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Let \( {A}_{i} \) be the event the \( i \) th card is drawn on the first selection and let \( {B}_{i} \) be the event the card numbered \( i \) is drawn on the second selection (from the box). Determine \( P\left( {B}_{5}\right), P\left( {{A}_{1}{B}_{5}}\right) \), and \( P\left( {{A}_{1} \mid {B}_{5}}\right) \). | From Example 3.4 (What is the conditioning event?), we have \( P\left( {A}_{i}\right) = 6/{10} \) and \( P\left( {{A}_{i}{A}_{j}}\right) = \) \( 3/{10} \) . This implies\n\n\[ P\left( {{A}_{i}{A}_{j}^{c}}\right) = P\left( {A}_{i}\right) - P\left( {{A}_{i}{A}_{j}}\right) = 3/{10} \]\n\n(3.22)\n\nNow we can draw card fiv... | Yes |
\( \gg \mathrm{{pm}} = \operatorname{minprob}\left( {{0.1} * \left\lbrack \begin{array}{lll} 4 & 7 & 6 \end{array}\right\rbrack }\right) \) | pm = 0.0720 0.1080 0.1680 0.2520 0.0480 0.0720 0.1120 0.1680 | No |
\( \gg \mathrm{{pm}} = {0.01} * \left\lbrack \begin{array}{llll} {15} & 5 & 2 & 1 \\ 8 & {25} & 5 & 1 \\ 8 & {12} & 1 & \end{array}\right\rbrack ;\;\% \) An arbitrary class \( \gg \operatorname{disp}\left( {\operatorname{imintest}\left( \mathrm{{pm}}\right) }\right) \) | The class is NOT independent\nMinterms for which the product rule fails \( \begin{array}{llll} 1 & 1 & 1 & 0 \end{array} \) \( \begin{array}{llll} 1 & 1 & 1 & 0 \end{array} \) | No |
Suppose the class \( \{ A, B, C\} \) is independent, with respective probabilities \( {0.4},{0.6},{0.8} \) . Determine \( P\left( {A \cup {BC}}\right) \) . | The minterm expansion is\n\n\[ A \cup {BC} = M\left( {3,4,5,6,7}\right) \text{, so that}P\left( {A \cup {BC}}\right) = p\left( {3,4,5,6,7}\right) \]\n\n(4.10)\n\nIt is not difficult to use the product rule and the replacement theorem to calculate the needed minterm probabilities. Thus \( p\left( 3\right) = P\left( {A}^... | Yes |
Consider the independent class \( \left\{ {{A}_{1},{A}_{2},{A}_{3},{B}_{1},{B}_{2},{B}_{3},{B}_{4}}\right\} \), with respective probabilities \( {0.4},{0.7},{0.3},{0.5},{0.8},{0.3},{0.6} \). Consider \( {M}_{3} \), minterm three for the class \( \left\{ {{A}_{1},{A}_{2},{A}_{3}}\right\} \), and \( {N}_{5} \), minterm f... | \[ P\left( {M}_{3}\right) = P\left( {{A}_{1}^{c}{A}_{2}{A}_{3}}\right) = {0.6} \cdot {0.7} \cdot {0.3} = {0.126}\text{and}P\left( {N}_{5}\right) = P\left( {{B}_{1}^{c}{B}_{2}{B}_{3}^{c}{B}_{4}}\right) = {0.5} \cdot {0.8} \cdot {0.7} \cdot {0.6} = {0.168} \] Also \[ P\left( {{M}_{3}{N}_{5}}\right) = P\left( {{A}_{1}^{c}... | Yes |
We wish to show that the pair \( \{ A, B\} \) is independent; i.e., the product rule \( P\left( {AB}\right) = P\left( A\right) P\left( B\right) \) holds. | COMPUTATION\n\n\[ P\left( A\right) = P\left( {A}_{1}\right) + P\left( {A}_{2}\right) + P\left( {A}_{3}\right) = {0.3} + {0.4} + {0.1} = {0.8}\text{and}P\left( B\right) = P\left( {B}_{1}\right) + \]\n\n\[ P\left( {B}_{2}\right) = {0.2} + {0.5} = {0.7} \]\n\nNow\n\n\[ {AB} = \left( {{A}_{1}\bigvee {A}_{2}\bigvee {A}_{3}}... | Yes |
Example 1.2.1 Suppose a human population is growing at \( 1\% \) per year and initially has 1,000,000 individuals. Let \( {P}_{t} \) denote the populations size \( t \) years after the initial population of \( {P}_{0} = 1,{000},{000} \) individuals. If one asks what the population will be in 50 years there are two opti... | Option 1. At \( 1\% \) per year growth, the dynamic equation would be\n\n\[ {P}_{t + 1} - {P}_{t} = {0.01}{P}_{t} \]\n\nand the corresponding iteration equation is\n\n\[ {P}_{t + 1} = {1.01}{P}_{t} \]\n\nWith \( {P}_{0} = 1,{000},{000},{P}_{1} = {1.01} \times 1,{000},{000} = 1,{010},{000},{P}_{2} = {1.011},{010},{000} ... | Yes |
Example 1.7.2 Simulation of chemical discharge into a lake. Begin with two one-liter beakers, a supply of distilled water and salt and a meter to measure conductivity in water. Place one liter of distilled water and \( {0.5}\mathrm{\;g} \) of salt in beaker \( \mathrm{F} \) (factory). Place one liter of distilled water... | In Exercise 1.7.4 you are asked to write and solve a mathematical model of this simulation and compare the solution with the data. ∎ | No |
Example 1.10.1 Jack A. Wolfe \( {}^{14} \) observed that leaves of trees growing in cold climates tend to be incised (have ragged edges) and leaves of trees growing in warm climates tend to have smooth edges (lacking lobes or teeth). He measured the percentages of species that have smooth margins among all species of t... | Figure for Example 1.10.0.2 Average temperature \( {\mathrm{C}}^{ \circ } \) vs percentage of tree species with smooth edge leaves in 33 forests in eastern Asia. The equation of the line is \( y = - {0.89} + {0.313x} \) .  ) and also recorded the air (ambient) temperature \( \left( T\right) \) in \( {\mathrm{F}}^{ \circ } \) for the night. The data were collected betw... | These data also appear linear and the line through (65,100) and (75,145), \[ \frac{R - {100}}{T - {65}} = \frac{{145} - {100}}{{75} - {65}}, \; R = {4.5T} - {192.5} \] (1.32) lies close to the data. We can use the line to estimate temperature to be about \( {69.5}{\mathrm{\;F}}^{ \circ } \) if cricket chirp rate is 120... | Yes |
Example 2.2.1 Data for the percentage of U.S. population in 1955 that had antibodies to the polio virus as a function of age is shown in Table 2.2.1.1. The data show an interesting fact that a high percentage of the population in 1955 had been infected with polio. A much smaller percentage were crippled or killed by th... | Although Table 2.2.1.1 is a function, it is only an approximation to a perhaps real underlying function. The order pair, \( \left( {{17.5},{72}}\right) \), signals that 72 percent of the people of age 17.5 years had antibodies to the polio virus. More accurately, \( \left( {{17.5},{72}}\right) \) signals that of a samp... | Yes |
For the function, \( R \), defined by\n\n\[ R\left( x\right) = x + \frac{1}{x}\;x \neq 0 \] | \[ R\left( {1 + 3}\right) = R\left( 4\right) = 4 + \frac{1}{4} = {4.25} \]\n\n\[ R\left( 1\right) = 1 + \frac{1}{1} = {2.0}\;\text{ and } \]\n\n\[ R\left( 3\right) = 3 + \frac{1}{3} = {3.3333}\cdots \]\n\n\[ R\left( 1\right) + R\left( 3\right) = 2 + {3.3333}\cdots = {5.3333}\cdots \neq {4.25} = R\left( 4\right) \]\n\nI... | Yes |
Show that polynomials are linear in their coefficients. | Consider the following.\n\nLet \( \;P\left( x\right) = 7 - {3x} + 5{x}^{2},\; \) and \( \;Q\left( x\right) = - 2 + {4x} - {x}^{2} + 6{x}^{3} \) .\n\n\[ P\left( x\right) + Q\left( x\right) = \left( {7 - {3x} + 5{x}^{2}}\right) + \left( {-2 + {4x} - {x}^{2} + 6{x}^{3}}\right. \]\n\n\[ = \left( {7 - 2}\right) + \left( {-3... | Yes |
Example 2.5.1 If we use these equations to fit a line to the cricket data of Example 1.10.1 showing a relation between temperature and cricket chirp frequency, we get\n\n\[ y = {4.5008x} - {192.008}, \\text{ close to the line } y = {4.5x} - {192} \]\n\nthat we ’fit by eye’ using the two points, \( \\left( {{65},{100}}\... | Explore 2.5.1 Technology. Your calculator or computer will hide all of the arithmetic of\n\nEquations 2.4 and give you the answer. The overall procedure is:\n\n1. Load the data. [Two lists, X and Y, say].\n\n2. Compute the coefficients of a first degree polynomial close to the data and store them in \( \\mathrm{P} \).\... | Yes |
Exercise 2.5.1 Use Equations 2.3 to find the linear function that is the least squares fit to the data: | \[ \left( {-2,5}\right) \;\left( {3,{12}}\right) \] | No |
The graph of the inverse of \( F \) is the reflection of the graph of \( F \) about the diagonal line, \( y = x \) . | The reflection of \( G \) with respect to the diagonal line, \( y = x \) consists of the points \( Q \) such that either \( Q \) is a point of \( G \) on the diagonal line, or there is a point \( P \) of \( G \) such that the diagonal line is the perpendicular bisector of the interval \( \overline{PQ} \) . | Yes |
To compute the equation for the inverse of the function \( \mathrm{S} \) , \[ S\left( x\right) = {x}^{2}\;\text{ for }\;x \geq 0 \] | let the equation for \( \mathrm{S} \) be written as \[ y = {x}^{2}\;\text{ for }\;x \geq 0 \] Then interchange \( y \) and \( x \) to obtain \[ x = {y}^{2}\;\text{ for }\;y \geq 0 \] and solve for \( \mathrm{y} \) . \[ x = {y}^{2}\;\text{ for }\;y \geq 0 \] \[ {y}^{2} = x\;\text{ for }\;y \geq 0 \] \[ {\left( {y}^{2}\r... | Yes |
Example 2.6.3 The graph of function \( F\left( x\right) = 2{x}^{2} - {6x} + 3/2 \) shown in Figure 2.12 has two invertible portions, the left branch and the right branch. We compute the inverse of each of them. | Let \( y = 2{x}^{2} - {6x} + 5/2 \), exchange symbols \( x = 2{y}^{2} - {6y} + 5/2 \), and solve for \( y \) . We use the steps of 'completing the square' that are used to obtain the quadratic formula.\n\n\[ x = 2{y}^{2} - {6y} + 5/2 \]\n\n\[ = 2\left( {{y}^{2} - {3x} + 9/4}\right) - 9/2 + 5/2 \]\n\n\[ = 2{\left( y - 3... | Yes |
Example 2.7.1 Two properties of the logarithm and exponential functions are\n\n\[ \n\\text{(a)}{\\log }_{b}{b}^{\\lambda } = \\lambda \\;\\text{and}\\;\\text{(b)}u = {b}^{{\\log }_{b}u} \n\] | The logarithm function, \( L\\left( x\\right) = {\\log }_{b}\\left( x\\right) \) is the inverse of the exponential function, \( E\\left( x\\right) = {b}^{x} \), and the properties simply state that\n\n\[ \nL \\circ E = I\\;\\text{ and }E \\circ L = I \n\] | Yes |
Is there an equation for such a signal? Yes, a very messy one! | The graph of the following equation is shown in Figure 2.18(b). It is similar to the typical electrocardiogram in Figure 2.18(a).\n\n\[ H\left( t\right) = {25000}\frac{\left( {t + {0.05}}\right) t\left( {t - {0.07}}\right) }{\left( {1 + {\left( {20}t\right) }^{10}}\right) {2}^{\left( {2}^{\left( {40}t\right) }\right) }... | Yes |
Find the period, frequency, and amplitude of \[ H\left( t\right) = 3\sin \left( {{5t} + \pi /3}\right) \] | Solution. Write \( H\left( t\right) = 3\sin \left( {{5t} + \pi /3}\right) \) as \[ H\left( t\right) = 3\sin \left( {\frac{2\pi }{{2\pi }/5}t + \pi /3}\right) . \] Then the amplitude of \( P \) is 3, and the period is \( {2\pi }/5 \) and the frequency is \( 5/\left( {2\pi }\right) \) . | Yes |
At what rate was the Vibrio natriegens population of Section 1.1 growing at time \( T = {40} \) minutes? | The average growth rate between times \( T = {32} \) and \( T = {48} \) is\n\n\[ \frac{{0.101} - {0.060}}{{48} - {32}} = {0.0026}\;\frac{\text{ Absorbance units }}{\text{ minute }}, \]\n\nand is a pretty good estimate of the growth rate at time \( T = {40} \), particularly because 40 is midway between 32 and 48. | Yes |
Example 3.1.2 At what rate was the world human population increasing in 1980? Shown in Figure 3.3 are data for the twentieth century and a graph of an approximating function, \( F \) . A tangent to the graph of \( F \) at \( \left( {{1980}, F\left( {1980}\right) }\right) \) is drawn and has a slope of \( {0.0781} \cdot... | Now,\n\n\[ \text{ slope is }\;\frac{\text{ rise }}{\text{ run }}\; = \;\frac{\text{ change in population }}{\text{ change in years }}\; \approx \;\frac{\text{ people }}{\text{ year }}. \]\n\nThe units of slope, then, are people/year. Therefore,\n\n\[ \text{ slope } = {78},{100},{000}\;\frac{\text{ people }}{\text{ year... | Yes |
Show that if \( a \) is a positive number, then\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}\sqrt{x} = \sqrt{a} \] | Suppose \( \epsilon > 0 \) . Let \( \delta = \epsilon \sqrt{a} \) . Suppose \( x \) is in the domain of \( \sqrt{x} \) and \( 0 < \left| {x - a}\right| < \delta \) . Then\n\n\[ \left| {\sqrt{x} - \sqrt{a}}\right| \;\overset{\left( i\right) }{ = }\;\frac{\left| x - a\right| }{\sqrt{x} + \sqrt{a}}\;\overset{\left( ii\rig... | Yes |
Theorem 3.2.1 : Limit is Unique Theorem. Suppose \( G \) is a function and\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}G\left( x\right) = {L}_{1}\;\text{ and }\;\mathop{\lim }\limits_{{x \rightarrow a}}G\left( x\right) = {L}_{2}. \]\n\nThen \( {L}_{1} = {L}_{2} \) . | 1. Suppose that \( {L}_{1} < {L}_{2} \).\n\n2. Let \( \epsilon = \left( {{L}_{2} - {L}_{1}}\right) /2 \). | No |
At what rate is the light decreasing when \( d = 5\mathrm{\;{cm}} \) ? | We find the derivative of \( L \) . For any value of \( d \) between 3 and \( {16}\mathrm{\;{cm}} \) ,\n\n\[ \n{L}^{\prime }\left( d\right) = \mathop{\lim }\limits_{{b \rightarrow d}}\frac{L\left( b\right) - L\left( d\right) \;\mathrm{{mW}}/{\mathrm{{cm}}}^{2}}{b - d\;\mathrm{\;{cm}}} \]\n\n\[ \n= \mathop{\lim }\limits... | Yes |
Example 3.3.2 Problem. The graphs of a function \( P \) and its derivative \( {P}^{\prime } \) are shown in Figure 3.16. Which graph is the graph of \( P \) ? | Solution. We claim that graph 1 is not \( P \), because every tangent to graph 1 has negative slope and some \( y \) -coordinates of graph 2 are positive. Therefore graph 2 must be the graph of \( P \) . | No |
Let \( P\left( t\right) = 3{t}^{4} - 5{t}^{2} + 2t + 7 \) . Then \( {P}^{\prime }\left( t\right) = {\left\lbrack P\left( t\right) \right\rbrack }^{\prime } \) | \[ {P}^{\prime }\left( t\right) = {\left\lbrack 3{t}^{4} - 5{t}^{2} + 2t + 7\right\rbrack }^{\prime } \] Definition of \( P \) \[ = {\left\lbrack 3{t}^{4}\right\rbrack }^{\prime } - {\left\lbrack 5{t}^{2}\right\rbrack }^{\prime } + {\left\lbrack 2t\right\rbrack }^{\prime } + {\left\lbrack 7\right\rbrack }^{\prime } \] ... | Yes |
Example 3.5.2 It is useful to write some fractions in forms so that the Constant Factor Rule obviously applies. For example\n\n\[ P\left( t\right) = \frac{{t}^{3}}{7} = \frac{1}{7}{t}^{3} \]\n\nand to compute \( {P}^{\prime } \) you may write | \n\[ {P}^{\prime }\left( t\right) = {\left\lbrack \frac{{t}^{3}}{7}\right\rbrack }^{\prime } = {\left\lbrack \frac{1}{7}{t}^{3}\right\rbrack }^{\prime } = \frac{1}{7}{\left\lbrack {t}^{3}\right\rbrack }^{\prime } = \frac{1}{7}3{t}^{2} = \frac{3}{7}{t}^{2} \] | Yes |
Example 3.5.3 The derivative of a quadratic function is a linear function. | Solution. Suppose \( P\left( t\right) = a{t}^{2} + {bt} + c \) where \( a, b \), and \( c \) are constants. Then\n\n\[ \n{P}^{\prime }\left( t\right) = {\left\lbrack a{t}^{2} + bt + c\right\rbrack }^{\prime } \n\]\n\n\[ \n= {\left\lbrack a{t}^{2}\right\rbrack }^{\prime } + {\left\lbrack bt\right\rbrack }^{\prime } + {\... | Yes |
In baseball, a 'pop fly' is hit and the ball leaves the bat traveling vertically at 30 meters per second. How high will the ball go and how much time does the catcher have to get in position to catch it? | Using a formula from Section 3.6.1, the ball will be at a height \( s\left( t\right) = - {4.9}{t}^{2} + {30t} \) meters \( t \) seconds after it is released, where \( s\left( t\right) \) is the height above the point of impact with the bat. The velocity, \( v\left( t\right) = {s}^{\prime }\left( t\right) \) is\n\n\[ \n... | Yes |
A farmer's barn is 60 feet long on one side. He has 100 feet of fence and wishes to build a rectangular pen along that side of his barn. What should be the dimensions of the pen to maximize the area? | Because there are 100 feet of fence, 2 * W + L = {100}. The area, A, is A = {LW}. Because 2W + L = {100}, L = {100} - {2W} and A = {LW} = \left( {{100} - {2W}}\right) W or A = {100W} - 2{W}^{2}. The graph of A vs W is a parabola with its highest point at the vertex. The tangent to the parabola at the vertex is horizont... | Yes |
Example 3.5.6 This problem is written on the assumption, to our knowledge untested, that spider webs have an optimum size. Seldom are they so small as \( 1\mathrm{\;{cm}} \) in diameter and seldom are they so large as \( 2\mathrm{\;m} \) in diameter. If they are one cm in diameter, there is a low probability of catchin... | Solution. Assume a circular spider web of diameter, \( d \) . It is reasonable to assume that the amount of food gathered by the web is proportional to the area, \( A \), of the web. Because \( A = \pi {d}^{2}/4 \) , the amount of food gather is proportional to \( {d}^{2} \) . We also assume that the energy required to... | Yes |
Example 3.6.2 Students measured height \( {vs} \) time of a falling bean bag using a Texas Instruments Calculator Based Laboratory Motion Detector, and the data are shown in Figure 3.26A. Average velocities were computed between data points and plotted against the midpoints of the data intervals in Figure 3.26B. Mid-ti... | An equation of the line fit by least squares to the graph of Average Velocity \( {vs} \) Midpoint of time interval is\n\n\[ \n{v}_{\text{ave }} = - {849}{t}_{\text{mid }} + {126}\;\mathrm{\;{cm}}/\mathrm{s}. \n\]\n\nIf we assume a continuous model based on this data, we have\n\n\[ \n{s}^{\prime }\left( t\right) = - {84... | Yes |
Theorem 4.1.1 Locally Positive Theorem. If a function, \( f \), is continuous at a number a in its domain and \( f\left( a\right) \) is positive, then there is a positive number, \( \delta \), such that \( f\left( x\right) \) is positive for every number \( x \) in \( \left( {a - \delta, a + \delta }\right) \) and in t... | 1. Suppose the hypothesis of the Locally Positive Theorem.\n2. Let \( \epsilon = f\left( a\right) \) .\n3. Use the hypothesis that \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = f\left( a\right) \) . | No |
Theorem 4.2.1 The Derivative Requires Continuity. If \( u \) is a function and \( {u}^{\prime }\left( t\right) \) exists at \( t = a \) then \( u \) is continuous at \( t = a \) . | Proof. In Exercise 4.2.2 you are asked to give reasons for the following steps, \( \left( i\right) - \left( v\right) \) . Suppose the hypothesis of Theorem 4.2.1.\n\n\[ \left( {\mathop{\lim }\limits_{{b \rightarrow a}}u\left( b\right) }\right) - u\left( a\right) = \mathop{\lim }\limits_{{b \rightarrow a}}\left( {u\left... | No |
Find the slope of the tangent to the circle, \( {x}^{2} + {y}^{2} = {13} \), at the point \( \left( {2,3}\right) \) . | Solution First check that \( {2}^{2} + {3}^{2} = 4 + 9 = {13} \), so that \( \left( {2,3}\right) \) is indeed a point of the circle. Then solve for \( y \) in \( {x}^{2} + {y}^{2} = {13} \) to get\n\n\[ \n{y}_{1} = \sqrt{{13} - {x}^{2}} \n\] \n\nThen the Power Chain Rule with \( n = \frac{1}{2} \) yields\n\n\[ \n{y}_{1... | Yes |
A forester needs to get from point \( A \) on a road to point \( B \) in a forest (see diagram in Figure 4.4). She can travel \( 5\mathrm{\;{km}}/\mathrm{{hr}} \) on the road and \( 3\mathrm{\;{km}}/\mathrm{{hr}} \) in the forest. At what point, \( P \) , should she leave the road and enter the forest in order to minim... | Solution. She might go directly from \( A \) to \( B \) through the forest; she might travel from \( A \) to \( C \) and then to \( B \) ; or she might, as illustrated by the dashed line, travel from \( A \) to a point, \( P \), along the road and then from \( P \) to \( B \) .\n\nAssume that the road is straight, the ... | Yes |
Find the slope of the graph of\n\n\\[ \sqrt{x} + \sqrt{5 - {y}^{2}} = 5\\;\\text{ at }\\left( {9,1}\\right) \\text{ and at }\\left( {4,2}\\right) \\] | First we check to see that \\( \\left( {9,1}\\right) \\) satisfies the equation:\n\n\\[ \sqrt{9} + \sqrt{5 - {1}^{2}} = \sqrt{9} + \sqrt{4} = 3 + 2 = 5.\\;\\text{It checks.} \\]\n\nThen we assume there is a function \\( y\\left( x\\right) \\) such that\n\n\\[ \sqrt{x} + \sqrt{5 - {\\left( y\\left( x\\right) \\right) }^... | No |
Theorem 5.1.1 If \( E\left( t\right) = {B}^{t} \) where \( B > 0 \), then\n\n\[ \n{E}^{\prime }\left( t\right) = {E}^{\prime }\left( 0\right) E\left( t\right) \n\] | Proof: For \( E\left( t\right) = {B}^{t}, B > 0 \) ,\n\n\[ \n{E}^{\prime }\left( t\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{B}^{t + h} - {B}^{t}}{h} = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{B}^{t}{B}^{h} - {B}^{t}}{h} = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{B}^{t}\left( {{B}^{h} - 1}\... | Yes |
Theorem 5.2.1 If \( {s}_{1} \leq {s}_{2} \leq {s}_{3} \leq \cdots \) is a bounded nondecreasing sequence of numbers there is a number \( s \) such that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{s}_{n} = s \) . | To prove Theorem 5.2.1 we need a clear definition of limit of a sequence.\n\nDefinition 5.2.2 A number \( s \) is the limit of a number sequence \( {s}_{1},{s}_{2},{s}_{3}\cdots \) means that if \( \left( {u, v}\right) \) is an open interval containing \( s \) there is a positive integer \( N \) such that if \( n \) is... | Yes |
The limit of the sequence \( \left\{ {1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\cdots }\right\} \) is zero. | If \( \left( {u, v}\right) \) contains \( 0, v \) is greater than zero and there is \( {}^{2} \) a positive integer \( N \) that is greater than \( 1/v \) . Therefore, if \( n > N, n > 1/v \) so that \( 1/n < v \), and \( u < 0 < 1/n < v \), so \( \frac{1}{n} \) is in \( \left( {u, v}\right) \) ). | No |
We can use Theorem 5.2.1 to show a useful result, that if \( a \) is a number and \( 0 < a < 1 \), then the sequence \( {x}_{n} = {a}^{n} \) converges to zero. | Proof. We assume the alternate version of Theorem 5.2.1 that If \( {s}_{1} \geq {s}_{2} \geq {s}_{3} \geq \cdots \) is a ounded nonincreasing sequence of numbers there is a number \( s \) such that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{s}_{n} = s \) .\n\nBecause \( 0 < a < 1 \) and \( {x}_{n} = {a}^{n},{x}... | Yes |
Example 5.4.1 We can also compute \( {E}^{\prime }\left( t\right) \) for \( E\left( t\right) = {2}^{t} \) . | \[ {\left\lbrack {2}^{t}\right\rbrack }^{\prime } = {\left\lbrack {\left( {e}^{\ln 2}\right) }^{t}\right\rbrack }^{\prime } = {\left\lbrack {e}^{\left( {\ln 2}\right) t}\right\rbrack }^{\prime } = {e}^{\left( {\ln 2}\right) t}\ln 2 = {2}^{t}\ln 2 \] | Yes |
Property 5.5.1 Proportional Growth or Decay. If \( P \) is a function defined by\n\n\[ P\left( t\right) = C{e}^{kt} \]\n\nwhere \( C \) and \( k \) are numbers, then\n\n\[ {P}^{\prime }\left( t\right) = {kP}\left( t\right) \] | Proof that \( P\left( t\right) = C{e}^{kt} \) implies that \( {P}^{\prime }\left( t\right) = {kP}\left( t\right) \) :\n\n\[ {P}^{\prime }\left( t\right) = {\left\lbrack C{e}^{kt}\right\rbrack }^{\prime } = C{\left\lbrack {e}^{kt}\right\rbrack }^{\prime } = C{e}^{kt}k = {kC}{e}^{kt} = {kP}\left( t\right) \] | Yes |
Property 5.5.2 Exponential Growth or Decay If \( P \) is a function and there is a number \( k \) for which\n\n\[ \n{P}^{\prime }\left( t\right) = {kP}\left( t\right) \;\text{ for all }t \geq 0 \n\]\n\nthen there is a number \( C \) for which\n\n\[ \nP\left( t\right) = C{e}^{kt} \n\]\n\nFurthermore,\n\n\[ \nC = P\left(... | In the preceding equations, \( k \) can be either positive or negative. When \( k \) is negative, it is more common to emphasize this and write \( - k \) and write \( P\left( t\right) = {e}^{-{kt}} \), where in this context it is understood that \( k \) is a positive number. | Yes |
Example 5.5.1 A distinction between discrete and continuous models. Suppose in year 2000 a population is at 5 million people and the population growth rate (excess of births over deaths) is 6 percent per year. One interpretation of this is to let \( P\left( t\right) \) be the population size in millions of people at ti... | The discrepancy lies with the model equation \( {P}^{\prime }\left( t\right) = {0.6P}\left( t\right) \) . Instead, we may write\n\n\[ P\left( 0\right) = 5\;{P}^{\prime }\left( t\right) = {kP}\left( t\right) \]\n\nwhere \( k \) is to be determined. Then from Exponential Growth or Decay 5.5.2 we may write\n\n\[ P\left( t... | Yes |
Mathematical Model 5.5.3 Potassium-40 decomposition. The rate of disintegration of \( {}^{40}\mathrm{K} \) is proportional to the amount of \( {}^{40}\mathrm{\;K} \) present. | If we let \( K\left( t\right) \) be the amount of \( {}^{40}\mathrm{\;K} \) present \( t \) years after deposition of rock of volcanic origin and \( {K}_{0} \) the initial amount of \( {}^{40}\mathrm{\;K} \) present, then\n\n\[ K\left( 0\right) = {K}_{0},\;{K}^{\prime }\left( t\right) = - {rK}\left( t\right) \]\n\nwher... | Yes |
Suppose a runner is exhaling at the rate of 2 liters per second. Then the amount of air in her lungs is decreasing at the rate of two liters per second. If, furthermore, the \( {\mathrm{{CO}}}_{2} \) partial pressure in the exhaled air is \( {50}\mathrm{\;{mm}}\mathrm{{Hg}} \) (approx \( {0.114}\mathrm{\;g}{\mathrm{{CO... | \( \blacksquare \) | No |
Example 5.5.4 Classical Washout Curve. A barrel contains 100 liters of water and 300 grams of salt. You start a stream of pure water flowing into the barrel at 5 liters per minute, and a compensating stream of salt water flows from the barrel at 5 liters per minute. The solution in the barrel is 'well stirred' so that ... | Solution. First let us analyze \( S \) . We use Primitive Concept 2. The concentration of salt in the water flowing into the barrel is 0 . The concentration of salt in the water flowing out of the barrel is the same as the concentration \( C\left( t\right) \) of salt in the barrel which is\n\n\[ C\left( t\right) = \fra... | Yes |
Problem. Suppose a 100 liter barrel is full of pure water and at time \( t = 0 \) minutes a stream of water flowing at 5 liters per minute and carrying \( 3\mathrm{\;g}/ \) liter of salt starts flowing into the barrel. Assume the salt is well mixed in the barrel and water overflows at the rate of 5 liters per minute. L... | Solution: We analyze \( S \) ; again we use Primitive Concept 2. The concentration of salt in the inflow is \( 3\mathrm{\;g}/ \) liter. The concentration \( C\left( t\right) \) of salt in the tank at time \( t \) minutes is\n\n\[ C\left( t\right) = \frac{S\left( t\right) }{100} \]\n\nThe salt concentration in the outfl... | Yes |
Assume that \( {1000}\mathrm{\;w}/{\mathrm{m}}^{2} \) of light is striking the surface of a lake and that \( {40}\% \) of that light is reflected back into the atmosphere. We first solve the initial value problem\n\n\[ I\left( 0\right) = {600} \]\n\n\[ {I}^{\prime }\left( x\right) = - {KI}\left( x\right) \] | to get\n\n\[ I\left( x\right) = {600}{e}^{-{Kx}} \]\n\nIf we have additional information that, say, the light intensity at a depth of \( {10}\mathrm{{meters}} \) is \( {400}\mathrm{\;W}/{\mathrm{m}}^{2} \) we can find the value of \( K \) . It must be that\n\n\[ I\left( {10}\right) = {600}{e}^{-K \times {10}} = {400} \... | Yes |
Suppose a patient is prescribed to take \( {80}\mathrm{{mg}} \) of Sotolol, a drug that regularizes heart beat, once per day. Sotolol has a half-life in the body of 12 hrs. Compute the daily fluctuations of sotolol. | Let \( {s}_{t}^{ - } \) be the amount of sotolol in the body at time, \( t \), just before the sotolol pill is taken and \( {s}_{t}^{ + } \) be the amount of sotolol in the body at time \( t \) just after the sotolol pill is taken. Then\n\n\[ \n{s}_{0}^{ - } = 0,\;{s}_{0}^{ + } = {80},\;{s}_{1}^{ - }\; = \;\frac{1}{4}{... | Yes |
Example 5.5.8 In Section 1.3 we showed the results of an experiment measuring the light decay as a function of depth. The data and a semilog graph of the data are shown in Figure 5.6. | As shown in the figure, \[ {\log }_{10}{I}_{d} = - {0.4} - {0.087d} \] is a good approximation to the data. Therefore \[ {I}_{d} \doteq {10}^{-{0.4} - {0.087d}} \] \[ = {0.4} \times {0.82}^{d} \] which is the same result obtained in Section 1.3. As shown in the previous subsection, the relation \[ {I}^{\prime }\left( d... | Yes |
Find the derivatives of\n\n\[ P\left( x\right) = {200}{e}^{-{3x}}\;Q\left( x\right) = \frac{1}{\sqrt{2\pi }}{e}^{-{x}^{2}/2} \] | Solutions:\n\n\[ {P}^{\prime }\left( x\right) = {\left\lbrack {200}{e}^{-{3x}}\right\rbrack }^{\prime }\;{Q}^{\prime }\left( x\right) = {\left\lbrack \frac{1}{\sqrt{2\pi }}{e}^{-{x}^{2}/2}\right\rbrack }^{\prime }\;\text{ Logical Identity } \]\n\n\[ = {200}{\left\lbrack {e}^{-{3x}}\right\rbrack }^{\prime }\; = \frac{1}... | Yes |
Logarithm functions to other bases have derivatives, but not as neat as \( 1/t \) . We compute \( {L}^{\prime }\left( t\right) \) for \( L\left( t\right) = {\log }_{b}t \) where \( b > 0 \) and \( b \neq 1 \) . | \[ {\left\lbrack {\log }_{b}t\right\rbrack }^{\prime } = {\left\lbrack \frac{\ln t}{\ln b}\right\rbrack }^{\prime } \] \[ \left( i\right) \] \[ = \frac{1}{\ln b}{\left\lbrack \ln t\right\rbrack }^{\prime } \] (ii) (5.39) \[ = \frac{1}{\ln b}\frac{1}{t} \] (iii) We summarize this as \[ {\left\lbrack {\log }_{b}t\right\r... | Yes |
Find the derivatives of a. \( y = \ln \left( \sqrt{1 - {t}^{2}}\right) \) b. \( y = \ln \left( \frac{1 - t}{1 + t}\right) \) | \[ {\left\lbrack \ln \left( \sqrt{1 - {t}^{2}}\right) \right\rbrack }^{\prime } = {\left\lbrack \frac{1}{2}\ln \left( 1 - {t}^{2}\right) \right\rbrack }^{\prime }\;\text{ Logarithm Property } \] \[ = \frac{1}{2}{\left\lbrack \ln \left( 1 - {t}^{2}\right) \right\rbrack }^{\prime }\;\text{Constant Factor} \] \[ = \frac{1... | Yes |
Suppose we are to differentiate\n\n\\[ y\\left( t\\right) = \\left( {t + 2}\\right) \\left( {t + 1}\\right) \\left( {t - 1}\\right) \\] | Proceeding indirectly, we first compute the derivative of the natural logarithm of \\( y \\) .\n\n\\[ \\ln \\left( {y\\left( t\\right) }\\right) = \\ln \\left( {\\left( {t + 2}\\right) \\left( {t + 1}\\right) \\left( {t - 1}\\right) }\\right) \\]\nLogical Identity\n\n\\[ \\ln \\left( {y\\left( t\\right) }\\right) = \\l... | Yes |
Theorem 6.1.1 Suppose \( u \) and \( v \) are two functions. Then for every number a for which \( {u}^{\prime }\left( a\right) \) and \( {v}^{\prime }\left( a\right) \) exist,\n\n\[{\left\lbrack u\left( t\right) \times v\left( t\right) \right\rbrack }_{t = a}^{\prime } = {u}^{\prime }\left( a\right) \times v\left( a\ri... | The proof uses Theorem 4.2.1, The Derivative Requires Continuity, which in symbols is:\n\n\[ \mathop{\lim }\limits_{{b \rightarrow a}}\frac{u\left( b\right) - u\left( a\right) }{b - a} = {u}^{\prime }\left( a\right) \;\text{ exists implies that }\;\mathop{\lim }\limits_{{b \rightarrow a}}u\left( b\right) = u\left( a\ri... | Yes |
Theorem 6.1.2 Suppose \( u \) and \( v \) are functions and \( {u}^{\prime }\left( a\right) \) and \( {v}^{\prime }\left( a\right) \) exist and \( v\left( a\right) \neq 0 \) . Then\n\n\[ \n{\left\lbrack \frac{u\left( t\right) }{v\left( t\right) }\right\rbrack }_{t = a}^{\prime } = \frac{{u}^{\prime }\left( a\right) \ti... | Proof of Theorem 6.1.2.\n\n\[ \n{\left\lbrack \frac{u\left( t\right) }{v\left( t\right) }\right\rbrack }_{t = a}^{\prime } = \mathop{\lim }\limits_{{b \rightarrow a}}\frac{\frac{u\left( b\right) }{v\left( b\right) } - \frac{u\left( a\right) }{v\left( a\right) }}{b - a}\n\]\n\n(i)\n\n\[ \n= \mathop{\lim }\limits_{{b \ri... | Yes |
The logistic function and its derivative. The logistic function\n\n\\[ \nP\\left( t\\right) = \\frac{{P}_{0}M{e}^{rt}}{M - {P}_{0} + {P}_{0}{e}^{rt}} \n\\]\n\ndescribes the size of a population of initial size \\( {P}_{0} \\) and low density relative growth rate \\( r \\) growing in an environment with limited carrying... | \[ \n{P}^{\\prime }\\left( t\\right) = {\\left\\lbrack \\frac{{P}_{0}M{e}^{rt}}{M - {P}_{0} + {P}_{0}{e}^{rt}}\\right\\rbrack }^{\\prime } \n\]\n\n\[ \n= {P}_{0}M{\\left\\lbrack \\frac{{e}^{rt}}{M - {P}_{0} + {P}_{0}{e}^{rt}}\\right\\rbrack }^{\\prime } \n\]\n\n\[ \n- {P}_{0}M\\frac{\\left( {M - {P}_{0} + {P}_{0}{e}^{r... | Yes |
\[ \text{a.}P\left( t\right) = {e}^{2t}\ln t\;{P}^{\prime }\left( t\right) = {\left\lbrack {e}^{2t}\ln t\right\rbrack }^{\prime } \] | \[ = {\left\lbrack {e}^{2t}\right\rbrack }^{\prime }\ln t + {e}^{2t}{\left\lbrack \ln t\right\rbrack }^{\prime } \] \[ = {e}^{2t}2\ln t + {e}^{2t}\frac{1}{t} \] | Yes |
Compute \( {F}^{\prime }\left( t\right) \) for \( F\left( t\right) = {\left( 1 - {t}^{2}\right) }^{3} \) . | Let\n\n\[ G\left( z\right) = {z}^{3}\;\text{ and }\;u\left( t\right) = 1 - {t}^{2}.\;\text{ Then }\;F\left( t\right) = G\left( {u\left( t\right) }\right) .\n\n\]\n\n\[ {G}^{\prime }\left( z\right) = 3{z}^{2}\;\text{ and }\;{\left\lbrack u\left( t\right) \right\rbrack }^{\prime } = - {2t} \]\n\n\[ {G}^{\prime }\left( {u... | Yes |
Theorem 6.2.1 Chain Rule. Suppose \( G \) and \( u \) are functions that have derivatives and \( G\left( {u\left( t\right) }\right) \) is defined for all numbers \( t \) . Then \( G\left( {u\left( t\right) }\right) \) has a derivative for all \( t \) and\n\n\[{\left\lbrack G\left( u\left( t\right) \right) \right\rbrack... | Proof: The argument is similar to that for the exponential chain rule. The difference is that we now have a general function \( G\left( u\right) \) rather than the specific functions \( {e}^{u} \) . We argue only for \( u \) an increasing function, and we need Theorem 4.2.1, The Derivative Requires Continuity.\n\nLet \... | Yes |
Compute the derivative of\n\n\\[ F\\left( t\\right) = {e}^{\\sqrt{\\ln t}} \\; t > 1 \\; \\text{ so that } \\; \\ln t > 0.\\]\n | We peel the layers off from the outside. \\( F\\left( t\\right) \\) can be thought of as\n\n\\[ F\\left( t\\right) = G\\left( {H\\left( {K\\left( t\\right) }\\right) }\\right) ,\\; \\text{where} \\; G\\left( z\\right) = {e}^{z}, \\; H\\left( x\\right) = \\sqrt{x}, \\; \\text{and} \\; K\\left( t\\right) = \\ln t \\]\n\n... | Yes |
Example 6.2.3 Find \( \frac{dy}{dt} \) for \( y\left( t\right) = {\left( 1 + {t}^{4}\right) }^{7}.y\left( t\right) \) is the composition of \( G\left( u\right) = {u}^{7} \) and \( u\left( t\right) = 1 + {t}^{4} \) . | \[ \frac{dG}{du} = \frac{d}{du}{u}^{7}\; = 7{u}^{6} \] \[ \frac{du}{dt} = \frac{d}{dt}\left( {1 + {t}^{4}}\right) = 4{t}^{3} \] \[ \frac{dG}{dt} = \frac{dG}{du}\frac{du}{dt}\; = 7{u}^{6} \times 4{t}^{3} = 7{\left( 1 + {t}^{4}\right) }^{6}4{t}^{3} \] | Yes |
The derivative of the inverse of a function. If \( g \) is an invertible function that has a nonzero derivative and \( h \) is its inverse, then for every number, \( t \), in the domain of \( g \) , | \[ {g}^{\prime }\left( {h\left( t\right) }\right) = \frac{1}{{h}^{\prime }\left( t\right) }\;\text{ and }\;{g}^{\prime }\left( t\right) = \frac{1}{{h}^{\prime }\left( {g\left( t\right) }\right) }.\]\n\nIf \( g \) is an invertible function and \( h \) is its inverse, then for every number, \( t \), in the domain of \( g... | Yes |
Example 6.3.3 The function, \( h\left( t\right) = \sqrt{t}, t > 0 \) is the inverse of the function, \( g\left( x\right) = {x}^{2}, x > 0 \) . | \[ {g}^{\prime }\left( x\right) = {2x} \] \[ {h}^{\prime }\left( t\right) = \frac{1}{{g}^{\prime }\left( {h\left( t\right) }\right) } = \frac{1}{{2h}\left( t\right) } = \frac{1}{2\sqrt{t}} \] a result that we obtained directly from the definition of derivative. | Yes |
We illustrate the use of the derivative formulas for sine, cosine, and tangent by computing the derivatives of\n\n\[ \text{a.}y = 3\sin t\cos t\;\text{b.}y = {\sin }^{4}t\;\text{c.}y = \ln \left( {\tan t}\right) \;\text{d.}y = {e}^{\sin t} \] | \[ \text{a.}\;{\left\lbrack 3\sin t\cos t\right\rbrack }^{\prime } = 3\left( {\sin t{\left\lbrack \cos t\right\rbrack }^{\prime } + {\left\lbrack \sin t\right\rbrack }^{\prime }\cos t}\right) \]\n\n\[ = 3\left( {\sin t\left( {-\sin t}\right) + \cos t\cos t}\right) \]\n\n\[ = - 3{\sin }^{2}t + 3{\cos }^{2}t \] | No |
Example 7.2.2 The function \( F\left( t\right) = {e}^{-t/{10}}\sin t \) is an example of ’damped oscillation,’ an important type of vibration. Its graph is shown in Figure 7.4. The peaks and valleys of the oscillation are marked by values of \( t \) for which \( {F}^{\prime }\left( t\right) = 0 \) . We find them by | \[ {\left\lbrack {e}^{-t/{10}}\sin t\right\rbrack }^{\prime } = {e}^{-t/{10}}{\left\lbrack \sin t\right\rbrack }^{\prime } + {\left\lbrack {e}^{-t/{10}}\right\rbrack }^{\prime }\sin t \] \[ = {e}^{-t/{10}}\cos t - \frac{1}{10}{e}^{-t/{10}}\sin t = {e}^{-t/{10}}\left( {\cos t - \frac{1}{10}\sin t}\right) \] Now \( {e}^{... | Yes |
For \( u\left( t\right) = {kt} \) where \( k \) is a constant, | \[ {\left\lbrack \sin \left( kt\right) \right\rbrack }^{\prime } = \cos \left( {kt}\right) \times {\left\lbrack kt\right\rbrack }^{\prime } = \cos \left( {kt}\right) \times k \] | Yes |
Example 7.3.2 With repeated use of the chain rule, derivatives of some rather difficult and exotic functions can be computed. For example, find \( {y}^{\prime } \) for\n\n\[ y\left( t\right) = \ln \left( {\sin \left( {e}^{\cos t}\right) }\right) \] | \[ {y}^{\prime }\left( t\right) = {\left\lbrack \ln \left( \sin \left( {e}^{\cos t}\right) \right) \right\rbrack }^{\prime }\;\text{Outside layer} \]\n\n\[ = \frac{1}{\sin \left( {e}^{\cos t}\right) } \times {\left\lbrack \sin \left( {e}^{\cos t}\right) \right\rbrack }^{\prime }\;G\left( u\right) = \ln u,\;{G}^{\prime ... | Yes |
We show that \( y\left( t\right) = {e}^{-t}\cos {6t} \) solves \( {y}^{\prime \prime }\left( t\right) + 2{y}^{\prime }\left( t\right) + {37y}\left( t\right) = 0 \) . | \[ {y}^{\prime }\left( t\right) = {\left\lbrack {e}^{-t}\cos t\right\rbrack }^{\prime } \] \[ = {\left\lbrack {e}^{-t}\right\rbrack }^{\prime }\cos {6t} + {e}^{-t}{\left\lbrack \cos 6t\right\rbrack }^{\prime } \] \( \left( i\right) \) \[ = \left( {-{e}^{-t}}\right) \cos {6t} + {e}^{-t}\left( {-\sin {6t}}\right) \times ... | Yes |
Example 7.3.4 A searchlight is \( {400}\mathrm{\;m} \) from a straight beach and rotates at a constant rate once in two minutes. How fast is the beam moving along the beach when the beam is \( {600}\mathrm{\;m} \) from the nearest point, A, of the beach to the searchlight. | Solution. See Figure 7.6. Let \( \theta \) measure the rotation of the light with \( \theta = 0 \) when the light is pointing toward A. Let \( x \) be the distance from A to the point where the beam strikes the beach.\n\nWe have\n\[ \tan \left( \theta \right) = \frac{x}{400} \]\n\nBoth \( \theta \) and \( x \) are func... | Yes |
Suppose a body of mass \( m \) is suspended from a spring with spring constant \( k \). If \( m = {20}\mathrm{{gm}} = {0.020}\mathrm{{Kg}} \) and \( k = {0.125} \) Newtons/meter and the initial displacement, \( {y}_{0} = 5 \) \( \mathrm{{cm}} = {0.05}\mathrm{\;m} \), then what is the angular frequency, the equation of ... | \[
\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{{0.125}\mathrm{{Kg}}}{{0.020}\mathrm{{Kg}} - \mathrm{m}/{\mathrm{s}}^{2}/\mathrm{m}}} = {2.5}/\mathrm{s}
\]
and
\[
y\left( t\right) = {0.05}\cos \left( {2.5t}\right)
\]
The period of oscillation is
\[
\frac{2\pi }{\omega } = \frac{2\pi }{{2.51}/\mathrm{s}} = {2.51}\mathr... | Yes |
Suppose \( m = {20}\mathrm{{gm}} = {0.020}\mathrm{\;{kg}},\mathrm{r} = {0.06} \) Newtons/(meter/sec) and \( k = {0.125} \) Newtons/meter. Then the equation of damped motion is\n\n\[ {0.02}{y}^{\prime \prime }\left( t\right) + {0.06}{y}^{\prime }\left( t\right) + {0.125y}\left( t\right) = 0 \] | We show that a solution to this equation is\n\n\[ y\left( t\right) = {e}^{-{1.5t}}\cos \left( {2t}\right) \]\n\n\[ {y}^{\prime }\left( t\right) = {\left\lbrack {e}^{-{1.5t}}\cos \left( 2t\right) \right\rbrack }^{\prime } \]\n\n\[ = {\left\lbrack {e}^{-{1.5t}}\right\rbrack }^{\prime }\cos \left( {2t}\right) + {e}^{-{1.5... | Yes |
Example 8.1.1 A graph of the cubic function, \( P\left( t\right) = {t}^{3} \) is shown in Figure 8.1.1.1 (and in Explore Figure 8.1.1), together with the tangent to the graph at the point \( \left( {0,0}\right) .P\left( t\right) = {t}^{3} \) is an increasing function: If \( a < b \) then \( {a}^{3} < {b}^{3} \) . | However, the tangent to the graph of \( P \) at \( \left( {0,0}\right) \) is horizontal. \( {\left. {P}^{\prime }\left( 0\right) = 3{t}^{2}\right| }_{t = 0} = {30}^{2} = 0 \) . So that \( {P}^{\prime } \) is not everywhere positive. | No |
Example 8.1.2 The natural logarithm, ln, is an increasing function. | By the previous theorem, ln is increasing if \( {\ln }^{\prime } \) is positive. \( \ln x \) is only defined for \( x > 0 \), and\n\n\[{\left\lbrack \ln x\right\rbrack }^{\prime } = \frac{1}{x} > 0\;\text{ for }\;x > 0\] | Yes |
During what time interval is the concentration of penicillin in the blood increasing? What is the maximum concentration of penicillin? | Compute\n\n\[ \n{C}^{\prime }\left( t\right) = {\left\lbrack 5{e}^{-{0.2t}} - 5{e}^{-{0.3t}}\right\rbrack }^{\prime } \n\]\n\n\[ \n= 5{e}^{-{0.2t}}\left( {-{0.2}}\right) - 5{e}^{-{0.3t}}\left( {-{0.3}}\right) \n\]\n\n\[ \n= 5{e}^{-{0.3t}}\left( {-{0.2}{e}^{0.1t} + {0.3}}\right) \n\]\n\nBecause \( \;5 > 0\; \) and \( \;... | Yes |
Theorem 5.2.2. If \( {a}_{1},{a}_{2},\cdots ,{a}_{n} \) is a sequence of \( n \) positive numbers then\n\n\[ \frac{{a}_{1} + {a}_{2} + \cdots {a}_{n}}{n} \geq \sqrt[n]{{a}_{1}{a}_{2}\cdots {a}_{n}} \]\n\nwith equality only when \( {a}_{1} = {a}_{2} = \cdots = {a}_{n} \) . | Proof of Theorem 5.2.2. We proceed by induction. First we prove that if \( {a}_{1} \) and \( {a}_{2} \) are two positive numbers then \( \left( {{a}_{1} + {a}_{2}}\right) /2 \geq \sqrt[2]{{a}_{1}{a}_{2}} \) .\n\n\[ {\left( {a}_{1} - {a}_{2}\right) }^{2} \geq 0 \]\n\n\[ {a}_{1}^{2} - 2{a}_{1}{a}_{2} + {a}_{2}^{2} \geq 0... | Yes |
Theorem 8.2.1 If \( \left( {c, f\left( c\right) }\right) \) is an interior local maximum for a function, \( f \) , and \[ \text{if}{f}^{\prime }\left( c\right) \text{exists, then}\;{f}^{\prime }\left( c\right) = 0\text{.} \] (Equivalently, if the graph of \( f \) has a tangent at an interior local maximum \( \left( {c,... | Proof. Suppose \( \left( {c, f\left( c\right) }\right) \) is an interior local maximum for \( f \), and \( \left( {p, q}\right) \) is an interval in D containing \( c \) for which \( f\left( x\right) \leq f\left( c\right) \) for all \( x \) in \( \left( {p, q}\right) \) . We wish to show that \( {f}^{\prime }\left( c\r... | Yes |
Suppose you are going to make a rectangular box with open top from a 3 meter by 4 meter sheet of tin. One procedure for doing so would be to cut squares of side \( x \) from each corner as shown in Figure 8.6A, and to fold the ’tabs’ up. Four pieces of area \( {x}^{2} \) would be discarded. What value of \( x \) will m... | After the corners are cut, the 'core' of the tin that will make the bottom of the box will be of length \( 4 - {2x} \) and width \( 3 - {2x} \) . The height of the box will be \( x \) and the volume of the box formed will be\n\n\[ V = \left( {4 - {2x}}\right) \left( {3 - {2x}}\right) x\;0 \leq x \leq \frac{3}{2} \]\n\n... | Yes |
Suppose a box with a square base and closed top and bottom is to have a volume of 8 cubic meters. What dimensions of the box will minimize the surface area of the box? | Solution. Let \( x \) be the length of one side of the square base and \( y \) be the height of the box. Then the volume and surface area of the box are\n\n\[ V = {x}^{2}y\;S = 2{x}^{2} + {4xy} \]\n\nBecause \( V \) is specified to be 8 cubic meters\n\n\[ 8 = {x}^{2}y\;\text{ so that }\;y = \frac{8}{{x}^{2}} \]\n\nWe s... | Yes |
Of all cans of volume equal to 1 , which has the smallest surface area? | Assume that the 'can' (cylinder) in Example Figure 8.2.3.3 has volume equal to 1. The area of the top end of the can is \( \pi {r}^{2} \) and the circumference of the top lid is \( {2\pi r} \) . The volume, \( V \), of the can is\n\n\[ V = \pi {r}^{2}h \]\n\nand the surface area, \( S \), of the can is\n\n\[ S = {2\pi ... | Yes |
Snell's Law When you see a fish in a lake it typically is below where it appears to be. A spear, arrow, bullet, rock or other projectile launched toward the image of the fish that you see will pass above the fish. The different speeds of light in air and in water cause the light beam traveling from the fish to your eye... | Pierre Fermat asserted in 1662 that the path of the light beam will be that path that minimizes the total time of travel in the two media. The speed of light in water, \( {v}_{2} \), is about 0.75 times the speed of light in air, \( {v}_{1} \) . Suppose \( d \) is the horizontal distance between your eye and the fish, ... | Yes |
A 10 meter ladder leans against a wall. The foot of the ladder slips horizontally at the rate of 1 meter per minute. At what rate does the top of the ladder descend when the top is 6 meters from the ground? | Draw a picture. See Figure 8.4.1.1. Let \( x \) be the distance from the wall to the foot of the ladder and let \( y \) be the distance from the ground to the top of the ladder. We are asked to find \( {y}^{\prime } \)\n\nat a certain instant. Because the top of the ladder is descending, we expect our answer \( \left( ... | Yes |
In an aqueous solution the concentrations of \( {\mathrm{H}}^{ + } \) and \( {\mathrm{{OH}}}^{ - } \) ions satisfy\n\n\[ \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \left\lbrack {\mathrm{{OH}}}^{ - }\right\rbrack = {10}^{-{14}} \]\n\nIf in a certain lake the \( \mathrm{{pH}} \) is \( 6\left( {\left\lbrack {\mathrm{H}}... | Solution: It is useful to take the \( {\log }_{10} \) of the two sides of the previous equation:\n\n\[ {\log }_{10}\left( {\left\lbrack {\mathrm{H}}^{ + }\right\rbrack \left\lbrack {\mathrm{{OH}}}^{ - }\right\rbrack }\right) = {\log }_{10}{10}^{-{14}} \]\n\n\[ {\log }_{10}\left\lbrack {\mathrm{H}}^{ + }\right\rbrack + ... | Yes |
Solve \( x{e}^{-x} = a \) for \( a = {0.2} \) and \( a = 2 \) . | Graphs help explore the problem. Shown in Figure 8.10A is a graph of \( y = x{e}^{-x} \) . It is apparent that the highest point of the graph has \( y \) -coordinate about \( y = {0.37} \) . We are relieved of solving \( x{e}^{-x} = 2 \) ; there is no solution. The dashed line at \( y = {0.2} \) does intersect the grap... | Yes |
Example 8.7.1 Usually more than one derivative formula may be used in a single step toward computing a derivative. It is important to know which steps are being used, and a slow but sure way to insure proper use is to use but a single formula for each step. We illustrate with \( P\left( t\right) = {\left\lbrack {e}^{{t... | \[ {P}^{\prime }\left( t\right) = {\left\lbrack {\left\lbrack {e}^{{t}^{2}} + \sin t\right\rbrack }^{4}\right\rbrack }^{\prime } \] Notational identity.\n\n\[ = 4{\left\lbrack {e}^{{t}^{2}} + \sin t\right\rbrack }^{3}{\left\lbrack {e}^{{t}^{2}} + \sin t\right\rbrack }^{\prime } \] Equation 8.31\n\n\[ = 4{\left\lbrack {... | Yes |
Find (approximately) the area of the region \( R \) in Figure 9.3B bounded by the graph of the normal distribution, \( f \), the \( \mathrm{X} \) -axis, the vertical line, \( x = - 1 \) and the vertical line, \( x = 1 \) . | 1. Because of the symmetry of the region \( R \) about the \( Y \) -axis, it is sufficient to find the area of that portion of \( R \) that is to the right of the \( Y \) -axis, and then to multiply by 2 . The portion \( R \) to the right of the \( Y \) -axis is shown in Figure 9.4A.\n\n2. Compute the sum of the areas ... | No |
Find the area of the region \( \mathbf{R} \) illustrated in Example Figure 9.2.1.1A that is bounded by the parabola \( \mathbf{y} = {\mathbf{t}}^{\mathbf{2}} \), the \( \mathbf{t} \) -axis, and the line \( \mathbf{t} = \mathbf{x} \) where \( x \) is a positive number. | Solution. The sum of the areas of the rectangles in Figure 9.2.1.1B approximates the area of R. The interval \( \left\lbrack {0, x}\right\rbrack \) on the horizontal axis has been partitioned into \( n \) subintervals by the numbers\n\n\[ 0,\;\frac{x}{n},\;2\frac{x}{n}\;\cdots \;\left( {n - 1}\right) \frac{x}{n},\;n\fr... | Yes |
Find approximately the area of the region \( \mathbf{R} \) illustrated in Example Figure 9.2.2.2A bounded by the cubic \( \mathbf{y} = {\mathbf{x}}^{\mathbf{2}}\left( {\mathbf{2} - \mathbf{x}}\right) \), and the \( \mathbf{X} \) -axis. | The cubic \( y = {x}^{2}\left( {2 - x}\right) = 2{x}^{2} - {x}^{3} \) intersects the \( x \) -axis at \( x = 0 \) and \( x = 2 \) . The rectangles in Figure 9.2.2 partition \( \left\lbrack {0,2}\right\rbrack \) into ten intervals and the sum of the areas of the rectangles\n\napproximates the area of the region \( R \) ... | Yes |
1. First, one may have computed the upper approximating sum to the area of the region bounded by the graphs of \( y = \sqrt{t}, y = 0 \), and \( t = 4 \) using 20 subintervals, as shown in Figure 9.3.1.1. The upper sum is \( {5.51557}\cdots \) . The error box for this computation is shown to the right of the region. Be... | 2. On the other hand, one may wish to compute the area of the region in Figure 9.3.1.1 correct to 0.01 , and need to know how many intervals are required to insure that accuracy. For any number, \( n \), of intervals, \[ \text{The error box is of height}\;\sqrt{4} - \sqrt{0} = 2\;\text{and width}\;\frac{4 - 0}{n} \] Th... | Yes |
Problem. How much work was done? | Solution. The force against the plunger on the \( k \) th step was\n\n\[ \n{F}_{k} \doteq \left( {{P}_{k} - {P}_{0}}\right) \;\mathrm{N}/{\mathrm{{cm}}}^{2} \times {5.5}\;{\mathrm{\;{cm}}}^{2}.\n\]\n\nThe total work done, \( W \), was approximately\n\n\[ \nW \doteq \mathop{\sum }\limits_{{k = 1}}^{9}\left( {{P}_{k} - {... | Yes |
We use Definition 9.4.1 to compute the area of the region \( R \) bounded by \( y = {x}^{2}\left( {2 - x}\right) = 2{x}^{2} - {x}^{3}, y = 0, x = 0 \), and \( x = 2 \) . | \[ {\int }_{0}^{2}2{x}^{2} - {x}^{3}{dx} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\sum }\limits_{{k = 1}}^{n}\left( {2{\left( k\left( 2/n\right) \right) }^{2} - {\left( k\left( 2/n\right) \right) }^{3}}\right) \times 2/n = 2\mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {\mathop{\sum }\limits_{{k... | Yes |
We use Definition 9.4.1 to compute the area of the region bounded by \( y = {e}^{t} \) , \( y = 0, t = 0 \), and \( t = x \) . The region for which we wish to compute the area is shown in Figure 9.4.3.3 and we wish to compute\n\n\[ \n{\int }_{0}^{x}{e}^{t}{dt} \n\] | Assume the interval \( \left\lbrack {0, x}\right\rbrack \) to be partitioned into \( n \) equal subintervals, each of length \( x/n \) and we let \( h = x/n \) . The \( {kth} \) such interval has endpoints on the \( t \) axis at \( \left( {k - 1}\right) \times h \) and \( k \times h \) . The area of the \( k \) th rect... | Yes |
Use the trigonometric identity Bolt out of the Blue! \( {}^{6} \)\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{n}\sin {k\theta } = \frac{-\cos \left( {{n\theta } + \frac{\theta }{2}}\right) + \cos \frac{\theta }{2}}{2\sin \frac{\theta }{2}}\;\text{ and }\;\mathop{\lim }\limits_{{h \rightarrow 0}}\frac{\sin \left( h\right) }{... | \n\n\( {}^{6} \) The Blue: Use \( 2\sin x\sin y = - \cos \left( {x + y}\right) + \cos \left( {x - y}\right) \) . Let \( S = \mathop{\sum }\limits_{{k = 1}}^{n}\sin {k\theta } \) . Then\n\n\[ {2S}\sin \frac{\theta }{2} = 2\mathop{\sum }\limits_{{k = 1}}^{n}\sin {k\theta }\sin \frac{\theta }{2} = \mathop{\sum }\limits_{{... | Yes |
Property 9.5.1 Linearity of the integral. Suppose \( f \) and \( g \) are integrable functions defined on an interval \( \left\lbrack {a, b}\right\rbrack \) and \( c \) is a number. Then\n\n\[ \text{A.}\;{\int }_{a}^{b}\left\lbrack {f\left( t\right) + g\left( t\right) }\right\rbrack {dt} = {\int }_{a}^{b}f\left( t\righ... | The reasons the integral has the two linearity properties is that the approximating sum also has the two properties. The integral being the limit of the approximating sums inherits the linearity properties of the approximating sums. For example, if \( f \) and \( g \) are functions defined on an interval \( \left\lbrac... | Yes |
\[ \text{A.}{\int }_{0}^{1}\left\lbrack {{t}^{2} + t}\right\rbrack {dt}\;\text{B.}{\int }_{0}^{2}2 \times {tdt} \] | A. By Equation 9.17 in Exercise 9.4.1, \( {\int }_{0}^{1}{t}^{2}{dt} = \frac{1}{3} \), and \( {\int }_{0}^{1}{tdt} = \frac{1}{2} \). From Property 9.5.1 A, \[ {\int }_{0}^{1}\left\lbrack {{t}^{2} + t}\right\rbrack {dt} = {\int }_{0}^{1}{t}^{2}{dt} + {\int }_{0}^{1}{tdt} = \frac{1}{3} + \frac{1}{2} = \frac{5}{6} \] B. B... | Yes |
For example, a ball thrown vertically may have a positive velocity as it ascends, but will then have a negative velocity as it descends. If one throws the ball with a vertical velocity of \( {19.6}\mathrm{\;m}/\mathrm{s} \), then the velocity \( t \) seconds later will be \( {19.6} - {9.8}\mathrm{\;{tm}}/\mathrm{s} \) ... | At \( t = 2 \) seconds the velocity is zero, the ball is at its maximum height, and that height is the area of the triangle marked ’ + ’ in Figure 9.16. Between \( t = 2 \) and \( t = 4 \) seconds the velocity is negative, the motion of the ball is downward and the displacement is the area of the triangle marked ’-’ in... | No |
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