); i += 1; } return ans; } // Driver code static void Main() { string st = "4x^3 + 3x^1 + 2x^2"; int val = 2; Console.WriteLine(integrationVal(st, val)); }} // This code is contributed by divyeshrabadiya07.. ## Javascript. Output:. 27. Previous. Next.
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## Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (include reading these from a table).. 94 Lesson(s). ### Writing Linear Equations (Day 1 of 2). Algebra I » Unit: Linear & Absolute Value Functions. Algebra I » Unit: Linear & Absolute Value Functions. Washington, DC. Environment: Urban. Big Idea:. Standards:. Favorites (32). Resources (19). Reflections (2). ### Linear Functions. Algebra I » Unit: Linear Functions. Algebra I » Unit: Linear Functions. Amherst, NY. Environment: Suburban. Big Idea:. Standards:. Favorites (10). Resources (16). ### Investigating Linear and Nonlinear Tile Patterns. 12th Grade Math » Unit: Linear and Nonlinear Functions. 12th Grade Math » Unit: Linear and Nonlinear Functions. Oakland, CA. Environment: Urban. Big Idea:. Standards:. Favorites (9). Resources (16). Reflections (1). ### The Skyscraper Problem. 12th Grade Math » Unit: Sequences and Series. 12th Grade Math » Unit: Sequences and Series. Troy, MI. Environment: Suburban. Big Idea:. Standards:. Favorites (8). Resources (14). Reflections (1). ### Explore the Rebound Height of A Ball. Algebra I » Unit: Exponential Functions. Algebra I » Unit: Exponential Functions. Rogers, AR. Environment: Rural. Big Idea:. Standards:. Favorites (10). Resources (12). ### Logs, Loans, and Life Lessons!. Algebra II » Unit: Exponential and Logarithmic Functions. Algebra II » Unit: Exponential and Logarithmic Functions. Huntington, IN. Environment: Suburban. Big Idea:. Standards:. Favorites (4). Resources (16). ### Organizing a List and Guess & Check. Algebra I » Unit: Systems of Equations. Algebra I » Unit: Systems of Equations. Worcester, MA. Environment: Urban. Big Idea:. #### In order for guessing and checking to be useful and efficient, it’s important to have an idea of how to organize a list of possibilities.. Standards:. Favorites (6). Resources (31). Reflections (1). Algebra I » Unit: Linear Functions. Rogers, AR.
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Environment: Rural. Big Idea:. #### Students reason about Perpendicular Lines with simple concrete examples using Geoboards and extend their reasoning to designing parking lines.. Standards:. Favorites (2). Resources (17). Reflections (1). Algebra I » Unit: Exponential Functions. Salem, MA. Environment: Urban. Big Idea:. #### Students compare linear and exponential functions, and, learn about actual and theoretical data.. Standards:. Favorites (12). Resources (28). Reflections (2). Algebra I » Unit: Graphing Linear Functions. Washington, DC. Environment: Urban. Big Idea:. Standards:. Favorites (29). Resources (17). Reflections (2). ### Arithmetic vs. Geometric Sequences. Algebra I » Unit: Exponential Functions. Algebra I » Unit: Exponential Functions. Rogers, AR. Environment: Rural. Big Idea:. Standards:. Favorites (9). Resources (11). ### "Demystifying e" Day #1. Algebra II » Unit: Exponential and Logarithmic Functions. Algebra II » Unit: Exponential and Logarithmic Functions. Huntington, IN. Environment: Suburban. Big Idea:. Standards:. Favorites (4). Resources (12). ### Sequences, Spreadsheets, and Graphs. Algebra I » Unit: Linear and Exponential Functions. Algebra I » Unit: Linear and Exponential Functions. Worcester, MA. Environment: Urban. Big Idea:. #### Writing formulas in a spreadsheet lays foundations for writing recursive rules for arithmetic and geometric sequences.. Standards:. Favorites (1). Resources (18). Reflections (2). Algebra I » Unit: Linear Functions. Rogers, AR. Environment: Rural. Big Idea:. #### For students to write the equation of a line parallel to or perpendicular to a line and passing through a given point using different representations of linear functions.. Standards:. Favorites (1). Resources (12). 12th Grade Math » Unit: Sequences and Series. Troy, MI. Environment: Suburban. Big Idea:. Standards:. Favorites (1). Resources (5). Common Core Math.
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https://math.berkeley.edu/~sethian/2006/Explanations/fast_marching_explain.html
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HOME
OVERVIEW
• Fast Introduction
• Level Set Methods
• Fast Marching Methods
• Ordered Upwind Methods
• APPLICATIONS
INTERACTIVE APPLETS
HISTORY OF THE METHODS/FLOW CHART
PUBLICATIONS
EDUCATIONAL MATERIAL
ACKNOWLEDGEMENTS
1996-2010
J.A. Sethian
# Fast Marching Methods: A boundary value formulation
### Tracking a moving boundary
Suppose you are given an interface separating one region from another, and a speed F that tells you how to move each point of the interface. In the figure below, a black curve separates a dark blue inside from a light blue outside, and at each point of the black curve the speed F is given. Furthermore, suppose that the speed F is always positive, that is, the front always moves outwards. of physical effects. For example
• Imagine that the dark blue is a substance moving into the light blue. For example, let the boundary be the edge of an acid eating its way into the exterior region. The speed of the acid depends on the resistance it meets in the underlying material; strong parts of the material resist the acid and slow it down more than more corrosive parts.
• Imagine that the dark blue is the edge of a disturbance that is propagating into the light blue. For example, suppose that there is an earthquake in the center of the box; the grey region represents parts of the earth that have heard about the earthquake; the light blue is still quiet. The speed of the earthquake depends on the type of rock it is going through, which can change from spot to spot.
Most numerical techniques rely on markers, which try to track the motion of the boundary by breaking it up into buoys that are connected by pieces of rope. The idea is to move each buoy under the speed F, and rely on the connecting ropes to keep things straight. The hope is that more buoys will make the answer more accurate.
Unfortunately, things get pretty dicey if the buoys try to cross over themselves, or if the shape tries to break into two; in these cases, it is very hard to keep the connecting ropes organized. In three dimensions, following a surface like a breaking ocean wave is particular tough.
### The Fast Marching Approach
Rather than follow the interface itself, the Fast Marching Method makes use of stationary approach to the problem. At first glance, this sounds counter-intuitive; we are going to trade a moving boundary problem for one in which nothing moves at all! To see how this is done, imagine a grid laid down on top of the problem:
Suppose that somebody is standing at each red grid point with a watch. When the front crosses each grid point, the person standing there writes down this crossing time T. This grid of crossing time values T(x,y) determines a function; at each grid point T, T(x,y) gives the time at which the front crosses the point (x,y).
As an example, suppose the initial disturbance is a circle propagating outwards. The original region (the blue one on the left below) propagates outwards, crossing over each of the timing spots. The function T(x,y) gives a cone-shaped surface, which is shown on the right. This surface has a great property; it intersects the xy plane exactly where the curve is initially. Better yet, at any height T the surface gives the set of points reached at time T. The surface on the right below is called the arrival time surface, because it gives the arrival time.
### Why is this called a "boundary value formulation"?
The reason it is called a "boundary value formulation" is because we have let the initial position of the front be the boundary for this arrival time surface T(x,y) that we would like to find. We have taken the idea of finding something that moves in time an exchanged it for a stationary problem in which the arrival surface contains the information about what is moving.
### Why is this a good idea?
The reason this is a good idea is two-fold. First, even if the evolving front changes shape, the arrival time surface T(x,y) is nice and well-behaved. By slicing it with a saw at height T (that is, finding the level set T), it will always give the position of the front at time T.
The second reason to do this, as we will see below, is that there is an incredibly fast way to solve this, known as the Fast Marching Method.
### The Fast Marching Method
How can this stationary surface be constructed? As a motivation, imagine scaffolding being erected around a house! One stands on one of the boards, puts a board above the head, and then moves to another board at the same level and put a board one level up. Once all the boards are placed at a given level, one then climbs up to the next level set repeat the process. The thing to remember is that the scaffolding is built from the ground up; each level must be completed before the next is begun.
The Fast Marching Method imitates this process. Given the initial curve (shown in red), stand on the lowest spot (which would be any point on the curve), and build a little bit of the surface that corresponds to the front moving with the speed F. Repeat the process over and over, always standing on the lowest spot of the scaffold, and building that little bit of the surface. When this process ends, the entire surface has been built.
(354K) Construction of stationary level set solution Green squares show next level to be built
The speed from this method comes from figuring out in what order to build the scaffolding; fortunately, there are lots of fast sorting algorithms that can do this job quickly.
### Background and History: The Fast Marching Method, Dijkstra's Method, Tsitsiklis' Method, and Ordered Upwind Methods
The Fast Marching Method is very closely related to Dijkstra's method, which is a very well-known method from the 1950's for computing the shortest path on a network. Dijkstra's method is ubiquitous: everywhere from internet routing to your car's navigation system. To explain the connection, consider a network in which there is cost assigned to entering each node.
Find the shortest path from Start to Finish Shortest path shown in Red
Dijkstra's method is an extraordinarily clever way to solve this problem. The basic idea is as follows:
1. Put the starting point in a set called "Accepted".
2. Call the grid points which are one link away from the Start "Neighbors".
3. Compute the cost of reaching each of these Neighbors.
4. The smallest of these Neighbors must have the correct cost. Remove it, and call it "Accepted." Add any new Neighbors to this point that are not already "Accepted", and find the cost of reaching all Neighbors. Repeat this step until there all points are Accepted.
This is an appealing algorithm. What it is doing is ordering the way in which the cost of reaching points are accepted, systematically working its way out from the known costs (the starting point) all the way to the finish. The method can be very made very fast: it is one-pass, because each point is touched essentially only once.
### Solving the Continuous Eikonal Problem:
It is not too hard to see that this method cannot converge to the solution of a continuous Eikonal problem. No matter how much you refine the mesh, you will still get stairstep patterns: you can never find the diagonal that you want. This is where the Fast Marching Method comes in: it uses upwind difference operators to approximate the gradient, but retains the Dijkstra idea of a one-pass algorithm. It allows this methodology to be used to compute a large class of continuous problems.
Tsitsiklis (see the reference below) was the first to develop a Dijkstra-like method for solving the Eikonal equation: his was a control theory approach (as opposed to the Fast Marching upwind finite difference perspective) to solving the Eikonal equation: it is also a viscosity solution with the same operation count. This leads to a causality relationship based on the optimality criterion rather than on the upwind finite difference operators employed in Fast Marching Methods. In the particular special case of a first order upwind finite difference for the Fast Marching Method on a square mesh, the resulting update equation at each grid point can be seen to be the same quadratic equation obtained through Tsitsiklis's control theoretic approach. There are similarities and differences between his approach and Fast Marching Methods, we refer the interested reader to the paper on Ordered Upwind Methods (see the reference below) which discusses his approach in detail.
Finally, the advent of Ordered Upwind Methods allowed the Dijkstra-like methodology to be extended to a far wider class of problems, including continuous anisotropic static convex Hamilton-Jacobi equations, all with the same Dijkstra-like one pass flavor.
### What is the difference between Fast Marching Methods and Level Set Methods?
Fast Marching Methods are designed for problems in which the speed function never changes sign, so that the front is always moving forward or backward. This allows us to convert the problem to a stationary formulation, because the front crosses each red grid point only once. This conversion to a stationary formulation, plus a whole bunch of numerical tricks, gives it its tremendous speed
Level Set Methods are designed for problems in which the speed function can be positive in some places are negative in others, so that the front can move forwards in some places and backwards in others. While significantly slower than Fast Marching Methods, embedding the problem in one higher dimension gives the method tremendous generality.
## Details
### Fast Marching Methods
The Fast Marching Method solves the general static Hamilton-Jacobi equation, which applies in the case of a convex, non-negative speed function. Starting with an initial position for the front, the method systematically marches the front outwards one grid point at a time, relying on entropy-satisfying schemes to produce the correct viscosity solution. The main idea is exploit a fast heapsort technique to systematically locate the proper grid point to update, so that one need never backtrack over previously evaluated grid points. The resulting technique sweeps through a grid of N total points in N log N steps to obtain the evolving time position of the front as it propagates through the grid. For details, see the technical explanations and flow chart.
References:
• A Note on Two Problems in Connection with Graphs, : Dijkstra, E.W., Numerische Mathematic, 1:269--271, 1959.
• Efficient Algorithms for Globally Optimal Trajectories : Tsitsiklis, J.N., IEEE Transactions on Automatic Control, Volume 40, pp. 1528-1538, 1995.
• A Fast Marching Level Set Method for Monotonically Advancing Fronts : Sethian, J.A., Proc. Nat. Acad. Sci., 93, 4, pp.1591-1595, 1996.
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HOME. OVERVIEW. • Fast Introduction. • Level Set Methods. • Fast Marching Methods. • Ordered Upwind Methods. • APPLICATIONS. INTERACTIVE APPLETS. HISTORY OF THE METHODS/FLOW CHART. PUBLICATIONS. EDUCATIONAL MATERIAL. ACKNOWLEDGEMENTS. 1996-2010. J.A. Sethian. # Fast Marching Methods: A boundary value formulation. ### Tracking a moving boundary. Suppose you are given an interface separating one region from another, and a speed F that tells you how to move each point of the interface. In the figure below, a black curve separates a dark blue inside from a light blue outside, and at each point of the black curve the speed F is given. Furthermore, suppose that the speed F is always positive, that is, the front always moves outwards. of physical effects. For example. • Imagine that the dark blue is a substance moving into the light blue. For example, let the boundary be the edge of an acid eating its way into the exterior region. The speed of the acid depends on the resistance it meets in the underlying material; strong parts of the material resist the acid and slow it down more than more corrosive parts.. • Imagine that the dark blue is the edge of a disturbance that is propagating into the light blue. For example, suppose that there is an earthquake in the center of the box; the grey region represents parts of the earth that have heard about the earthquake; the light blue is still quiet. The speed of the earthquake depends on the type of rock it is going through, which can change from spot to spot.. Most numerical techniques rely on markers, which try to track the motion of the boundary by breaking it up into buoys that are connected by pieces of rope. The idea is to move each buoy under the speed F, and rely on the connecting ropes to keep things straight. The hope is that more buoys will make the answer more accurate.. Unfortunately, things get pretty dicey if the buoys try to cross over themselves, or if the shape tries to break into two; in these cases, it is very hard to keep the connecting ropes organized. In three dimensions, following a surface like a breaking ocean wave is particular tough.. ### The Fast Marching Approach. Rather than follow the interface itself, the Fast Marching Method makes use of stationary approach to the problem. At first glance, this sounds counter-intuitive; we are going to trade a moving boundary problem for one in which nothing moves at all! To see how this is done, imagine a grid laid down on top of the problem:. Suppose that somebody is standing at each red grid point with a watch. When the front crosses each grid point, the person standing there writes down this crossing time T. This grid of crossing time values T(x,y) determines a function; at each grid point T, T(x,y) gives the time at which the front crosses the point (x,y).. As an example, suppose the initial disturbance is a circle propagating outwards. The original region (the blue one on the left below) propagates outwards, crossing over each of the timing spots. The function T(x,y) gives a cone-shaped surface, which is shown on the right. This surface has a great property; it intersects the xy plane exactly where the curve is initially. Better yet, at any height T the surface gives the set of points reached at time T. The surface on the right below is called the arrival time surface, because it gives the arrival time.. ### Why is this called a "boundary value formulation"?. The reason it is called a "boundary value formulation" is because we have let the initial position of the front be the boundary for this arrival time surface T(x,y) that we would like to find. We have taken the idea of finding something that moves in time an exchanged it for a stationary problem in which the arrival surface contains the information about what is moving.. ### Why is this a good idea?. The reason this is a good idea is two-fold. First, even if the evolving front changes shape, the arrival time surface T(x,y) is nice and well-behaved. By slicing it with a saw at height T (that is, finding the level set T), it will always give the position of the front at time T.. The second reason to do this, as we will see below, is that there is an incredibly fast way to solve this, known as the Fast Marching Method.. ### The Fast Marching Method. How can this stationary surface be constructed? As a motivation, imagine scaffolding being erected around a house! One stands on one of the boards, puts a board above the head, and then moves to another board at the same level and put a board one level up. Once all the boards are placed at a given level, one then climbs up to the next level set repeat the process. The thing to remember is that the scaffolding is built from the ground up; each level must be completed before the next is begun.. The Fast Marching Method imitates this process.
|
Given the initial curve (shown in red), stand on the lowest spot (which would be any point on the curve), and build a little bit of the surface that corresponds to the front moving with the speed F. Repeat the process over and over, always standing on the lowest spot of the scaffold, and building that little bit of the surface. When this process ends, the entire surface has been built.. (354K) Construction of stationary level set solution Green squares show next level to be built. The speed from this method comes from figuring out in what order to build the scaffolding; fortunately, there are lots of fast sorting algorithms that can do this job quickly.. ### Background and History: The Fast Marching Method, Dijkstra's Method, Tsitsiklis' Method, and Ordered Upwind Methods. The Fast Marching Method is very closely related to Dijkstra's method, which is a very well-known method from the 1950's for computing the shortest path on a network. Dijkstra's method is ubiquitous: everywhere from internet routing to your car's navigation system. To explain the connection, consider a network in which there is cost assigned to entering each node.. Find the shortest path from Start to Finish Shortest path shown in Red. Dijkstra's method is an extraordinarily clever way to solve this problem. The basic idea is as follows:. 1. Put the starting point in a set called "Accepted".. 2. Call the grid points which are one link away from the Start "Neighbors".. 3. Compute the cost of reaching each of these Neighbors.. 4. The smallest of these Neighbors must have the correct cost. Remove it, and call it "Accepted." Add any new Neighbors to this point that are not already "Accepted", and find the cost of reaching all Neighbors. Repeat this step until there all points are Accepted.. This is an appealing algorithm. What it is doing is ordering the way in which the cost of reaching points are accepted, systematically working its way out from the known costs (the starting point) all the way to the finish. The method can be very made very fast: it is one-pass, because each point is touched essentially only once.. ### Solving the Continuous Eikonal Problem:. It is not too hard to see that this method cannot converge to the solution of a continuous Eikonal problem. No matter how much you refine the mesh, you will still get stairstep patterns: you can never find the diagonal that you want. This is where the Fast Marching Method comes in: it uses upwind difference operators to approximate the gradient, but retains the Dijkstra idea of a one-pass algorithm. It allows this methodology to be used to compute a large class of continuous problems.. Tsitsiklis (see the reference below) was the first to develop a Dijkstra-like method for solving the Eikonal equation: his was a control theory approach (as opposed to the Fast Marching upwind finite difference perspective) to solving the Eikonal equation: it is also a viscosity solution with the same operation count. This leads to a causality relationship based on the optimality criterion rather than on the upwind finite difference operators employed in Fast Marching Methods. In the particular special case of a first order upwind finite difference for the Fast Marching Method on a square mesh, the resulting update equation at each grid point can be seen to be the same quadratic equation obtained through Tsitsiklis's control theoretic approach. There are similarities and differences between his approach and Fast Marching Methods, we refer the interested reader to the paper on Ordered Upwind Methods (see the reference below) which discusses his approach in detail.. Finally, the advent of Ordered Upwind Methods allowed the Dijkstra-like methodology to be extended to a far wider class of problems, including continuous anisotropic static convex Hamilton-Jacobi equations, all with the same Dijkstra-like one pass flavor.. ### What is the difference between Fast Marching Methods and Level Set Methods?. Fast Marching Methods are designed for problems in which the speed function never changes sign, so that the front is always moving forward or backward. This allows us to convert the problem to a stationary formulation, because the front crosses each red grid point only once. This conversion to a stationary formulation, plus a whole bunch of numerical tricks, gives it its tremendous speed. Level Set Methods are designed for problems in which the speed function can be positive in some places are negative in others, so that the front can move forwards in some places and backwards in others. While significantly slower than Fast Marching Methods, embedding the problem in one higher dimension gives the method tremendous generality.. ## Details. ### Fast Marching Methods. The Fast Marching Method solves the general static Hamilton-Jacobi equation, which applies in the case of a convex, non-negative speed function. Starting with an initial position for the front, the method systematically marches the front outwards one grid point at a time, relying on entropy-satisfying schemes to produce the correct viscosity solution. The main idea is exploit a fast heapsort technique to systematically locate the proper grid point to update, so that one need never backtrack over previously evaluated grid points. The resulting technique sweeps through a grid of N total points in N log N steps to obtain the evolving time position of the front as it propagates through the grid. For details, see the technical explanations and flow chart.. References:. • A Note on Two Problems in Connection with Graphs, : Dijkstra, E.W., Numerische Mathematic, 1:269--271, 1959.. • Efficient Algorithms for Globally Optimal Trajectories : Tsitsiklis, J.N., IEEE Transactions on Automatic Control, Volume 40, pp. 1528-1538, 1995.. • A Fast Marching Level Set Method for Monotonically Advancing Fronts : Sethian, J.A., Proc. Nat. Acad. Sci., 93, 4, pp.1591-1595, 1996.
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