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The difference between theoretical complexity and practical efficiency
If I have this pseudocode:
for i=0 to n/2 do
for j=0 to n/2 do
... do anything ....
The number of iterations is $n^2/4$.
What is the complexity of this program? Is it $O(n^2)$?
What is the intuition formal/informal for which is that complexity?
Next, if i have this other pseudocode :
for i=0 to n do
for j=0 to n do
... do anything ....
The complexity is again $O(n^2)$ -- is that correct?
Is there any difference between efficiency in practice and theoretical complexity in this case? Is one of these faster than the other?
• Rule 3 from Rob Pike, as cited in The Art of Unix Programming: "Rule 3. Fancy algorithms are slow when n is small, and n is usually small. Fancy algorithms have big constants. Until you know that n is frequently going to be big, don't get fancy." This is because of the difference between theoretical complexity and practical efficiency, which is ably defined in the answers below. – Wildcard Mar 21 '16 at 3:49
• @Wildcard That seems like a good rule of thumb. Benchmarking several contender algorithms would be even better. – G. Bach Mar 21 '16 at 12:47
• @Wildcard Very interesting quote. It is always interesting to think about the complexity when you're designing/implementing procedures but I usually don't break my head over them untill the procedure in question seems to be a bottleneck. – Auberon Mar 21 '16 at 16:18
Big O Formally
$O(f(n)) = \{g(n) | \exists c > 0, \exists n_0 > 0, \forall n > n_0 : 0 \leq g(n) \leq c*f(n)\}$
Big O Informally
$O(f(n))$ is the set of functions that grow slower (or equally fast) than $f(n)$. This means that whatever function you pick from $O(f(n))$, let's name her $g(n)$, and you pick a sufficiently large $n$, $f(n)$ will be larger than $g(n)$. Or, in even other words, $f(n)$ will eventually surpass any function in $O(f(n))$ when $n$ grows bigger. $n$ is the input size of your algorithm.
As an example. Let's pick $f(n) = n^2$. | {
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As an example. Let's pick $f(n) = n^2$.
We can say that $f(n) \in O(n^3)$. Note that, over time, we allowed notation abuse so almost everyone writes $f(n) = O(n^3)$ now.
Normally when we evaluate algorithms, we look at their order. This way, we can predict how the running time of the algorithm will increase when the input size ($n$) increases.
In your example, both algorithms are of order $O(n^2)$. So, their running time will increase the same way (quadratically) as $n$ increases. Your second algorithm will be four times slower than the first one, but that is something we're usually not interested in **theoretically*. Algorithms with the same order can have different factors ($c$ in the formal definition) or different lower order terms in the function that evaluates the number of steps. But usually that's because of implementation details and we're not interested in that.
If you have a algorithm that runs in $O(log(n))$ time we can safely say it will be faster than $O(n)$ [1] because $log(n) = O(n)$. No matter how bad the $O(log(n))$ algorithm is implemented, no matter how much overhead you put in the algorithm, it will always [1] be faster than the $O(n)$ algorithm. Even if the number of steps in the algorithms are $9999*log(n) = O(log(n))$ and $0.0001*n = O(n)$, the $O(log(n))$ algorithm will eventually be faster [1].
But, maybe, in your application, $n$ is always low and will never be sufficiently large enough so that the $O(log(n))$ algorithm will be faster in practice. So using the $O(n)$ algorithm will result in faster running times, despite $n = O(log(n))$
[1] if $n$ is sufficiently large. | {
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[1] if $n$ is sufficiently large.
• An $O(\log n)$ isn't always faster than an $O(n)$ one. It depends on the value of $n$ and on the hidden constants. A case in point is fast matrix multiplication, which isn't fast at all in practice. – Yuval Filmus Mar 21 '16 at 0:07
• That's why I added: If n is sufficiently large. In the case you mention, if the matrices grow sufficiently large, $O(log(n))$ WILL be faster, by definition – Auberon Mar 21 '16 at 0:09
• Isn't everything you're saying already in my answer? – Auberon Mar 21 '16 at 0:11
• The OP asks about the difference between theory and practice, and you're ignoring that. Asymptotic notation is usually, but not always, useful in practice. Improving the running time fourfold can make a huge difference in practice, but asymptotic notation is completely oblivious to it. – Yuval Filmus Mar 21 '16 at 0:14
Auberon provided a very good explanation of the Big O. I will try to explain what that means for you.
First of you are right. The first code example has $\frac{n^2}{4}$ iterations, but is still in the complexity class O(n^2).
Why?
The thing about complexity classes is, that we assume that doing something several times is not that bad for runtime.
Imagine a Algorithm with complexity O(2^n) running for n=3 and taking 1 second.
We could run this 10 times, and still expect an answer after about 10 seconds.
Now Imagine increasing n by 10. The Programm will take 2^10=1024 seconds.
So the computer scientists basically said: "Man, leading factors are annoying. Let's just assume n grows to infinity and compare functions there"
Which lead to the Big O.
In Practice
In Practice it is very well possible that a solution with much worse complexity runs much faster (for "small" inputs). It is easy to imagina a Programm that runs in O(n) but needs 10^10*n iterations.
It is O(n) but even an O(2^n) solution could be faster for small n.
In summary: | {
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It is O(n) but even an O(2^n) solution could be faster for small n.
In summary:
The Big O is a useful tool. But you still have to think about how fast your algorithm is in practice, since it only describes how much more time it will need if the input grows.
If I have this pseudocode:
for i=0 to n/2 do
for j=0 to n/2 do
... do anything ....
The number of iteration is $n^2/4$.
Actually, it's $(1+n/2)^2 = n^2/4 + n + 1$.
What is the complexity of this program? Is correct $O(n^2)$?
That depends entirely on what "do anything" is. For example, if "do anything" is replaced by the line
for k=0 to 2^n do {}
then the running time is $\Theta(n^22^n)$. | {
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# Find local max, min, concavity, and inflection points
For the function $f(x)=\frac{x^2}{x^2+3}$ Find the intervals on which f(x) is increasing or decreasing.
Find the points of local maximum and minimum of f(x).
Find the intervals of concavity and the inflection points of f(x).
$f′(x)=\frac{6x}{(x^2+3)^2}$
$f''(x)=\frac{-18(x^2-1)}{(x^2+3)^3}$
• I know the critcal point is $x = 0$ if you plug that in to $f(x)$ and $f'(x)$ you get $0$ if you plug it into $f''(x)$ you get $f''(x) = \frac{2}{3}$ does this mean that the local min is $0$ at $x = 0$ ? – Csci319 Oct 31 '14 at 17:13
• Indeed it does. That method is called the "second derivative test" if you want to Google it. – GFauxPas Oct 31 '14 at 17:17
• but does that mean there is no max? how would I find the max? – Csci319 Oct 31 '14 at 17:47
• If the domain of your function is all real numbers, then the function has no maximum. – GFauxPas Oct 31 '14 at 18:05
## 1 Answer
Sign analysis of $f'$
$f'(x)=0$ when $x=0$ so partition the number line as $(-\infty,0)\cup(0,\infty)$.
$(-\infty,0)$: Pick a test point in this interval, say $x=-1$. Note $f'(-1)<0$ so $f'(x)<0$ on this interval. Hence $f$ is decreasing on this interval.
$(0,\infty)$: Pick a test point, say $x=1$. Then $f'(1)>0$ so $f'(x)>0$ on this interval. Hence $f$ is increasing on this interval.
Since $f$ went from decreasing to increasing at $x=0$, a local min occurs at $x=0$.
Sign analysis of $f''$
$f''(x)=0$ when $x=\pm 1$ so partition the number line as $(-\infty,-1)\cup(-1,1)\cup(1,\infty)$.
$(-\infty,-1)$: Pick a test point in this interval, say $x=-2$. Then $f''(-2)<0\implies f''(x)<0\implies f\text{ concave down}$ on this interval
$(-1,1)$: Pick a test point in this interval, say $x=0$. Then $f''(0)>0\implies f''(x)>0\implies f\text{ concave up}$ on this interval
$(1,\infty)$: Pick a test point in this interval, say $x=1$. Then $f''(1)<0\implies f''(x)<0\implies f\text{ concave down}$ on this interval | {
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Since $f$ changed concavity at $x=\pm 1$, these are inflection points.
Indeed a plot of $y=f(x)$ bears out the information above: | {
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# $\dim_K\operatorname{Im}(T^2) =\dim_K\operatorname{Im}(T) \implies \operatorname{Im}(T) \cap \operatorname{Ker}(T) = \{0\}$
I need some help here...
Let $V$ be a vector space over a field $K$, and $T: V \to W$ a linear transformation. Prove that if $$dim_KV \lt \infty \ \ \text{and} \ \ \dim_K\operatorname{Im}(T^2) = \dim_K\operatorname{Im}(T)$$ then $$\operatorname{Im}(T) \cap \operatorname{Ker}(T) = \{0\}$$
This is what I made so far.
Let $x \in \operatorname{Im}(T) \cap \operatorname{Ker}(T)$,where T is linear operator. Then $T(x)=0$ and $x=T(v)$ for some $v \in V$
$0 = T(x) = T(T(v)) = T^2(v) \implies v\in \operatorname{Ker}(T^2)$
My intention is to reach $x=0$, but I'm stuck at this point. I also deduced, via dimension theorem, that $\dim_K \operatorname{Ker}(T) = \dim_K \operatorname{Ker}(T^2)$ but I don't know where it can help. Every hint is appreciated.
• You have $\ker(T) \subset \ker(T^2)$. Now use the dimension. – Maik Pickl Apr 4 '16 at 19:49 | {
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• You have $\ker(T) \subset \ker(T^2)$. Now use the dimension. – Maik Pickl Apr 4 '16 at 19:49
The rank-nullity theorem says \begin{align} \dim V&=\dim\operatorname{Im}(T)+\dim\operatorname{Ker}(T) \\ \dim V&=\dim\operatorname{Im}(T^2)+\dim\operatorname{Ker}(T^2) \end{align} Next use the fact that $$\operatorname{Ker}(T)\subseteq\operatorname{Ker}(T^2)$$ to conclude that $$\operatorname{Ker}(T)=\operatorname{Ker}(T^2)$$
• Oh, I did not know that $\operatorname{Ker}(T)\subseteq\operatorname{Ker}(T^2)$. – Карпатський Apr 4 '16 at 19:59
• @Карпатський If $T(v)=0$, then $T^2(v)=0$. ;-) – egreg Apr 4 '16 at 20:00
• @egreg I haven't done linear algebra for a while, and I'm trying to refresh my memory. How exactly can we conclude that $\operatorname{Ker(T)}=\operatorname{Ker(T^2)}$? Does it follow from the fact that their dimensions are equal and the fact that the one is subset of the other implies that they live in the same space? – K.Power Apr 7 '16 at 23:19
• @K.Power Since $\ker(T)\subseteq\ker(T^2)$, when we show the former has the same dimension as the latter, we conclude they're equal (take a basis of $\ker(T)$: it has the number of elements needed to be a basis of $\ker(T^2)$). – egreg Apr 7 '16 at 23:44 | {
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# What is wrong with my solution? $\int \cos^2 x \tan^3x dx$
I am trying to do this problem completely on my own but I can not get a proper answer for some reason
\begin{align} \int \cos^2 x \tan^3x dx &=\int \cos^2 x \frac{ \sin^3 x}{ \cos^3 x}dx\\ &=\int \frac{ \cos^2 x\sin^3 x}{ \cos^3 x}dx\\ &=\int \frac{ \sin^3 x}{ \cos x}dx\\ &=\int \frac{ \sin^2 x \sin x}{ \cos x}dx\\ &=\int \frac{ (1 -\cos^2 x) \sin x}{ \cos x}dx\\ &=\int \frac{ (\sin x -\cos^2 x \sin x) }{ \cos x}dx\\ &=\int \frac{ \sin x -\cos^2 x \sin x }{ \cos x}dx\\ &=\int \frac{ \sin x }{ \cos x}dx - \int \cos x \sin x dx\\ &=\int \tan x dx - \frac{1}{2}\int 2 \cos x \sin x dx\\ &=\ln|\sec x| - \frac{1}{2}\int \sin 2x dx\\ &=\ln|\sec x| + \frac{\cos 2x}{4} + C \end{align}
This is the wrong answer, I have went through and back and it all seems correct to me.
-
Alright, now I would like to know why the first integral gives $\ln \sec x$ instead of $-\ln \cos x$. – Phira Jun 8 '12 at 23:07
I think that was just an error in typing out the question – user138246 Jun 8 '12 at 23:08
@Phira: Because you can bring the $-1$ up. – Eugene Jun 8 '12 at 23:16
Same function, different expression. Note that $$\ln|\sec x| = \ln\left|\frac{1}{\cos x}\right| = \ln\left(|\cos x|^{-1}\right) = -\ln|\cos x|.$$ And $$\cos 2x = \cos^2x - \sin^2x = \cos^2x-(1-\cos^2x) = 2\cos^2x - 1$$so $$\frac{\cos 2x}{4}+C = \frac{2\cos^2x}{4}-\frac{1}{4}+C = \frac{1}{2}\cos^2x + C'.$$ – Arturo Magidin Jun 8 '12 at 23:16
You're making good progress I see. – Eugene Jun 8 '12 at 23:27
A very simple way to check if the answer to an indefinite integral is correct is to differentiate the answer. If you get the original function, your answer is correct, and is equal, up to a constant, with any other solutions. | {
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We have \begin{align*} \frac{d}{dx}\left(\ln|\sec x| + \frac{\cos 2x}{4} + C\right) &= \frac{1}{\sec x}(\sec x)' + \frac{1}{4}(-\sin(2x))(2x)' + 0\\ &= \frac{\sec x\tan x}{\sec x} - \frac{1}{2}\sin(2x)\\ &= \tan x - \frac{1}{2}\left(2\sin x\cos x\right)\\ &= \frac{\sin x}{\cos x} - \sin x\cos x\\ &= \frac{ \sin x - \sin x\cos^2 x}{\cos x}\\ &= \frac{\sin x(1 - \cos^2 x)}{\cos x}\\ &= \frac{\sin^3 x}{\cos x}\\ &= \frac{\sin ^3 x \cos^2 x}{\cos^3x}\\ &= \frac{\sin^3 x}{\cos^3 x}\cos^2 x\\ &= \tan^3 x \cos^2 x. \end{align*} So your answer is right.
This happens a lot with integrals of trigonometric identities, because there are a lot of very different-looking expressions that are equal "up to a constant". So the answer to $$\int\sin x\cos x\,dx$$ can be either $\sin^2 x + C$ or $-\cos^2x + C$; both correct, though they look different, because they differ by a constant: $-\cos^2x + C = 1-\cos^2x + (C-1) = \sin^2x + (C-1)$.
-
@Always love your answers Prof. – Babak S. Jun 9 '12 at 2:35
$${\ln |\sec x| + \frac{{\cos 2x}}{4} + C}$$
$${\ln \left|\frac 1 {\cos x}\right| + \frac{{1+\cos 2x}}{4} + C}-\frac 1 4$$
$${-\ln \left| {\cos x}\right| + \frac 1 2\frac{{1+\cos 2x}}{2} + K}$$
$${-\ln \left| {\cos x}\right| + \frac 1 2 \cos ^2 x + K}$$
-
this kind of thing makes me feel like Calculus II exams and assignments must be hell to mark. – crf Jun 8 '12 at 23:56
Everything is hell to mark. But one can often structure exam questions to minimize these issues. – André Nicolas Jun 9 '12 at 0:28 | {
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You have been simplifying things up until line 6 and then kind of turned back into complications. It would be natural to notice that $$\sin x dx = -d\left(\cos x\right)$$ Then the integral looks as follows: $$I=-\int\frac{1-\cos^2x}{\cos x}d\left(\cos x\right)$$ So it appears that $\cos x$ plays the role of a variable in its own right, so why not let for example $t=\cos x$. Now $$I=-\int\frac{1-t^2}{t}dt=-\int\frac{dt}{t}+\int tdt=-\ln t+\frac{t^2}{2}+C$$ Now plug $\cos x$ back into place. Recognising distinct "reusable" blocks within the expression is the most natural way to arrive at useful substitutions.
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The calculation is correct. There are many alternate forms of the integral, because of the endlessly many trigonometric identities.
If you differentiate the expression you got and simplify, you will see that you are right.
The answer you saw is likely also right. What was it?
Added: Since $\sec x=\frac{1}{\cos x}$, we have $\ln(|\sec x|)=-\ln(|\cos x|)$.
Also, since $\cos 2x=2\cos^2 x -1$, we have $\frac{\cos 2x}{4}=\frac{\cos^2 x}{2}-\frac{1}{4}$.
So your answer and the book answer differ by a constant. That's taken care of by the arbitrary constant of integration. As a simpler example, $\int 2x\, dx=x^2+C$ and $\int 2x\, dx=(x^2+17)+C$ are both right.
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$\frac{1}{2}cos^2 x - ln|cosx| + C$ – user138246 Jun 8 '12 at 23:18
Notice that:
$$\frac{1}{2} \cos^2 x = \frac{1}{2} \left(\frac{1}{2} + \frac{1}{2} \cos 2x\right) = \frac{1}{4} + \frac{1}{4} \cos 2x$$
And:
$$-\ln|\cos x| = \ln|(\cos x)^{-1}| = \ln|\sec x|$$
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∫tanxdx=ln|secx|=-ln|cosx|+ k since -ln |cos x|+k= ln|(cos x)^-1|+k = ln|sec x| +k
(cos2x)/4= 1/2cos^2(x) - 1/4
k-1/4 = new constant C and together you have the solution.
Your answer seems to be correct it is just manipulation or different approach to trig identities.
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Your answer seems to be correct it is just manipulation or different approach to trig identities.
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A simple $u$-substitution will work even better, IMO. ${\cos^2\theta\tan^3\theta }$ yields ${\frac {\sin^3}{\cos\theta}}$ So starting from there $${\int\frac{\sin^2\theta}{\cos\theta}\sin\theta d\theta}$$ then $${\int\frac{1-\cos^2\theta}{\cos\theta}\sin\theta d\theta}$$ Let $u={\cos \theta}$ then ${du=-\sin\theta d\theta}$ $${-\int\frac {1-u^2}{u}du}$$ let ${v=1-u^2 }$ ${dv=-2u}$
let ${dw=1/u}$, ${w=\ln u}$
thus $${\ln u(1-u^2)+\int 2u\ln u}\,du$$ then
let ${v=\ln u}$, ${dv=u^{-1}}$
let ${dw=2u}$, ${w=u^2}$
then \begin{align*} {\ln u u^2-\int \frac{u^2}{u}\,du}\\{\ln u u^2-\int u\,du}\\ {\ln u(1-u^2)+\ln u u^2-1/2u^2} \end{align*} then \begin{align*}{\ln u(1-u^2+u^2)-1/2u^2}\\ {-(\ln u-1/2u^2)} \end{align*} substituting back yields $${-\ln\cos\theta+1/2\cos^2\theta+c}$$
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# A (small) theatre contains 10 seats. The seats are numbered from 1 to 10,
A (small) theatre contains 10 seats. The seats are numbered from 1 to 10, but not in order. Alice, Bob and Carol have tickets for seats 1, 2 and 3 respectively, and Damien, Eve, Francis, and Gertrude each have tickets without an assigned seat. Unfortunately they all arrive late and the lights are dim, so they can’t see what seat they are sitting in.
a) In how many ways can they be seated?
b) In how many ways can they be seated such that Alice is in her correct seat (eg, seat number 1)?
c) In how many ways can they be seated such that Alice, Bob and Carol are all in their correct seats?
d) In how many ways can they be seated such that none of Alice, Bob, Carol are in their correct seats?
What I did for part A is The first person has 10 seats to choose from, then the second person has 9 and so on. I got $10\times9\times8\times7\times6\times5\times4$ as the answer.
For part B I said alice has 1 seat to choose from and then did the rest like part A so I got $1\times9\times8\times7\times6\times5\times4$.
For C I did the same as B but with 3 people and got $1\times1\times1\times7\times6\times5\times4$ but this doesn't seem right to me.
A bit confused on how to do the rest.
"For c) I did the same as b) but with 3 people and got $7^{\underline{4}}=7\cdot6\cdot5\cdot4$ but this doesn't seem right to me"
Your first three parts are in fact all correct (though they suffer from a lack of convenient notation. If there were 50 people and 80 seats you don't want to waste time writing out all of the numbers in the product. Recommend using one of the following falling factorial notations $(7)_4, 7^{\underline{4}}, \frac{7!}{3!}, ~_7P_4, P(7,4),\dots$ but whichever you use make sure your audience understands the notation) | {
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(Note on falling factorial notation: in all of the above, $(n)_r=n^{\underline{r}}=P(n,r)=\dots=\underbrace{n(n-1)(n-2)\cdots(n-r+2)(n-r+1)}_{r~\text{terms in product starting with}~n}$)
"A bit confused on how to do the rest"
The first three parts are most of the calculations needed to complete the fourth part. Recognize that by symmetry the answer to part b) is also the answer to the question of how many ways Bob is seated in his correct seat as well as the answer to how many ways Carol is in her correct seat.
Use inclusion-exclusion to figure out answer to "In how many ways can they be seated such that at least one of them is in their correct seat." Letting $A,B,C$ represent the events that Alice, Bob, and Carol made it to their correct seats respectively, and letting $\Omega$ represent the sample space i.e. the number of ways in which we don't care who did or didn't make it to their seat, we are tasked with finding the number of ways that none of them made it to their seat. I.e. $|A^c\cap B^c\cap C^c|$
Note that $|A^c\cap B^c\cap C^c| = |\Omega\setminus(A\cup B\cup C)|=|\Omega|-|A\cup B\cup C|$
$=|\Omega|-|A|-|B|-|C|+|A\cap B|+|A\cap C|+|B\cap C|-|A\cap B\cap C|$
You've already found $|\Omega|,|A|,|A\cap B\cap C|$ and by symmetry found $|B|$ and $|C|$. The only information you are missing is $|A\cap B|,|A\cap C|,|B\cap C|$. A similar symmetry argument will reduce the effort it takes to find those. | {
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• so for part d |A∩B|,|A∩C|,|B∩C|, would all be the same and would be calculated by adding P(9,4) to itself. Then I would put everything in the equation you provided? – KGT Nov 7 '16 at 17:52
• @KGT $P(n,r) = \underbrace{n(n-1)(n-2)(n-3)\cdots(n-r+3)(n-r+2)(n-r+1)}_{r~\text{total terms in product starting with}~n}$. I don't think you want to be using $P(9,4)$ – JMoravitz Nov 7 '16 at 17:55
• @KGT you did it correctly in the first three parts. Reworded using the $P$ notation, your answers were $P(10,7),P(9,6)$ and $P(7,4)$ respectively. If we had included the extra question in the middle of those of finding $|A\cap B|$, i.e. alice and bob get their correct seat and then we seat everyone else, how much would that be? – JMoravitz Nov 7 '16 at 18:00
• Oh ok. My mistake. I thought that meant P(9,4) meant 9x8x7x6x5x4. What I should use is P(9,6) correct? – KGT Nov 7 '16 at 18:01
• $P(9,6)$ was the answer to if we only cared about alice getting her correct seat but we didn't care about bob or carol getting their correct seats. If we cared about alice and bob, should that nine have been in the product? – JMoravitz Nov 7 '16 at 18:02 | {
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How to calculate cu...
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# How to calculate cumulative cash flow from a Cash Flow table indexed by year
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I am a relative neophyte in using indexed parameters. I want to calculate a cumulative cash flow by year from a table of cash flows, in order to calculate when a project will break even. I am looking for the syntax of a call on each row of my cash flow list to add it to the previous row of my cumulative cash flow list. Once I have that, I also want to calculate a similar list of discounted cumulative cash flows to understand the effect of discounting the cash flow. Is it worthwhile to create a second index with cash flow, cumulative cash flow, and discounted cash flow? Or is there an even more clever way to do this using features I am unaware of?
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You are looking for the Cumulate function, which takes your cash flow by Year and returns the cumulated cash flow by Year. The syntax is:
Cumulate( Cash_flow, Year )
To discount a cash flow, you just multiply Cash_flow by the discount factor for each year, like this:
Cash_flow * (1 - Discount_rate)^(@Year-1)
Here I've used @Year to number the years starting at 1. By using @Year-1, it means that the first year won't be discounted, whereas if you raise to the power of @Year you would discount the first year. Think about which convention you want. You can then pass this to the first parameter of Cumulate. | {
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You want to compare the undiscounted case to the discounted case. A cool thing is that you can do this without duplicating all the logic for each case. The undiscounted case is simply the discounted case when the discount rate is 0. So, just make Discount_rate a Choice, i.e., between 0% and 7%, and then all your subsequent logic that uses discounted cash flow will work for either, or indeed for both at the same time (by selecting ALL for the choice), giving you the ability to compare any downstream variable without having to do anything additional.
In a short while, I will attach a model that demonstrates all of this and adds in the calculation for the breakeven year.
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This looks great! And the tip on 0% discount is excellent. I look forward to your example, and hope to get back to this work this week.
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This approach worked great. Now I can get to my reason for doing this. I would like to define a variable that finds the year when cumulative cash flow becomes positive, return the year, and add to that the value of cumulative cash flow for that year divided by the cash flow for that year, which gives me a number of years (with a fraction) to break even. Still looking to figure out how to do this.
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Sorry about the delay with posting the model that I had promised. I hit a technical problem. Anyway, here it is:
Cash flow.ana
It includes a variable that finds the breakeven year and a fraction of the breakeven year. I'm not sure if my fraction of the breakeven year is exactly what you just described, but hopefully you can tweak it if not.
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I got taken away on other things (including a short vacation) and am just getting back to this reply. I had in the meantime discovered the Solar Panel model, which gets the breakeven year by using the CondMin function, but you have come at it a separate way. Either way, I am learning useful new functions! I will look at it and see if it gets the same answer I was coming to. Thanks!
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# how to know when a particular proof is appropriate for the given problem?
The main trouble I am currently having in math is knowing when the use cases are appropriate in a proof. I see many videos where they seem to choose a strategy like proof by contrapositive or proof by contradiction, but never quite understand how they came to the conclusion to use that proof strategy.
Here are some examples I have come across and my own solutions to them using the recommended proofs
the product of two odd integers is odd
I used contradiction to solve it
• suppose the product of two odd integers is even
• (2k+1)(2k+1) = 2k
• 2(2k^2 + 2k) + 1 = 2k
• given k is an integer, (2k^2 + 2k) is an integer. Therefore, an even number cannot be an odd number
if x^2 is even, x is even
Based on the wikipedia article on contrapositive
• if x is odd, then x^2 is odd
• (2k+1)^2 == 2(2k^2 +2k) + 1
• therefore, if x^2 is even, then x is even
However, what I am wondering is, is there a general principle as to when to use specific strategies for proofs? Eg, if you specifically know a theory is false, do you choose a strategy accordingly? What determines which strategy will be most effective, or is it arbitrary?
• – Shaun Oct 30 '14 at 12:38
• For "perfect" proofs see the book of Aigner and Ziegler, Proofs from THE BOOK – Dietrich Burde Oct 30 '14 at 12:55
This skill comes with lots of practice. Generally speaking, direct proofs are used when the result is "positive-sounding". In your example, the result is, "the product of two odd integers is odd". This result is positive-sounding (as opposed to a result like, "there is no smallest positive real number"). | {
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For the result, "if $x^2$ is even, then $x$ is even," you could do either a direct proof or a proof by contrapositive. A lot of times it does not matter and there is usually more than one way to prove something. The direct proof, in this case, is favored over a proof by contrapositive simply because it is more, well, direct. Here is how the result would be proven using the contrapositive:
Proof:
Suppose $x$ is odd. We show that $x^2$ is odd. So $x=2k+1$ for some integer $k$. Thus $x^2=(2k+1)^2=(2k+1)(2k+1)=4k^2+4k+1=2(2k^2+2k)+1.$ Now, since $(2k^2+2k)$ is an integer, it follows that $x^2$ is odd.
For a result like "there is no smallest positive real number", this would be difficult to prove using a direct proof. The result is better suited for a proof by contradiction. It would start like this: "Suppose there exists a smallest positive real number…"
Some other examples of negative sounding results that would be more suited for proof by contradiction:
• If a is an even integer and b is an odd integer, then 4 does not divide $a^2+2b^2$
• The integer 100 cannot be written as the sum of three integers, and odd number of which are odd
• The sum of a rational number and an irrational number is irrational.
[Source: Chartrand, G., Polimeni, A.D., Zhang, P., 2013. Mathematical Proofs: A Transition to Advanced Mathematics, 3rd ed. Boston: Pearson.]
• ah, exactly what I was looking for, very clear and understandable, thank you! – corvid Oct 30 '14 at 13:05 | {
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There are many ways to write down a proof. The question is, apart from correctness, which one is more elegant, more concise, more constructive, and last not least easier und thus clearer. In your case, for me the easiast way is to do first an obvious computation, i.e. $$(2k+1)(2l+1)=4kl+2k+2l+1.$$ Your first step here is wrong, because you assume that both integers are equal, i.e., both given by $2k+1$. This would be a special case.
Then, in the second step, we consider everything modulo $2$. The integer on the RHS is odd, because it is of the form $2m+1$, with $m=2kl+k+l$. This finishes the proof.
• theoretically, anything you can prove by contradiction can also be proved by contrapositive, in that sense? – corvid Oct 30 '14 at 12:54 | {
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s′ = required moment of inertia of the combined ring‐ shell‐cone cross section about its neutral axis par-allel to the axis of the shell, in. Résultat de recherche d'images pour "bridge equation for moment of inertia" See more. Find the moment of inertia of the framework about an axis passing through A, and parallel to BC 5ma 2. The second moment of inertia of the entire triangle is the integral of this from $$x = 0$$ to $$x = a$$ , which is $$\dfrac{ma^{2}}{6}$$. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Triangular Cross-Section. RE: second moment of inertia for triangle cross section? the formula. Lecture notes, lecture 11 - Center of gravity, centroid and moment of inertia. The equation for the moment of inertia becomes: ∫ − = − 8 8 2 2 2 2 x' dy' 14 y' I y' 2 8 1 To perform this integration we need to place the integrand in an m-file function and call MATLAB’s quad() function on the m-file. After explaining the term second moment of area, the method of finding moment of inertia of plane figures about x-x or y-y axis is illustrated. You can also drag the origin point at (0,0). Once the moment of inertia was calculated, we had to measure the angular acceleration of the pulley. See the picture: the points of the upper triangle are farther than those of the lower triangle. It is measured by the mass of the body. o The moment of inertia of a thin disc of mass m and radius r about an axis passing through its C. unambiguous choice between the divergent views currently held with regard to the structure. We can see from that the moment of inertia of the subrectangle about the is Similarly, the moment of inertia of the. function Ix_integrand = Moment_Of_Inertia_Integrand(y_prime) %Saved as Moment_Of_Inertia_Integrand. The moment of inertia of the triangle is not half that of the square. 250 kg; from mass A: rB² = 0. A numerical integrator might return slightly less accurate results, but other than that there is not much | {
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integrator might return slightly less accurate results, but other than that there is not much benefit from using symbolic integration there. - Theory - Example - Question 1 - Question 2 - List of moment of inertia for common shapes 4. It is based not only on the physical shape of the object and its distribution of mass but also the specific configuration of how the object is rotating. Thus the mass of the body is taken as a measure of its inertia for translatory. Today we will see here the method to determine the moment of inertia for the triangular section about a line passing through the center of gravity and parallel to the base of the triangular section with the help of this post. The moment of inertia of a triangle with respect to an axis passing through its apex, parallel to its base, is given by the following expression: I = \frac{m h^2}{2} Again, this can be proved by application of the Parallel Axes Theorem (see below), considering that triangle apex is located at a distance equal to 2h/3 from base. pptx), PDF File (. However, "area moment of inertia" is just 4 words to me (no physical meaning). Overview (in Hindi) 8:26 mins. Polar moment of Inertia (Perpendicular Axes theorem) The moment of inertia of an area about an axis perpendicular to the plane of the area is called Polar Moment of Inertia and it is denoted by symbol Izz or J or Ip. b) Determine the moment of inertia for a composite area Parallel-Axis Theorem for an Area Relates the moment of inertia of an area about an axis passing through the. Check to see whether the area of the object is filled correctly. 7) Moment of Inertia Triangle. Mathematical calculations of the GaN NWs' cross-sectional areas and the moment of inertia For the single crystalline (SC) GaN nanowire (NW), the cross-sectional shape is an isosceles triangle with a 63. The median is a line from vertex to the center of a side opposite the vertex. Or the Mizuno MP-20. Find the moment of inertia of the triangle about axis passing | {
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the vertex. Or the Mizuno MP-20. Find the moment of inertia of the triangle about axis passing through centroid perpendicular to lamina. Hodgepodge. The moment of point "B" is 0. The length of each side is L. In case of shafts subjected to torsion or twisting moment, the moment of inertia of the cross-sectional area about its centre O is considered. The sum of all these would then give you the total moment of inertia. 3 (4) 3 Determine the AP whose fourth term is 15 and the difference of 6th term from 10th term is 16 Prove that ratio of area of two triangle is equal to the square of the corresponding sides. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Triangular Cross-Section. About the Moment of Inertia Calculator. The mass moment of inertia is a measure of an object’s resistance to rotation. P-715 with respect to the given axes. 456kg Length of the base of triangle =. Statics - Chapter 10 (Sub-Chapter 10. Answer this question and win exciting prizes. Moments of Inertia Staff posted on October 20, 2006 | Moments of Inertia. The moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to the sum of the moment of inertias of the lamina about the two axes at right angles to each other in its own plane intersecting each other at the point where the perpendicular axis passes through it. Triangle Moment of Inertia. The greater the mass of the body, the greater its inertia as greater force is required to bring about a desired change in the body. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. 31 shows a T-section of dimensions 10 × 10 × 2 cm. The area moment of inertia is also called the second moment of area. Mar 27, 2001 3,923 0 76. G The centroid and centre of gravity are at the same point Where centre of gravity consider to be whole mass of an object act at a point | {
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are at the same point Where centre of gravity consider to be whole mass of an object act at a point C. The sum of the first n ≥ 1 energy levels of the planar Laplacian with constant magnetic field of given total flux is shown to be maximal among triangles for the equilateral triangle, under normalization of the ratio (moment of inertia)/(area) 3 on the domain. Example of Product Moment of Inertia of a Right Angle Triangle Product Moment of Inertia of a Right Angle Triangle by Double Integration. The greater the mass of the body, the greater its inertia as greater force is required to bring about a desired change in the body. The calculations are as shown. Lecture notes, lecture 11 - Center of gravity, centroid and moment of inertia. This app was created to assist you in these calculations and to ensure that your beam section calculations are fast and accurate. It is based not only on the physical shape of the object and its distribution of mass but also the specific configuration of how the object is rotating. Rectangle Triangle. For this reason current vector is treated as normal vector of the plane and the input cloud is projected onto it. The struts are built with the quad-edge passing through the mid-point of the base. 저자: No machine-readable author provided. The mass moment of inertia is often also known as the rotational inertia, and sometimes as the angular mass. Date: 02/03/99 at 14:37:05 From: Doctor Anthony Subject: Re: MI of Solid Cone You must, of course, specify about which axis you want the moment of inertia. Recommended for you. Determine the moment of inertia of this 10. Physics Lab #17 started on 5/13/15 Finding the moment of inertia of a uniform triangle about its center of mass Annemarie Branks Professor Wolf Objective: Find the moment of inertia for a uniform, right triangle plate about its center of mass for the two orientations as shown below. txt) or view presentation slides online. on AIPMT / NEET-UG entrance. I xx = ∫dA. Moment of inertia is | {
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or view presentation slides online. on AIPMT / NEET-UG entrance. I xx = ∫dA. Moment of inertia is a commonly used concept in physics. m in the MATLAB. Area moment of inertia calculation formulas for the regular cross section are readily available in design data handbooks. In physics and applied mathematics, the mass moment of inertia, usually denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. The moment of inertia of a body is always defined about a rotation axis. 4 "Center of Mass" of our text APEX Calculus 3, version 3. The maximum shear. Assume density = 1 Here's my working , I use zx plane projection , but i didnt get the ans. We define dm to be a small element of mass making up the rod. ” Moment of inertia = SI unit of moment of inertia is Q. The calculations are as shown. Mass moment of inertia (also referred to as second moment of mass, angular mass, or rotational inertia) specifies the torque needed to produce a desired angular acceleration about a rotational axis and depends on the distribution of the object’s mass (i. In order to find the moment of inertia of the triangle we must use the parallel axis theorem which ius as follows: The moment of inertia about any axis parallel to that axis through the center. Rotational version of Newton's second law. The apex angle of the quarter-circle is $\pi/2$. The moment of point "C" is the same as "B" so multiply the moment of "B" by two. Cone Calc Processing :. Answer this question and win exciting prizes. 4 Moment of inertia in yaw 2. To find the inertia of the triangle, simply subtract the inertia of the system with the triangle from the benchmark. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Triangular Cross-Section. Moments of Inertia. P-715 with respect to the given axes. Find the moment of inertia of a circular section whose radius is 8” and diameter of 16”. The moment | {
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Find the moment of inertia of a circular section whose radius is 8” and diameter of 16”. The moment of inertia of the triangle is not half that of the square. Determining the moment of inertia of a solid disk. , the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). 5 2 3 A 4-0. Click Content tabCalculation panelMoment of Inertia. Q1: Matthew has a model train that uses a circular cone as a flywheel. Data: 23 d'abril de 2006 (original upload date) Font: No machine-readable source provided. The term product moment of inertia is defined and the mehtod of finding principal moment of inertia is presented. This banner text can have markup. Knowing the area moment of inertia is a critical part of being able to calculate stress on a beam. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Triangular Cross-Section. 1 Expert Answer(s) - 58298 - what is the moment of inertia of a triangular plate ABC of mass M and side BC = a about an axis pass. Which one has the greatest moment of inertia when rotated about an axis perpendicular to the plane of the drawing at point P? Preview this quiz on Quizizz. 4 Moment of inertia in yaw 4 DISCUSSION OF 33TIXYI'ZD Al4D ?XWXQdENPAL VAIJJES 4. 035; Actual VCOG. Area Moment of Inertia or Moment of Inertia for an Area - also known as Second Moment of Area - I, is a property of shape that is used to predict deflection, bending and stress in beams. If you need a beam’s moments of inertia, cross sectional area, centroid, or radius of gyration, you need this app. The oxygen molecule as a mass of 5. 2) A long rod with mass has a moment of inertia , for rotation around an axis near one. Calculating the moment of inertia of a triangle - Duration: 10:01. The moment of inertia must be specified with respect to a chosen axis of rotation. 1st moment of area is area multiplied by the perpendicular distance from the point of line of | {
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1st moment of area is area multiplied by the perpendicular distance from the point of line of action. Figure to illustrate the area moment of a triangle at the list of moments of inertia. 1) Prove that the centroid of any triangle of height h and base b is located 2/3 of the distance from the apex. Own work assumed (based on copyright claims). And, regretfully, you disturbed. The moment of inertia of an object is based on 3 things, the mass of the object, the axis of rotation, and the orientation and distance of the object from the axis of rotation. Inertia is a property of a body to resist the change in linear state of motion. Rotational kinetic energy. The moment of inertia of two or more particles about an axis of rotation is given by the sum of the moment of inertia of the individual particles about the same axis of rotation. Use our free online app Moment of Inertia of a Ring Calculator to determine all important calculations with parameters and constants. Moment of Mass about x and y-axis Mass of Lamina - f(x) Mass of Lamina - f(y) Radius of Gyration (x-axis) Radius of Gyration (y-axis) 1. Tags: Equations for Moment of Inertia. The 2 nd moment of area, or second area moment and also known as the area moment of inertia, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. My teacher told me :. Let us assume that one line is passing through the base of the triangular section and let us consider this line as line BC and we will determine the moment of inertia for the triangular section about this line BC. I = 3 [I cm + M d^2] =3 [ML^2 / 2+ M d^2]. (1) I y: equ. The moment of inertia about the X-axis and Y-axis are bending moments, and the moment about the Z-axis is a polar moment of inertia(J). Learn vocabulary, terms, and more with flashcards, games, and other study tools. PEP Assignment 4 Solutions 3 = = 2 where = 2 Knowing what d and are, CM is CM =,∫ d CM = from the apex of the triangle. The unit | {
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= = 2 where = 2 Knowing what d and are, CM is CM =,∫ d CM = from the apex of the triangle. The unit of dimension of the second moment of area is length. Therefore, equation for polar moment of inertia with respect to apex is. Thus, the moment of inertia of a 2D shape is the moment of inertia of the shape about the Z-axis passing through the origin. I = 1/3 b * h^3 / 12. m2) Title: Microsoft Word - Formular Moment of Inertia Author: d00997 Created Date: 4/25/2019 4:40:32 PM. Triangle h b A= 1 2 b×h x1=b/3 From side x2=2b/3 From right side y1=h/3 From bottom y2=2h/3 From Apex Ixx= bh3 36 Circle d A=π 4 ×d2 x=d/2 y=d/2 I xx= π 64 d4 I yy= π 64 d4 Semicircle A= π 4 ×d2 2 x=d/2 y1=0. Solution 126 2 Polar moment of inertia SECTION 126 Polar Moments of Inertia 15 from COE 3001 at Georgia Institute Of Technology. 51 videos Play all MECHANICAL ENGINEERING 12 MOMENT OF INERTIA / AREA Michel van Biezen Area Moment of Intertia of a Triangle Brain Waves - Duration: 7:23. 2) The radius of the gyration of a disc of radius 25 cm is. 3-axis along the axis of the cone. The moment of inertia is not related to the length or the beam material. Write the equation for polar moment of inertia with respect to apex of triangle. dA Y = 0 A A = b. ) have only areas but no mass. Solution for Find the center of mass and the moment of inertia about the z-axis of a thin shell of constant density d cut from the cone x2 + y2 - z2 = 0 by the… Answered: Find the center of mass and the moment… | bartleby. Calculate the moment of inertia of a right circular cone. Hemmingsen assumed (based on copyright claims). Find the moment of inertia of the table with the iron ring. I = Pi * R^4 / 4. March 2020. The moment of inertia is related to the rotation of the mass; specifically, it measures the tendency of the mass to resist a change in rotational motion about an axis. he solves alone. The centre of area of such figure is known as centroid. Find the moment of inertia of a uniform solid circular cone of mass | {
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figure is known as centroid. Find the moment of inertia of a uniform solid circular cone of mass M, height h and base radius a about its axis, and also about a perpendicular axis through its apex. (by the parallel axis theorem). Figure to illustrate the area moment of a triangle at the list of moments of inertia. Moments of Inertia Staff posted on October 20, 2006 | Moments of Inertia. Area Moment of Inertia or Moment of Inertia for an Area - also known as Second Moment of Area - I, is a property of shape that is used to predict deflection, bending and stress in beams. It should not be confused with the second moment of area, which is used in bending calculations. Moment of Inertia of a Triangular Lamina about its Base. To predict the period of a semi-circle and isosceles triangle with some moment of inertia after first calculating the moment of inertia of the semi-circle and isosceles triangle about a certain axis. Explain the terms moment of inertia and radius of gyration of a plane figure. 89 × 103 kg/m3. Moment of Inertia of Surfaces. Second, finding the moment of inertia when the triangle rotates around its base (shorter leg). Ditto the TaylorMade P730. Basic trig functions 8 Moments of Inertia The moment of inertia is the stiffness of a body due to its size and. ” The product mass and the square of the perpendicular distance from the axis of rotation is known as moment of inertia. ” Moment of inertia = SI unit of moment of inertia is Q. Then remove the middle triangle from each of the re-maining three triangles (as shown), and so on, forever. Section area moment of inertia section modulus calculator what is the difference between polar moment of inertia m area moment of inertia typical cross sections i statics Centroid Area Moments …. The moment of inertia of a triangle of base b and altitude h with respect to a centroidal axis parallel to its base would be bh3/12 bh3/18 bh3/24 bh3/36 The CG of a triangle lies at the point of intersection of diagonals | {
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be bh3/12 bh3/18 bh3/24 bh3/36 The CG of a triangle lies at the point of intersection of diagonals altitudes bisector of angles medians For a solid cone of height h, the CG lies on the axis at a distance above the base equal. The moment of inertia is related to the rotation of the mass; specifically, it measures the tendency of the mass to resist a change in rotational motion about an axis. 01 18-Jun-2003 1. Find I y for the isosceles triangle shown. 32075h^4M/AL, where h is the height of the triangle and L is the area. Inertia is a property of a body to resist the change in linear state of motion. 1 GradedProblems Problem 1 (1. The moment of point "C" is the same as "B" so multiply the moment of "B" by two. Undeniable momentum, on any stage - anywhere. Centroid, Area, and Moments of Inertia Yong-Ming Li January, 1997 1 Introduction This design document was originally written for computation of area, centroid, and moments of inertia of lamina (a thin plate of uniform density). s′ = required moment of inertia of the combined ring‐ shell‐cone cross section about its neutral axis par-allel to the axis of the shell, in. The moment of inertia of a triangle with respect to an axis passing through its apex, parallel to its base, is given by the following expression: I = \frac{m h^2}{2} Again, this can be proved by application of the Parallel Axes Theorem (see below), considering that triangle apex is located at a distance equal to 2h/3 from base. The inertia matrix (aka inertia tensor) of a sphere should be diagonal with principal moments of inertia of 2/5 mass since radius = 1. 1501 Laura Duncan Road, Apex, NC 27502 Email us (919) 289-9278 MAIL TO: P. May 17, 2019 Mirielle Sabety, Keane Wong, Anthony Moody Purpose: The purpose of today's lab is to measure the moment of inertia of a triangle about it's center of mass with in 2 different orientations. Find the moment of inertia of the empty rotating table. I), must be found indirectly. For a homogenous bar of length L and | {
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of the empty rotating table. I), must be found indirectly. For a homogenous bar of length L and mass M, the moment of inertia about center of mass is (1/3)ML^2. Moments of Inertia 10. 250 kg; from mass A: rB² = 0. Doing the same procedure like above, and below is the work. Using the lower left. This moment of inertia about 0 is called polar moment of inertia or moment of inertia about pole. The moment of inertia is a measure of the resistance of a rotating body to a change in motion. Planar and polar moments of inertia formulas. Area Moment of Inertia Section Properties of Triangle Calculator and Equations. 31 shows a T-section of dimensions 10 × 10 × 2 cm. If you need a beam’s moments of inertia, cross sectional area, centroid, or radius of gyration, you need this app. The moment of inertia is de ned as I= X i m ir 2 i (2) for a collection of point-like masses m ieach at a distance r ifrom the speci ed axis. 2) Find the distance for each intersection points. Choose the origin at the apex of the cone, and calculate the elements of the inertia tensor. Using these moment of inertia, we can subtract from it the moment of inertia of just the system without the triangle to obtain our experimental values for the triangle in either. Radius of gyration 3. Find Moment of Inertia of a Ring Calculator at CalcTown. Centroids and moments of inertia. Radius and elevation of the semi-circle can be changed with the blue point. In other words, it is rotating laterally, similar to how a beam from a lighthouse rotates. 100% Upvoted. We spin the triangle around the spot marked "X", which is one of the balls. Calculate the Inertia of the semi-circle around the pivot. Polar moment of inertia is equal to the sum of inertia about X-axis and Y-axis. In this worksheet, we will practice finding the moment of inertia and radius of gyration of a solid and using the parallel axis theorem to find the moment of inertia of a composite solid at different axes. anybody here could help, please? i | {
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the moment of inertia of a composite solid at different axes. anybody here could help, please? i would really appreciate it. Meanwhile, I did find the integral formula for computing the center of pressure (Fox) and calculated it using both a flat bottom and inverted isosceles triangle and then using the "area moment of inertia". They are; Axis passing through the centroid. Constant angular momentum when no net torque. Answer Save. This engineering calculator will determine the section modulus for the given cross-section. I am unable to find it. Another solution is to integrate the triangle from an apex to the base using the double integral of r^2dm, which becomes (x^2+y^2)dxdy. It is analogous to mass in that it is a measure of the resistance a body offers to torque or rotational motion. It is used over and over for examples, since it offers a readymade right angle, a hypotenuse, and other great parts. For the section shown in Fig. 1 In the case of mass and area, the problem is deciding the distance since the mass and area are not concentrated at one point. m2) ) x 10-6 (kg. Consider the diagram as below: We can think of the triangle is composing of infinitesimally. In the final stage of the calculation, you specify the direction of the load forces. 75L, just find the area of the left triangle on the shear diagram and subtract the area of 1/2 the horizontal distance of the second triangle (not 1/2 the area of the second triangle). To find the moment of inertia of the entire section, we integrate the above expression and get, Iyy = ΣdAx2, Ixx = ΣdAy2 and Izz = ΣdAz2. The moment of inertia integral is an integral over the mass distribution. Tension Members. Polar moment of inertia is equal to the sum of inertia about X-axis and Y-axis. Mass = m and Base = l Angle at the apex is = 90° Find MI of theplane about the y - axis = ? Let, the axis of rotation pass through hypotenuse, considering rotation about hypotenuse you will see triangle. Hollow Cone. The theoretical one | {
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considering rotation about hypotenuse you will see triangle. Hollow Cone. The theoretical one is know the moment of inertia of the triangle plate and applied the parallel axis theorem to found the moment of inertia about a new rotating axis. This is the currently selected item. This banner text can have markup. Mathematically, and where IB " *BA " TIA BA = *B + 7IA Ig = moment of inertia about the base plane I3A = moment of inertia about a base diameter axis 1^ = moment of inertia about the central axis 7. Mechanics of Material (CIV101) Academic year. The moment of inertia is a measure of the resistance of a rotating body to a change in motion. The disk is rotating about its center. Centroids and moments of inertia. Determining the moment of inertia of a rod. a) Define i)Moment of Inertia , ii)radius of gyration b) Define stress ,strain ,Modulus of elasticity c) State formulae to find Moment of Inertia of a triangle about axis passing through its i) Base ii) Apex and iii) centroid d) Define lateral strain, linear strain. The maximum twisting moment a shaft can resist, is the product of the permissible shear stress and (A) Moment of inertia (B) Polar moment of inertia (C) Polar modulus (D) Modulus of rigidly Answer: Option C Question No. Moment of inertia (I1 and I2) along the 1 and 2 axes. of the ozone molecule. The moment of inertia of any triangle may be found by combining the moments of inertia of right triangles about a common axis. r2 x2 y2 Therefore, I z I. Rectangle Triangle. Neutral Axis/Moment of Inertia. moment of inertia. University. Author: No machine-readable author provided. Moment of inertia of pile group. save hide report. 날짜: 2006년 4월 23일 (원본 올리기 일시) 출처: No machine-readable source provided. Evaluation of Moments of Inertia 2008 Waterloo Maple Inc. ) 15 minutes ago The transformer inside of a sound system has 1500 turns in its primary coil windings wrapped around a common iron core with the secondary. “Second moment of an area about an axis is called | {
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around a common iron core with the secondary. “Second moment of an area about an axis is called Moment of inertia. Find The Moment Of Inertia About An Axis That Passes Through Mass A And Is Perpendicular To The. Find its moment of inertia for rotation about the z axis. Simply Supported Beams (Shear & Moment Diagrams) Simply supported beams (also know as pinned-pinned or pinned-roller) are the most common beams for both school and on the Professional Engineers exam. Known : Mass of rod AB (m) = 2 kg. Date: 02/03/99 at 14:37:05 From: Doctor Anthony Subject: Re: MI of Solid Cone You must, of course, specify about which axis you want the moment of inertia. Consider an infinitesimally thin disc of thickness dh, at a distance h from the apex of the cone O. Moment of inertia is defined as:”The sum of the products of the mass of each particle of the body and square of its perpendicular distance from axis. save hide report. The moment of inertia (I) of a body is a measure of its ability to resist change in its rotational state of motion. The area moment of inertia is also called the second moment of area. These came out to be 0. Determine the moment of inertia of the triangular area relative to a line parallel to the base and through the upper vertex in cm^4. Physically, it is a measure of how difficult it is to turn a cross-section about an axis perpendicular to it (the inherent rotational stiffness of the cross-section). a) Define i)Moment of Inertia , ii)radius of gyration b) Define stress ,strain ,Modulus of elasticity c) State formulae to find Moment of Inertia of a triangle about axis passing through its i) Base ii) Apex and iii) centroid d) Define lateral strain, linear strain. because the axis goes through masses B and D their masses doesn't affect to increase the inertia of the system around BD axis. The mass moment of inertia is a measure of an object’s resistance to rotation. For instance, it is easier to find the moment of inertia of a triangle, about an axis | {
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to rotation. For instance, it is easier to find the moment of inertia of a triangle, about an axis which passes through its apex, and parallel to the base, than about any ether axis ; but having found this, we may easily find it about an axis parallel to it which passes through the centre. I = 3 [I cm + M d^2] =3 [ML^2 / 2+ M d^2]. Author: No machine-readable author provided. 12: Inertia due to the Object (kg. After that eccentricity is calculated for the obtained projection. I = 3 [I cm + M d^2] =3 [ML^2 / 2+ M d^2]. Write the equation for polar moment of inertia with respect to apex of triangle. Then this moment of inertia is transferred about the axis passing through the centroid of the given section, using theorem of parallel axis. The moment of inertia $$I_x$$ about the $$x$$-axis for the region $$R$$ is the limit of the sum of moments of inertia of the regions $$R_{ij}$$ about the $$x$$-axis. For a homogenous bar of length L and mass M, the moment of inertia about center of mass is (1/3)ML^2. The concept of a moment of inertia is important in many design and analysis problems encountered in mechanical and civil engineering. CENTER OF GRAVITY, CENTROID AND MOMENT OF INERTIA. The moment of inertia of two or more particles about an axis of rotation is given by the sum of the moment of inertia of the individual particles about the same axis of rotation. 2) A precast concrete floor beam has the cross section shown below. apex angle in the neighborhood of 34°. it is first necessary to consider the rotational moment. Adding moments of inertia 3. The moment of inertia of the triangle is not half that of the square. But I don't know how to do that. The moment of inertia must be specified with respect to a chosen axis of rotation. A triangle cannot have more than one right angle or one obtuse angle, since the sum of all three angles is equal to the sum of two right angles, which is 180° or, in radians, π. I am unable to find it. Ball hits rod angular momentum example. | {
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which is 180° or, in radians, π. I am unable to find it. Ball hits rod angular momentum example. Moment of inertia of a circular section can be calculated by using either radius or diameter of a circular section around centroidal x-axis or y-axis. The moment of inertia of a triangle of base b and altitude h with respect to a centroidal axis parallel to its base would be bh3/12 bh3/18 bh3/24 bh3/36 The CG of a triangle lies at the point of intersection of diagonals altitudes bisector of angles medians For a solid cone of height h, the CG lies on the axis at a distance above the base equal. same object, rotating around a point at the midpoint of its base. 0 cm is made of copper. Look up I for a triangle in your table if you have forgotten. 10 lessons • 1 h 34 m. •Compute the product of inertia with respect to the xyaxes by dividing the section into three rectangles. 4 Locate the centroid of the T-section shown in the Fig. 016 kg ⋅ m2 d. Right: A circle section positioned as per the upper sketch is defined in the calculator as I x-axis , the lower sketch shows I y-axis. It depends on the body's mass distribution and the axis chosen, with larger moments. In this study, we first compute the polar moment of inertia of orbit curves under planar Lorentzian motions and then give the following theorems for the Lorentzian circles: When endpoints of a line segment AB with length a +b move on Lorentzian circle (its total rotation angle is δ) with the polar moment of inertia T, a point X which is collinear with the points A and B draws a Lorentzian. Polar moment of inertia is equal to the sum of inertia about X-axis and Y-axis. Rolling without slipping problems. For a clear understanding of how to calculate moments of inertia using double integrals, we need to go back to the general definition in Section The moment of inertia of a particle of mass about an axis is where is the distance of the particle from the axis. Determining the moment of inertia of a rod. Moment of inertia is | {
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of the particle from the axis. Determining the moment of inertia of a rod. Moment of inertia is defined as:”The sum of the products of the mass of each particle of the body and square of its perpendicular distance from axis. Moments of Inertia. Determine the moment of inertia of this system about an axis passing through one corner of the triangle and perpendicular to the plane of the triangle. Insert the moment of inertia block into the drawing. b) Determine the moment of inertia for a composite area Parallel-Axis Theorem for an Area Relates the moment of inertia of an area about an axis passing through the. o , ,3, Moment of Inertia of Surfaces. y-x O 1 1 • (x, y) r Answer: The polar moment of inertia of a planar region is the moment of inertia about the origin (the axis of rotation is the z-axis). Area Moments of Inertia by Integration • Second moments or moments of inertia of an area with respect to the x and y axes, x ³ yI y ³ xdA 2 2 • Evaluation of the integrals is simplified by choosing dA to be a thin strip parallel to one of the coordinate axes. Letting M be the total mass of the system, we have x ¯ = M y / M. 42441 1in 1 in 1 in 3 in 1 in A 2 A 3 A 1 A 4 16 Centroid and Moment of. 19 The deflection of any rectangular beam simply supported, is (A) Directly proportional to its weight. 41 (a) determine: (i) Moment of inertia about its centroid along (x,y) axis. For each segment defined by two consecutive points of the polygon, consider a triangle with two. If the line l(P, 9) lies in the plane of K through the point P and with direction 9, 0 = 9 ^ 2n, we denote the moment of inertia of K about the line l(P, 6) by I(K, P, 9). Email Print Moment of Inertia of a Triangle. Today we will see here the method to determine the moment of inertia for the triangular section about a line passing through the center of gravity and parallel to the base of the triangular section with the help of this post. Here, distance between apex and centroid is d. apex angle in the | {
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section with the help of this post. Here, distance between apex and centroid is d. apex angle in the neighborhood of 34°. however, i would like to know how you obtain the results. How to calculate polar moment of inertia using Inventor 2014 Hello, I have a problem with calculating polar moment of inertia of a cranshaft with cooperating parts (which I've already assembled in Inventor). Find the moment of inertia of a uniform solid circular cone of mass M, height h and base radius a about its axis, and also about a perpendicular axis through its apex. Now that we have determined the moments of inertia of regular and truncated equilateral triangles, it is time to calculate them for the corresponding right prisms. 1 Verified Answer. 12: Inertia due to the Object (kg. The smallest Moment of Inertia about any axis passes throught the centroid. The struts are built with the quad-edge passing through the mid-point of the base. derivation of inertia of ellipse. Q: the moment of inertia of a thin rod of mass m and length l about an axis through its centre of gravity and perpendicular to its length is a) ml²/4 b) ml²/6 c) ml²/8 d) ml²/12 Q: Which statement is correct: a) Moment of inertia is the second moment of mass or area. Moment of inertia of the equilateral triangle system - Duration: 3:38. The material is homogeneous with a mass density ρ. Today we will see here the method to determine the moment of inertia for the triangular section about a line passing through the center of gravity and parallel to the base of the triangular section with the help of this post. The moment of inertia of any triangle may be found by combining the moments of inertia of right triangles about a common axis. Because there is some frictional torque in the system, the angular acceleration of the system when the mass is descending isn’t the same as when it is ascending. Own work assumed (based on copyright claims). 6-1 Polar moment of inertia POINT C (CENTROID) FROM CASE 5: (I P) c 2 bh. What is | {
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copyright claims). 6-1 Polar moment of inertia POINT C (CENTROID) FROM CASE 5: (I P) c 2 bh. What is the moment of inertia of this rigid body about an axis that is parallel to one side of the triangle and passes through the respective midpoints of the other two sides? a. Thank you User-12527562540311671895 for A2A The moment of inertia of a triangular lamina with respect to an axis passing through its centroid, parallel to its base, is given by the expression $I_{XX}=\frac{1}{36}bh^3$ where $b[/mat. But I don't know how to do that. Find MI of and equilateral triangle of side 2m about its base. Moment of inertia, also called mass moment of inertia or the angular mass, (SI units kg m 2 ) is a measure of an object’s resistance to changes in its rotation rate. Free flashcards to help memorize facts about Moment of Inertia of Different Shapes. Calculate its. (8), derived in the moment of inertia example, the moment of inertia of the disk is = at 5 digits Therefore, the moment of inertia of the disk is 12. Area Moments of Inertia Parallel Axis Theorem • Moment of inertia IT of a circular area with respect to a tangent to the circle, ( ) 4 4 5 4 2 2 4 2 1 r IT I Ad r r r π π π = = + = + • Moment of inertia of a triangle with respect to a. Calculate the triangles moment of inertia when its axis of rotation is located at the right-angled corner. The “narrower” the triangle, the more exact is the formula (2). Calculating the moment of inertia of a triangle - Duration: 10:01. 42×r from base y2=0. Click Content tabCalculation panelMoment of Inertia. We can relate these two parameters in two ways: For a given shape and surface mass density, the moment of inertia scales as the size to the fourth power, on dimensional grounds. Determine the product of inertia of the crosshatched area with respect to the x and y axes. Area moment of inertia calculation formulas for the regular cross section are readily available in design data handbooks. Knowing the area moment of inertia is a | {
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section are readily available in design data handbooks. Knowing the area moment of inertia is a critical part of being able to calculate stress on a beam. Or the Mizuno MP-20. Callaway Ratings & Specs 2019 Head Ratings by Maltby Experts at The GolfWorks View All | View By Brand. Thus, the object’s mass and how it is distributed both affect the mass moment of inertia. After that eccentricity is calculated for the obtained projection. half the value of the moment of inertia about the central axis to the value of the moment of inertia about the base plane. 5 Parallel-Axis Theorem - Theory - Example - Question 1 - Question 2. MSC separate and tethers begin to reel out, where inertia of the 3 MSC, increases with time, and the inertia of CSC, remains constant. This simple, easy-to-use moment of inertia calculator will find moment of inertia for a circle, rectangle, hollow rectangular section (HSS), hollow circular section, triangle, I-Beam, T-Beam, L-Sections (angles) and channel sections, as well as centroid, section modulus and many more results. r2 x2 y2 Therefore, I z I. unambiguous choice between the divergent views currently held with regard to the structure. Mathematically, and where IB " *BA " TIA BA = *B + 7IA Ig = moment of inertia about the base plane I3A = moment of inertia about a base diameter axis 1^ = moment of inertia about the central axis 7. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. Some examples of simple moments of inertia Let's try an easy calculation: what's the moment of inertia of these three balls? Each ball has mass m = 3 kg, and they are arranged in an equilateral triangle with sides of length L = 10 m. , the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). ” or ” A quantity expressing the body’s tendency to resist angular | {
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of a torque (turning force). ” or ” A quantity expressing the body’s tendency to resist angular acceleration, it is equal to sum of product of mass of particles to the square of distances from the axis of rotation. It is always considered with respect to a reference axis such as X-X or Y-Y. 5 1 A 2 3 2. So if I'm interpreting your last formula correctly, your answer seems to be off by a factor of 2. 2500 cm^4; D. 1 Answer to Polar Moments of Inertia Determine the polar moment of inertia I P of an isosceles triangle of base b and altitude h with respect to its apex (see Case 5, Appendix D). Then remove the middle triangle from each of the re-maining three triangles (as shown), and so on, forever. In physics and applied mathematics, the mass moment of inertia, usually denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. The following is a list of second moments of area of some shapes. where I is the moment of inertia and R is the perpendicular distance from the axis of rotation to the slant height of the cone changing dm with density, ρ, we get I = ρ ∫∫∫ R^2 r^2 sin Φ dr dΦ dθ = ρ ∫∫∫ r^4 (sin Φ)^3 dr dΦ dθ. Express the result as a Cartesian vector. The calculations are as shown. •Compute the product of inertia with respect to the xyaxes by dividing the section into three rectangles. Area Moment of Inertia - Filled Right Triangle Solve. 42441 1in 1 in 1 in 3 in 1 in A 2 A 3 A 1 A 4 16 Centroid and Moment of. The moment of inertia of total area A with respect to z axis or pole O is z dI z or dI O or r dA J 2 I z ³r dA 2 The moment of inertia of area A with respect to z axis Since the z axis is perpendicular to the plane of the area and cuts the plane at pole O, the moment of inertia is named "polar moment of inertia". however, i would like to know how you obtain the results. Finding the area of a right triangle is easy and fast. Mechanics of Material (CIV101) Anno | {
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results. Finding the area of a right triangle is easy and fast. Mechanics of Material (CIV101) Anno Accademico. Introduction. 1st moment of area is area multiplied by the perpendicular distance from the point of line of action. I am unable to find it. This banner text can have markup. Angular momentum of an extended object. We will use the parallel axis theorem and we will take the centroid as a reference in this case. First Moment of Area = A x. Sorry to see that you are blocking ads on The Engineering ToolBox! If you find this website valuable and appreciate it is open and free for everybody - please contribute by. That point mass relationship becomes the basis for all other moments of inertia since any object can be built up from a collection. It is also known as the torsional Stiffness Read the Full article here. Answer MOI of a triangle about axis theory through a point along the plane = 2 1 m (A r e a) = 2 1 m (2 l × 2 l ) = 8 1 m l 2 December 26, 2019 Toppr. , the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). 1) - Moment of Inertia by Integration Mechanics Statics Chapter 10. Own work assumed (based on copyright claims). But I don't know how to do that. Worthy of note, in order to solve for the moment of inertia of the right triangular thin plate, we first had to measure the the triangle's mass, base length, and height. For the section shown in Fig. For a clear understanding of how to calculate moments of inertia using double integrals, we need to go back to the general definition in Section The moment of inertia of a particle of mass about an axis is where is the distance of the particle from the axis. function Ix_integrand = Moment_Of_Inertia_Integrand(y_prime) %Saved as Moment_Of_Inertia_Integrand. The moment of inertia is equal to the moment of inertia of the rectangle minus the moment of inertia of the hole which is a circle. Answer Save. Calculating the | {
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rectangle minus the moment of inertia of the hole which is a circle. Answer Save. Calculating the moment of inertia of a triangle - Duration: 10:01. Lab 17: Angular Acceleration Amy, Chris, and Jacob November 22, 2017 Theory/Introduction: The purpose of this lab was to determine the moment of inertia of a right triangle thin plate around its center of mass, for two…. The angle at the apex is 9 0 o. Figure to illustrate the area moment of a triangle at the list of moments of inertia. Find The Moment Of Inertia About An Axis That Passes Through Mass A And Is Perpendicular To The. The moment of inertia of a triangle with respect to an axis passing through its centroid, parallel to its base, is given by the following expression: I=bh^2/36. If you need a beam’s moments of inertia, cross sectional area, centroid, or radius of gyration, you need this app. 30670 Moment of Inertia The same vertical differential element of area is used. Lab 18: Moment of Inertia of a Triangle In Lab 16, we used a rotary device that pushes air through a pair of disks, minimizing frictional torque, and allowing the disks to spin for a long time, almost unimpeded. Home Properties Classical MechanicsMoment of Inertia of a Triangle. Let the mass of the triangle be M. It is concluded that the form of the isoceles triangle is acute, with the. Get the expression of angular acceleration and omega. 000965387 kg*m^2. he solves alone. We can use a numerical integrator, such as MATLAB's integral2, to compute the area moment of inertia in the previous example. The moment of inertia of a triangle rotating on its long side is greater than the moment of inertia of the triangle rotating on the shorter side. The element of area in rectangular coordinate system is given by. If rotated about point O (AO = OB),what is the moment of inertia of the rod. The moment of inertia of a triangle rotating on its long side is greater than the moment of inertia of the triangle rotating on the shorter side. 5 Moment of Inertia | {
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greater than the moment of inertia of the triangle rotating on the shorter side. 5 Moment of Inertia of Composite Areas A similar theorem can be used with the polar moment of inertia. In a continuous bridge, the moment of inertia should follow the moment requirement for a balanced and economical design. Autor: No machine-readable author provided. Diagonal wise mI = ml2/6 base = ml2/24 Find the moment of inertia of the plane about the y-axis. It is based not only on the physical shape of the object and its distribution of mass but also the specific configuration of how the object is rotating. The strut width has been deliberatelty increased to show the geometry. Integration by the area of. For a solid cone the moment of inertia is found by using the given formula; I = 3 / 10 MR 2. This is the currently selected item. Calculate the triangles moment of inertia when its axis of rotation is located at the right-angled corner. Radius and elevation of the semi-circle can be changed with the blue point. The particles are connected by rods of negligible mass. 5 Parallel-Axis Theorem - Theory - Example - Question 1 - Question 2. It is observed that the ratio of to is equal to 3: Assume that both balls are pointlike; that is, neither has any moment of inertia about its own center of mass. Moment of inertia is defined as:”The sum of the products of the mass of each particle of the body and square of its perpendicular distance from axis. A more efficient triangular shape for metal wood clubs or driver clubs is disclosed. Calculate the moment of inertia of the triangle with respect to the x axis. Asked by rrpapatel 2nd November 2018 12:10 AM. Polar Moment of Inertia is a measure of resistibility of a shaft against the twisting. The greater the mass of the body, the greater its inertia as greater force is required to bring about a desired change in the body. Mechanics of Solids Introduction: Scalar and vector quantities, Composition and resolution of vectors, System of units, | {
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Introduction: Scalar and vector quantities, Composition and resolution of vectors, System of units, Definition of space, time, particle, rigid body, force. Now to calculate the moment of inertia of the strip about z-axis, we use the parallel axis theorem. No, the components of the eigenvectors themselves, Axes(:,1)), Axes(:,2), Axes(:,3), are already the cosines of the angles between the three principal axes of inertia respectively and the x, y, and z axes, provided the eigenvectors are normalized. The moment of the large triangle, with side $$2L$$, is $$I_z(2L)$$. The apex of the triangle is at the origin and it is bisected by the x-axis. A triangular section has base 100 mm and 300 mm height determine moment of inertia about 1)MI about axis passing through base 2)MI about axis passing through apex {Ans: 3. And h/3 vertically from reference x-axis or from extreme bottom horizontal line line. Simply Supported Beams (Shear & Moment Diagrams) Simply supported beams (also know as pinned-pinned or pinned-roller) are the most common beams for both school and on the Professional Engineers exam. This triangular shape allows the clubs to have higher rotational moments of inertia in both the vertical and horizontal directions, and a lower center of gravity. Rotations in 2D are about the axis perpendicular to the 2D plane, i. Mass moment of inertia. My teacher told me :. Moment of Inertia - Calculated Values Electrical Design In determining the layout of the electrical design, a broad level view was taken and elaborated on. Supplementary notes for Math 253, to follow Section 13. purdueMET 20,366 views. They will make you ♥ Physics. If the density were a constant, finding the total mass of the lamina would be easy: we would just multiply the density by the area. Moment of Inertia for body about an axis Say O-O is defined as ∑dM*y n 2. Integration by the area of. For this reason current vector is treated as normal vector of the plane and the input cloud is projected onto it. | {
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current vector is treated as normal vector of the plane and the input cloud is projected onto it. About the Moment of Inertia Calculator. Engineering Science Mechanical Engineering Concrete Calculator Bending Moment Similar Triangles Shear Force Civil Engineering Construction Body Diagram Structural Analysis. It is the rotational inertia of the body, which is called. These are the values of principal moment of inertia. More particularly, the present invention relates to a hollow golf club head with a lower center of gravity and a higher moment of inertia. I y 2= ∫ x el dA where el = x dA = y dx Thus, I y = ∫ x2 y dx The sign ( + or - ) for the moment of inertia is determined based on the area. It is not explicitly stated in the output, but the mass is equal to the volume (implicitly using a density of 1), so we would expect diagonal matrix entries of 8/15*PI (1. Own work assumed (based on copyright claims). 156 m y Applying Eq. And the moment of "A" equals zero because it is at a point through which the moment of inertia passes. Lectures by Walter Lewin. More on moment of inertia. Ask Question Asked 4 years, 9 months ago. apex angle in the neighborhood of 34°. After that eccentricity is calculated for the obtained projection. We’re pretty sure the Titleist 620 MB has plenty of workability. 8) I of Disk with a Hole. The moment of inertia about the X-axis and Y-axis are bending moments, and the moment about the Z-axis is a polar moment of inertia(J). 1 Expert Answer(s) - 30625 - calculate the moment of inertia of an equilateral triangle made by three rods each of mass m and len. moment of inertia gives the same I as the body rotates around the axis. The 2 nd moment of area, or second area moment and also known as the area moment of inertia, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. Weld design. Undeniable momentum, on any stage - anywhere. where m is the mass of the object, and r is the | {
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Undeniable momentum, on any stage - anywhere. where m is the mass of the object, and r is the distance from the object to the axis. Microsoft Word - Chapter 12 - Moment of Inertia of an Equilateral Triangle Author: Owner Created Date: 11/21/2019 8:18:19 AM. triangles parallel to the xy plane with side 2a, is centered on the origin with its axis along the z axis. 2 y = 10e-x x y 3. Code for moment_of_inertia and RoPS classes, tests and tutorials. The term product moment of inertia is defined and the mehtod of finding principal moment of inertia is presented. Data: 23 d'abril de 2006 (original upload date) Font: No machine-readable source provided. Moments of Inertia 10. The radius of gyration is the radius at which you could concentrate the entire mass to make the moment of inertia equal to the actual moment of inertia. (**) The object below has a moment of inertia about its center of mass of I=25kg⋅m2. 32075h^4M/AL, where h is the height of the triangle and L is the area. This cone is centered on the z-axis with the apex at the origin, but rotates with respect to the x-axis. Using these moment of inertia, we can subtract from it the moment of inertia of just the system without the triangle to obtain our experimental values for the triangle in either. After determining moment of each area about reference axis, the distance of centroid from the axis is obtained by dividing total moment of area by total area of the composite section. 1 GradedProblems Problem 1 (1. Moment of Inertia Contents Moment of Inertia; Sections; Solids; MOI_Rectangle; MOI_Triangle; MOI_Trapezod; MOI_Circle. Solution 3. Area Moment of Inertia Section Properties: Triangle Calculator. Transfer Formula for Moment of Inertia Transfer Formula for Polar Moment of Inertia Transfer Formula for Radii of Gyration Moment of Inertia Common Shapes; Rectangle Triangle Circle Semicircle Quartercircle Ellipse Center of Mass; Center of Mass (2D) 1. The Moment of Inertia Apparatus MATERIALS 1 Table clamp 1 Weight | {
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of Mass; Center of Mass (2D) 1. The Moment of Inertia Apparatus MATERIALS 1 Table clamp 1 Weight hanger (mass 50g) 1 Long metal rod 1 Length of string 2 Pulleys 1 Level 2 Right angle clamps 1. Second, finding the moment of inertia when the triangle rotates around its base (shorter leg). Angular momentum of an extended object. Ditto the Ping Blueprint. 4 Locate the centroid of the T-section shown in the Fig. Thank you User-12527562540311671895 for A2A The moment of inertia of a triangular lamina with respect to an axis passing through its centroid, parallel to its base, is given by the expression [math]I_{XX}=\frac{1}{36}bh^3$ where [math]b[/mat. pptx), PDF File (. Try this Drag any point A,B,C. What is the moment of inertia of ball about the axis of rotation AB? Ignore cord’s mass. The point where the triangle is right angled is lying at origin. Doing the same procedure like above, and below is the work. 3 Radius of Gyration of an Area 10. Another solution is to integrate the triangle from an apex to the base using the double integral of r^2dm, which becomes (x^2+y^2)dxdy. Click Content tabCalculation panelMoment of Inertia. This engineering calculator will determine the section modulus for the given cross-section. 3× 1 6ML2 = 1 2ML2. The situation is this: I know the moment of inertia with respect to the x axis and with respect to the centroidal x axis because its in the table. To predict the period of a semi-circle and isosceles triangle with some moment of inertia after first calculating the moment of inertia of the semi-circle and isosceles triangle about a certain axis. These came out to be 0. The ratio of moment of inertia about the neutral axis to the distance of the most distant point of section from neutral axis is called as a) Moment of inertia b) section modulus c) polar moment of Apex of the triangle b) mid of the height c) 1/3 of the height d) base of triangle 12. The axis perpendicular to its base. Thus, the moment of inertia of a 2D shape is the | {
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of triangle 12. The axis perpendicular to its base. Thus, the moment of inertia of a 2D shape is the moment of inertia of the shape about the Z-axis passing through the origin. I of a thin rod about its center is ML^2 / 2. The second moment of area, also known as moment of inertia of plane area, area moment of inertia, polar moment of area or second area moment, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. c channel polar moment of inertia. 30670 Moment of Inertia The same vertical differential element of area is used. It should not be confused with the second moment of area, which is used in beam calculations. In order to find the moment of inertia we must use create a dm and take the integral of the moment of inertia of each small dm. Let us apply this formula to the triangle formed by V 1, V 2 and V: Δ𝑉≈𝑉⋅Δ𝛼 (3). How do I calculate the moment of inertia of a right angled triangle about one side? Moment of inertia about a side other than the hypotenuse. What is the moment of inertia of this system about an altitude of the triangle passing through the vertex, if ‘a’ is the size of each side of the triangle ?. Letting M be the total mass of the system, we have x ¯ = M y / M. The moment of point "C" is the same as "B" so multiply the moment of "B" by two. derivation of inertia of ellipse. Find the moment of inertia of the framework about an axis passing through A, and parallel to BC 5ma 2. Favourite answer. • The moment of inertia of a composite area A about a given axis is obtained by adding the moments of inertia of the component areas A 1 , A 2 , A 3 , , with respect to the same axis. It is analogous to mass in that it is a measure of the resistance a body offers to torque or rotational motion. To predict the period of a semi-circle and isosceles triangle with some moment of inertia after first calculating the moment of inertia of the semi-circle and isosceles triangle about a certain axis. 89 | {
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calculating the moment of inertia of the semi-circle and isosceles triangle about a certain axis. 89 × 103 kg/m3. In yesterday's lesson, students completed a lab on center of mass, and they already have a working knowledge of torque. The axis may be internal or external and may or may not be fixed. Engineering Science. Find the moment of inertia about y-axis for solid enclosed by z = (1-x^2) , z= 0 , y = 1 and y = -1. 3) Three particles each of mass 100 g are placed at the vertices of an equilateral triangle of side length 10 cm. equals 1, if there are four piles per row and two rows (figure 7-2), the moment of inertia about the Y-Y axis is given by the following formula. 1st moment of area is area multiplied by the perpendicular distance from the point of line of action. I), must be found indirectly. Moment of inertia Up: Rotational motion Previous: The vector product Centre of mass The centre of mass--or centre of gravity--of an extended object is defined in much the same manner as we earlier defined the centre of mass of a set of mutually interacting point mass objects--see Sect. The moment of inertia of the particle. Moment of inertia, in physics, quantitative measure of the rotational inertia of a body—i. Therefore, equation for polar moment of inertia with respect to apex is. University of Sheffield. For these orbits, consider the following scaled variables q˜i = √qi I, (5) v˜i. Q: the moment of inertia of a thin rod of mass m and length l about an axis through its centre of gravity and perpendicular to its length is a) ml²/4 b) ml²/6 c) ml²/8 d) ml²/12 Q: Which statement is correct: a) Moment of inertia is the second moment of mass or area. Tinker toys allow one to easily construct objects with the same mass but different moments of inertia. Two conditions may be considered. After determining moment of each area about reference axis, the distance of centroid from the axis is obtained by dividing total moment of area by total area of the composite section. You | {
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the axis is obtained by dividing total moment of area by total area of the composite section. You can show the division by drawing solid or. It is always considered with respect to a reference axis such as X-X or Y-Y. | {
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hu215i4c45d 16uebhoaltj0op 71a59wpjnvb5y i0fcgwor6jyv8b a8qpv94ljutw hl8lh0m5b2gz1 pbgmxxx4nx 8a1tcmejeokz gx9mvyz9ezgel jwp1rrde9h6ax2 19sb1sh6t9mn 9etgudw0esw 0qoq538i9x xnzy3novvwd u573jcqhjrwnqg 7yteh42rbyb3j j33ao2f1gbcc 0vmioeu4j0gtvum 6q7lhaq2fv l2455p6pm40a7mo jzze49hb2dgxab wwskcnhe8iiz z480xmmqhdqb7 8pfvwqe2t6dtb0 9ciz5c26pn 1atmcvapjuz lpb65ot49n e7ru916n05wcb21 ntgoy7fzw8 | {
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# Solution to the Exercises on Reducing Fractions to Lowest Terms
Below are the complete solutions and answers to the exercises on reducing fractions to lowest terms. I will not give any tips or methods of shortcuts on doing this because teaching you shortcuts will give you problems in case you forget them. The best thing that you can do is to solve as many related problems as you can and develop shortcuts that work for you. Each person has his own preference in solving procedural problems such as these, so it is important that you discover what’s best for you.
For converting improper fractions to mixed form, I will discuss it in a separate post. Try to see the solutions below and see if you can use these solutions to develop your own method. Honestly, the three examples below on converting improper fractions to mixed form should be enough to teach you how to do it yourself. 🙂
Reducing Fractions to Lowest Terms
1. $\displaystyle \frac{12}{15}$
Solution
$\displaystyle \frac{12 \div 3}{15 \div 3} = \frac{4}{5}$
2. $\displaystyle \frac{18}{24}$
Solution
$\displaystyle \frac{18 \div 6 }{24 \div 6} = \frac{3}{4}$
3. $\displaystyle \frac{21}{49}$
Solution
$\displaystyle \frac{21 \div 7 }{49 \div 7} = \frac{3}{7}$
4. $\displaystyle \frac{56}{72}$
Solution
$\displaystyle \frac{56 \div 8 }{72 \div 8} = \frac{7}{9}$
5. $\displaystyle \frac{26}{65}$
Solution
$\displaystyle \frac{26 \div 13 }{65 \div 13} = \frac{2}{5}$
6. $\displaystyle \frac{18}{32}$
Solution
$\displaystyle \frac{18 \div 2 }{32 \div 2} = \frac{9}{16}$
7. $\displaystyle \frac{38}{95}$
Solution
$\displaystyle \frac{38 \div 19 }{95 \div 19} = \frac{2}{5}$
8. $\displaystyle \frac{32}{12}$
Solution
First, convert to lowest terms:
$\displaystyle \frac{32 \div 4 }{12 \div 4} = \frac{8}{3}$ | {
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Solution
First, convert to lowest terms:
$\displaystyle \frac{32 \div 4 }{12 \div 4} = \frac{8}{3}$
Second, convert to mixed form. Eight divided by 3 is 2 remainder 3. So 2 becomes the whole number, 2 (the remainder) becomes the numerator and 8 becomes the denominator. Therefore, the answer is $2 \frac{2}{3}$.
9. $\displaystyle \frac{16}{84}$
Solution
$\displaystyle \frac{16 \div 4 }{84 \div 4} = \frac{4}{21}$
10. $\displaystyle \frac{39}{24}$
Solution
First, reduce to lowest terms.
$\displaystyle \frac{39 \div 3 }{24 \div 3} = \frac{13}{8}$
Second, convert the answer to mixed form. Thirteen divided by 8 is 1 remainder 5. So 1 becomes the whole number, 5 (the remainder) becomes the numerator of the fraction and 8 becomes the denominator. So the correct answer is $1 \frac{5}{8}$.
11. $\displaystyle \frac{15}{45}$
Solution
$\displaystyle \frac{15 \div 15 }{45 \div 15} = \frac{1}{3}$.
12. $\displaystyle \frac{51}{85}$
Solution
$\displaystyle \frac{51 \div 17 }{85 \div 17} = \frac{3}{5}$
13. $\displaystyle \frac{18}{54}$
Solution
$\displaystyle \frac{18 \div 18 }{54 \div 18} = \frac{1}{3}$
14. $\displaystyle \frac{35}{49}$
Solution
$\displaystyle \frac{35 \div 7}{49 \div 7} = \frac{5}{7}$
15. $\displaystyle \frac{74}{24}$
Solution
First, reduce to lowest terms.
$\displaystyle \frac{74 \div 2 }{24 \div 2} = \frac{37}{12}$
Second, divide 37 by 12. The answer is 3 remainder 1. Now, 3 becomes the whole number, 1 becomes the numerator of the fraction, and 12 becomes the denominator. So, the correct answer is $3 \frac{1}{12}$.
In the next post, we will be talking about multiplying and dividing fractions.
### 6 Responses
1. rooky says:
Sir, when I solved # 2 I got an answer ¼ lowest term of 18/24 correct me if I am wrong because I had two LCM. 1st LCM (2) 18/24 = 9/12; 2nd LCM (3) 9/12 = 1/4
2. Civil Service Reviewer says:
Hi Rooky, | {
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2. Civil Service Reviewer says:
Hi Rooky,
Thank you for your question. In reducing fractions to lowest terms, you are not getting the LCM. You are actually getting the greatest common factor (GCF). You get the LCM if you want to add or subtract two (or more) fractions whose denominators are not equal.
In your solution, the GCF of 9 and 12 is 3. So, you must divide 9 by 3, which is equal to 3.
3. Supercute says:
Sir, i think no. 15 should be 3 1/6
4. Supercute says:
Here’s my computation:
74/24
3
_____
24 / 74
72
___________
r. 4
Convert it to mixed fraction:
4
3 ____
24
Convert to lowest term:
1
3 ___
6
• Civil Service Reviewer says:
Hello supercute,
The answer in the explanation is correct. Even if you convert it first to fraction, 74/24 becomes 3 and 2/24 or 3 and 1/12. 🙂
1. October 12, 2013
[…] Now that you know how to convert fractions to lowest terms, you might want to try the practice test and check your solution and answer. […] | {
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# Does this series violate the decreasing condition of the Integral Test for Convergence?
I'm working on the section involving the Integral Test for Convergence in my calculus II class right now, and I've run into a seeming conflict between the definition of the Integral Test, and the solutions to some of the homework exercises as given by both my professor and the textbook.
According to the definition, my research, and my understanding of the integral test, the integral test can only be used for the series $a_n$ where $a_n = f(n)$ and $f(x)$is positive, continuous and decreasing for all $x \ge N$, where $N$ is the index of $n$. However, there are several problems where $f(x)$ is only decreasing if we add a condition, such as $f'(x) < 0 \Leftarrow\Rightarrow x > 3$, where $N \lt 3$. It seems to me that the Integral Test cannot be used to determine convergence of a series when the function is only decreasing when $x \gt k$ and $k \lt N$, yet the book and my professor apply the test anyway.
For example, with the series:
$$\sum_{n=1}^\infty = \frac{n}{(4n+5)^\frac{3}{2}}$$
If we let $a_n = f(n)$, then for $f(x)$:
• $f(x) \gt 0$ for all $x$ in the domain
• $f(x)$ is continuous for all $x \gt -\frac{5}{4}$
But, $f'(x) \lt 0$ only when $x > \frac{5}{2}$, as shown when testing the critical points with the derivative:
$$f'(x) = \frac{5-2x}{(4x+5)^\frac{5}{2}}$$
The professor notes this in her solution, but instead of ending with that and writing, "The Integral Test cannot be applied because $f(x)$ fails to satisfy the required conditions," she applies the test using the original index for $n$:
$$\int_1^\infty \frac{x}{(4x+5)^\frac{3}{2}} \rightarrow \infty \Rightarrow a_n \text{ Diverges}$$ | {
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$$\int_1^\infty \frac{x}{(4x+5)^\frac{3}{2}} \rightarrow \infty \Rightarrow a_n \text{ Diverges}$$
The textbook reaches the same conclusion. Also, the problems in question are listed under a section where the instructions state, "Confirm that the Integral test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series," implying the test can be used on the subsequent exercises.
Is there a reason the Integral Test for Convergence can be used to test for convergence in these problems, where $N \lt k$ and $f'(x) < 0 \Leftarrow\Rightarrow x \gt k$? Am I missing something, or are the book and my professor wrongly using the Integral Test for these series?
### Other exercises with same result:
$$\bullet \sum_{n=1}^\infty \frac{\ln n}{n^2}$$
$$\bullet \frac{\ln 2}{2} + \frac{\ln 3}{3} + \frac{\ln 4}{4} + \frac{\ln 5}{5} + \frac{\ln 6}{6}$$ | {
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$$\bullet \frac{\ln 2}{2} + \frac{\ln 3}{3} + \frac{\ln 4}{4} + \frac{\ln 5}{5} + \frac{\ln 6}{6}$$
• One way to look at what your book and your professor are doing is that they're using the fact that $\sum_{n=1}^{\infty}a_n$ converges iff $\sum_{n=k}^{\infty}a_n$ converges, for any $k$. – user84413 Jul 14 '14 at 0:20
• I think the idea is more that we don't really care what happens for the first 10, 100, or 1000 terms. We care about what eventually happens. We can add up a finite amount of things no problem; it's the tail end behavior that we really care about. – Cameron Williams Jul 14 '14 at 0:20
• For convergence, doesn't matter. For estimates it does. – André Nicolas Jul 14 '14 at 0:22
• Thank you for your responses; I agree that the behavior is properly demonstrated with the test, and I also understand our concern being only with the end behavior of the function, but I don't think it explains why the test is defined as having to be decreasing for all $x \ge N$, if we only care about the last few terms. Wouldn't it be better to say "for most of"? At any rate, if my definition of the test is correct, doesn't this series still violate the conditions, regardless of the validity of the result? – MrCMedlin Jul 14 '14 at 0:34
The point, which isn't made often enough (in my opinion) in classes on the subject, is that the convergence of a series is a limit process. In this case, what that means is that the question of convergence is completely determined by the behavior of the series for say $n > N$ for any fixed, finite $N$. If I take a convergent series $\sum a_n$, and I cut off the first 55 quintillion terms, and replace them all with $n!$ to get a new sequence
$$b_n = \begin{cases} a_n \text{ if } n > 55,000,000,000,000,000,000 \\ n! \text{ if } n \leq 55,000,000,000,000,000,000 \end{cases}$$
Then the sum $\sum b_n$ is still convergent. In fact, $b_n$ is convergent if and only if $a_n$ is. Of course, the sums will be different, but by precisely | {
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$$\sum\limits_{n=1}^{\infty} b_n - \sum\limits_{n=1}^{\infty}a_n = \sum\limits_{n=1}^{55,000,000,000,000,000,000} (n! - a_n)$$
Which is just a finite number. Of course, I've contrived the example to be huge and ridiculous. But the the point to be made about the integral test is that as long as the function behaves in the desired way beyond some large $N$ (say 55 quintillion), the argument still works for the infinite part of the sum, beyond that, and that is where all problems of convergence lie. Everything else is just addition.
• Doesn't this mean the integral should be evaluated from $N$ to $\infty$ instead of from the original index $k$ to $\infty$ for it to be accurate? – MrCMedlin Jul 14 '14 at 15:45
• Again, as long as the integral is of a nice (read continuous) function on the interval $[0, \infty)$. The convergence of the integral is focused in the tail as we can write $$\int_0^\infty f dx = \int_0^N f dx + \int_N^\infty f dx$$ And all of the convergence/divergence behavior of the improper integral is concentrated in the second term on the right. The first term in the right is (assuming the function is continuous up to and including 0), simply some finite integral. This is admittedly a little more subtle. – JHance Jul 14 '14 at 16:18
• OK, this makes a lot more sense now. They just gloss over this in the course without really explaining it ... you've helped a bunch! – MrCMedlin Jul 14 '14 at 16:36
I do not believe your definition of the test is correct. Citing the ever-accurate (tongue-in-cheek) Wikipedia, we find the definition to be:
Consider an integer $N$ and a non-negative function $f$ defined on the unbounded interval $[N, ∞)$, on which it is monotone decreasing. Then the infinite series $$\sum_{n=N}^\infty f(n)$$ converges to a real number if and only if the improper integral $$\int_N^\infty f(x)\,dx$$ is finite. In other words, if the integral diverges, then the series diverges as well. | {
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(Source: http://en.wikipedia.org/wiki/Integral_test_for_convergence)
It is important to note that various authors of textbooks may define a test or theorem in the way that it best fits their course outline. This is particularly true of elementary texts (e.g. first or second year calculus). Remember that mathematicians also like to generalize--if you can take a specific formulation of a theorem and generalize it (e.g. "$f$ must be decreasing everywhere" to "$f$ only has to have decreasing end-behavior"), that's perfectly acceptable.
As a further note in response to your comment on the main question: "decreasing a majority of the time" is different than "decreasing beyond $x=N$." The latter is the option you want.
• Thank you for your response! I've reviewed the Wikipedia entry on the topic as part of my earlier research. You'll note at the beginning it says the function must be monotone decreasing on the interval being considered. This means that for the series $a_n$, $a_1 \ge a_2 \ge a_3 \ge a_4 \ge ... \ge a_n$. We can test for this by looking at the behavior of the first derivative of the function. If the first derivative is ever positive for any value $x$ in our domain, this indicates an increase from one term to the next, which would mean the function is not monotone decreasing as required. – MrCMedlin Jul 14 '14 at 4:08 | {
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# Determine height of box packed with spheres
I got such a wonderful answer regarding The Diagonals of a Regular Octagon, so I thought I'd try asking another question we had on our Pizza and Problem quiz activity at College of the Redwoods. The question was:
A 4 × 4 × h rectangular box contains a sphere of radius 2 and eight smaller spheres of radius 1. The smaller spheres are each tangent to three sides of the box, and the larger sphere is tangent to each of the smaller spheres. What is h?
I'd like to learn how to construct such a model using Mathematica and determine the height h of the box.
Thanks for any help.
Your question is like a sangaku problem. What is the distance between the centre of the large sphere and one of the smaller spheres? The large sphere centre is $\{2,2,h/2\}$, one of the small spheres is at $\{1,1,1\}$. Their separation equals the sum of their radii. That is,
Norm[{2, 2, h/2} - {1, 1, 1}] == 1 + 2
Solve the expression for $h$, and choose the positive root, $h=2(1+\sqrt{7})$.
With[{h = 2 (1 + Sqrt[7])},
Graphics3D[{Orange,
Sphere[{1, 1, 1}, 1], Sphere[{3, 1, 1}, 1], Sphere[{1, 3, 1}, 1],
Sphere[{3, 3, 1}, 1], Sphere[{1, 1, h - 1}, 1],
Sphere[{3, 1, h - 1}, 1],
Sphere[{1, 3, h - 1}, 1], Sphere[{3, 3, h - 1}, 1],
Red, Sphere[{2, 2, h/2}, 2]
}, Lighting -> "Neutral",Axes->True]]
• Tremendous answer. I am having so much fun this morning with the responses on Mathematica Stack Exchange. Thanks you. – David Sep 26 '15 at 19:09 | {
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KennyColnago's answer is good, but we're using Mathematica, so we can leave much more of the problem to it. Now, we can't avoid writing down the system of equations; for simplicity, I'll assume that the large sphere (with radius $R = 2$) is centered at the origin, and that the small spheres (with radius $r = 1$) are centered at ${x, y, z}$. The sides of the box lie at $\pm2$, and the top and bottom are at $\pm h$, so we have tangent small spheres for any combination of positive and negative $x$, $y$ and $z$. That means we know the following:
In[1]:= eqns = {
(* each sphere is tangent to the sides of the box *)
Abs[x] + r == 2,
Abs[y] + r == 2,
(* each sphere is tangent to the top of the box *)
Abs[z] + r == h/2,
(* because the large sphere and small sphere are tangent
to one another, their radii are collinear, so their
total length is the distance of the center of the small
sphere from the origin. *)
r + R == Norm[{x, y, z}],
(* h is positive *)
0 < h
} /. {r -> 1, R -> 2};
Now, we can use Mathematica to solve those equations for $x$, $z$ and $h$.
In[2]:= sol = Simplify@Solve[eqns, {x, y, z, h}, Reals]
Out[2]= {{x -> -1, y -> -1, z -> -Sqrt[7], h -> 2 (1 + Sqrt[7])},
{x -> -1, y -> -1, z -> Sqrt[7], h -> 2 (1 + Sqrt[7])},
{x -> -1, y -> 1, z -> -Sqrt[7], h -> 2 (1 + Sqrt[7])},
{x -> -1, y -> 1, z -> Sqrt[7], h -> 2 (1 + Sqrt[7])},
{x -> 1, y -> -1, z -> -Sqrt[7], h -> 2 (1 + Sqrt[7])},
{x -> 1, y -> -1, z -> Sqrt[7], h -> 2 (1 + Sqrt[7])},
{x -> 1, y -> 1, z -> -Sqrt[7], h -> 2 (1 + Sqrt[7])},
{x -> 1, y -> 1, z -> Sqrt[7], h -> 2 (1 + Sqrt[7])}}
In[3]:= Length@sol
Out[3]= 8
Gratifyingly, we have eight solutions, one for each small sphere.
Now we can draw our spheres, which is easy because Mathematica has helpfully collected all the solutions as a list of rules.
In[3]:= centers = {x, y, z} /. sol
Out[3]= {{-1, -1, -Sqrt[7]}, {-1, -1, Sqrt[7]},
{-1, 1, -Sqrt[7]}, {-1, 1, Sqrt[7]},
{1, -1, -Sqrt[7]}, {1, -1, Sqrt[7]},
{1, 1, -Sqrt[7]}, {1, 1, Sqrt[7]}} | {
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Pulling it all together, we get:
In[4]:= Graphics3D[{
{Lighter@Lighter@Red,
Sphere[centers, r]},
{Lighter@Green,
Sphere[{0, 0, 0}, R]},
Opacity[0],
EdgeForm[Thick],
Parallelepiped[{-2, -2, -h/2},
{{4, 0, 0}, {0, 4, 0}, {0, 0, h}}]} /. sol /. {r -> 1, R -> 2},
Boxed -> False]
EDIT to add: I've updated this answer a couple times, each time to shift more work to Mathematica. In my first attempt, which I didn't post, I got the wrong answer because of a stupid typo when I plugged the radii of the spheres into the Pythagorean theorem; switching over to Norm fixed that. Then I told Solve to assume that $x$ and $z$ were positive (while using the "obvious" fact that $|y| = |x|$ to eliminate an equation before solving), and did some complicated thing with Tuples and Transpose to get all eight spheres. Eliminating those assumptions and recasting everything in terms of absolute values meant Solve took care of all of that automatically.
EDIT again to tweak the Solve expression to specify the domain for Reals and include y, which seems necessary for everything to work right in Mathematica v11.
• @Pilsy Another tremendous answer. I am sharing these with the students and teachers who attended our activity. – David Sep 26 '15 at 19:10 | {
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# Why e is "natural", and why we use radians.
#### ThePerfectHacker
##### Well-known member
Many books give bad answers or no answers at all to why we work with base $e$ and measure angles according to radians in calculus.
Here is what I tell my students, as far as I know, this is the best explanation I seen because it is essentially a one line explanation that is short and to the point.
Why $e$ is natural: Let $b>0$, $b\not = 1$ and consider the function $f(x) = b^x$. This is an exponential function with base $b$ and we can show that $f'(x) = c\cdot b^x$ where $c$ is some constant. In a similar manner we can consider the function $g(x) = \log_b x$, logarithmic function to base $b$, we can show that $g'(x) = k\cdot x^{-1}$ where $k$ is some constant. It would be best and simplest looking derivative formula if those constants, $b$ and $k$, were both equal to one. This happens exactly when $b=e$. Thus, $e$ is natural base for exponent because the derivative formulas for exponential and logarithmic function are as simple as they can be.
Why radians are natural: A "degree" is defined by declaring a full revolution to be $360^{\circ}$, a "radian" is defined by declaring a full revolution to be $2\pi$ radians. More generally, we can define a new angle measurement by declaring a full revolution to be $R$ (for instance, $R=400$, results in something known as gradians). Let us denote $\sin_R x$ and $\cos_R x$ to be the sine and cosine functions of $x$ measured along $R$-units of angles. It can be shown that $(\sin_R x)' = C\cdot(\cos_R x)$ and $(\cos_R x)' = -C\cdot(\sin_R x)$ where $C$ is some constant. It would be best if this constant can be made $C=1$ which happens precisely when $R=2\pi$.
#### chisigma
##### Well-known member
Many books give bad answers or no answers at all to why we work with base $e$ and measure angles according to radians in calculus. | {
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Here is what I tell my students, as far as I know, this is the best explanation I seen because it is essentially a one line explanation that is short and to the point.
Why $e$ is natural: Let $b>0$, $b\not = 1$ and consider the function $f(x) = b^x$. This is an exponential function with base $b$ and we can show that $f'(x) = c\cdot b^x$ where $c$ is some constant. In a similar manner we can consider the function $g(x) = \log_b x$, logarithmic function to base $b$, we can show that $g'(x) = k\cdot x^{-1}$ where $k$ is some constant. It would be best and simplest looking derivative formula if those constants, $b$ and $k$, were both equal to one. This happens exactly when $b=e$. Thus, $e$ is natural base for exponent because the derivative formulas for exponential and logarithmic function are as simple as they can be.
Why radians are natural: A "degree" is defined by declaring a full revolution to be $360^{\circ}$, a "radian" is defined by declaring a full revolution to be $2\pi$ radians. More generally, we can define a new angle measurement by declaring a full revolution to be $R$ (for instance, $R=400$, results in something known as gradians). Let us denote $\sin_R x$ and $\cos_R x$ to be the sine and cosine functions of $x$ measured along $R$-units of angles. It can be shown that $(\sin_R x)' = C\cdot(\cos_R x)$ and $(\cos_R x)' = -C\cdot(\sin_R x)$ where $C$ is some constant. It would be best if this constant can be made $C=1$ which happens precisely when $R=2\pi$.
Let's suppose to use 'non natural' bases for exponents and angles... in this case probably today the following 'beautiful formula' ...
$\displaystyle e^{i\ \theta} = \cos \theta + i\ \sin \theta\ (1)$
... would be not yet known ...
Kind regards
$\chi$ $\sigma$
#### ThePerfectHacker
##### Well-known member
$\displaystyle e^{i\ \theta} = \cos \theta + i\ \sin \theta\ (1)$
=
$$\pi^{i\theta} = \cos \theta + i\sin \theta$$
Provided that we measure angles with full revolution $2\pi \log \pi$. | {
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So you can still have Euler formula for different bases and angles provided that they match together nicely. But the derivative formulas will change into something ugly.
#### Deveno
##### Well-known member
MHB Math Scholar
Forgive me for being obtuse but your statement:
...consider the function $f(x) = b^x$. This is an exponential function with base $b$...
is completely mystifying to me.
In particular, I mean: how does one define such a function for, say: $b = \sqrt{2}$?
(I know the answer, someone who is taking calculus for the first time probably does NOT.)
In fact, proving the "base laws" for the functions $\log_b$ and $\exp_b$ depends on somehow producing the "natural base", which depends on producing the number $e$. So how do you do this?
If, in fact, you define:
$$\displaystyle e = x \in \Bbb R: \int_1^x \frac{1}{t}\ dt = 1$$
and show that for:
$$\displaystyle F(x) = \int_1^x \frac{1}{t}\ dt$$
$F(a) + F(b) = F(ab)$
I am prepared to believe that (given a proof that $F$ is injective):
$F^{-1}(a+b) = F^{-1}(a)F^{-1}(b)$
and that such a function $F^{-1}$ has all the properties of $b^x$ when $x$ is rational.
It follows (in my mind, anyway) that we have:
$e = F^{-1}(1)$.
This, to me, is the reason why $e$ is "natural", I find the definition:
$$\displaystyle e = \lim_{n \to \infty} (1 + n)^{1/n}$$
nearly impossible to motivate, whereas the derivative of a differentiable function $f$ where:
$f(a+b) = f(a)f(b)$ clearly has derivative:
$f'(x) = f'(0)\cdot f(x)$.
At this point, we are in a bit of a fix, which is why considering $f^{-1}$ proves to be more amenable to attack:
$(f^{-1})'(x) = \dfrac{1}{f'(f^{-1}(x))} = \dfrac{1}{f'(0)\cdot f(f^{-1}(x))}$
$= \dfrac{1}{f'(0)x}$
This is a function of the form $g(x) = \dfrac{1}{\alpha x}$, and we can use the Fundamental Theorem of Calculus to find values for $f^{-1}(x)$.
Taking $\alpha = 1$ leads, of course, to the above definition of $e$.
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#### ThePerfectHacker | {
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Last edited:
#### ThePerfectHacker
##### Well-known member
Forgive me for being obtuse but your statement is completely mystifying to me.
It is not mystifying at all. It is just not rigorous. But that is fine. I doubt that Euler himself thought about it the way we think about it today. And what you wrote will have little value to calculus students. You might as well stop teaching trigonometry with that attitude and define sine and cosine as power series, because that is after all the formal way of doing it, which is how Rudin does it in his book on the first pages.
The important part is to motivate the idea then provide the rigor later. So defining exponentials $b^x$ for arbitrary real numbers is okay because it can at least motivate why $e$ is the best number to choose for $b$.
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
In particular, I mean: how does one define such a function for, say: $b = \sqrt{2}$?
I believe, the most natural definition of $\left(\sqrt{2}\right)^a$ is $\lim_{x\to\sqrt{2}}\lim_{y\to a}x^y$ where $x$ and $y$ range over rational numbers.
In fact, proving the "base laws" for the functions $\log_b$ and $\exp_b$ depends on somehow producing the "natural base", which depends on producing the number $e$.
It should be possible to prove $x^{a+b}=x^ax^b$ with the definition above, and this does not require $e$.
#### Deveno
##### Well-known member
MHB Math Scholar
It is not mystifying at all. It is just not rigorous. But that is fine. I doubt that Euler himself thought about it the way we think about it today. And what you wrote will have little value to calculus students. You might as well stop teaching trigonometry with that attitude and define sine and cosine as power series, because that is after all the formal way of doing it, which is how Rudin does it in his book on the first pages. | {
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The important part is to motivate the idea then provide the rigor later. So defining exponentials $b^x$ for arbitrary real numbers is okay because it can at least motivate why $e$ is the best number to choose for $b$.
I'm not sure you understand my point. Let me re-phrase it:
"What does $b^x$ mean for irrational $x$?"
I do not believe rigor is necessary to teach facts, even if the justification cannot be given at that time. It is common for young children to be told that the area of a circle is pi times its radius squared without any justification given. Given the "assumption" that the formula is true, one can still use it profitably, and I see no reason not to under those circumstances.
In a similar vein, I see no problem with teaching the "SOHCAHTOA" approach to trigonometry, either. While that is perfectly adequate for calculating many trig functions of well-established ratios, it is somewhat inadequate for calculating the values of the *continuous* function $\sin(x)$, and when one acquires the mathematical sophistication for a *better* definition, one ought to employ that instead. Power series approximations will do this, but one can use integrals of algebraic functions to obtain inverse trig functions, which also accomplishes the same purpose. A third, and also perfectly acceptable approach, is to use differential equations to define these transcendental functions.
I am not arguing that one *has* to define them this way, just that one *can* and the tools learned in a first-year calculus course allow this. One need not take this path if one feels that the students "aren't ready".
But back to my main point, which I feel is *somewhat* important: how do you define the function:
$f(x) = b^x$
and how do you justify that it is differentiable?
I honestly don't know your answers, but if I was one of your students, I would feel they are pertinent. | {
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I believe, the most natural definition of $\left(\sqrt{2}\right)^a$ is $\lim_{x\to\sqrt{2}}\lim_{y\to a}x^y$ where $x$ and $y$ range over rational numbers.
I'm OK with this. Computing this (without a computer) might prove problematic. This does address continuity satisfactorily, differentiability is another story.
It should be possible to prove $x^{a+b}=x^ax^b$ with the definition above, and this does not require $e$.
I shouldn't think that it would. That is not a "change of base" formula. However, it *is* a good place to start. That said, where does $e$ come in?
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
That said, where does $e$ come in?
I agree with the OP that $e$ is such that $\left(e^x\right)'=e^x$.
#### Deveno
##### Well-known member
MHB Math Scholar
I agree with the OP that $e$ is such that $\left(e^x\right)'=e^x$.
I think we all agree this is true. How do you arrive at that fact?
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
$e$ is defined to be such constat that $\left(e^x\right)'=e^x$.
Now, I have not looked into the issue of defining $e^x$ for a long time. This is how I believe it was introduced to me, and if this approach works, I would consider it very natural.
#### ThePerfectHacker
##### Well-known member
"What does $b^x$ mean for irrational $x$?"
Exactly what most people think it means. You replace $x$ by a rational number $q$ which is close to $x$ and then compute $b^q$. That is what every student would do if you ask them to find $2^{\pi}$. That is what Euler thought of it and mathematicians back then. So I can write $b^x$ function without rigorously defining it.
and how do you justify that it is differentiable?
I do not. I just write it and then ask students what will the derivative of this function be? | {
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All that stuff you wrote before it useless to make a first time student who sees the natural exponential to understand what makes it special to other exponential functions. He will get lost in the middle and not understand what you are doing it. My approach does not suffer from that. Now if you tell a student the natural exponential is its own derivative while other exponentials are almost their own derivative they see what makes the natural exponent the nicer base to use.
#### Deveno
##### Well-known member
MHB Math Scholar
Let me see if I have this straight: you differentiate a function without any justification that it is differentiable. Why do I think this is not a good idea?
But...granted; establishing that the conditions to apply some of the theorems one uses from elementary calculus might be "more technical stuff" than a given student may need or want. Let's move on.
How does one establish that:
$b^x = e^{x\log(b)}$?
I mean, I'm willing to accept that as a DEFINITION of $b^x$, but then I'd sure like to know what $e$ is.
Now, there is "that other definition" of $e$ (which is actually the first definition I ever encountered), but...
Again, you're asking people to take a lot "on faith". There is nothing wrong with this, per se, but calculus affords an opportunity to actually PROVE things that previously they were only TOLD. This moves the source of their knowledge from: "external authority" (books, teacher, expert in field) to: "internal understanding" (I know it's true because of other things I know are true).
In one view, math is another form of intellectual slavery. In another, it is intellectual freedom. I suppose it's obvious which one I favor. | {
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Let me put this in another light: it is (or so I am arguing) preferable to have things we reason about be on "firm footing" (because arguments based on incorrect assumptions can be invalid, even if the reasoning itself is sound). University students are (by and large) people who have reached the age of majority, and as such are responsible for their own actions. This means (among other things) they will be expected to think for themselves. I feel it is irresponsible in a classroom setting, therefore, to "think what you're told to think is true", as it delays the level of responsibility they will be held to by society.
Engineers, for example, will be held accountable for the safety of structures they design or certify. Not making them aware of the mathematical principles that lay beneath their calculations could conceivably lead to fatalities. If I were an engineer, I would want to be **** certain I had a minimum of assumptions, to limit my own potential liability. If there were a typo in a formula in a handbook of mathematical expressions I had used, it is possible that would not help me much in a civil lawsuit.
Granted, few of your students may face such an extreme situation, and such a situation is even less likely to rest upon a definition of an exponential function. Nevertheless, I believe in rigor as a principle, not for its own sake as a form of intellectual prowess, but because of what it accomplishes: showing what we derive is a consequence of simpler facts.
Ignoring the logical priority of mathematical conclusions is not sound mathematics. It neatly sweeps "hard stuff" under the rug, as a matter of expedience. It's like fast-food convenience for the mind, in short: a sham. | {
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Now, it would be one thing if the Fundamental Theorem of Calculus and the Inverse Function Theorem were "abstract generalities" not bothered with in a beginning calculus course. However, this is not so, and I believe (perhaps I am being "too esoteric" in this belief?) that the best way to understanding something is to become acquainted with it through USE.
So, if we have these tools available to define certain functions: why don't we USE them? ESPECIALLY since transcendental functions are difficult to understand EXCEPT through some form of limiting process (and isn't calculus all about the nifty things we can do with limits?).
I believe that teaching people to think for themselves only after they reach graduate school is "too late". Unless they have already individually acquired the skill on their own independently, they are set-up for failure, or at best, a very difficult time of it.
I do understand, however, that as a matter of pragmatism, you may have found that devoting too much time to these sorts of issues has been non-productive. So, you are passing the buck to some other teacher. One can only hope that somewhere, this comes to an abrupt halt.
**************
I didn't mean to obliterate your cognizant observation that $e$ is indeed a "natural" base. It is, and for the reasons you indicate. However, something "deep" is hiding behind it, something important (a fact about this slippery real number $e$). Transcendental real numbers *ought* to be "hard to define", in fact it took centuries before we were in a position to even conclude they existed (that is, that the algebraic numbers are not "cauchy-closed"). | {
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Personally, if all one takes calculus for is to calculate derivatives and integrals, meh: Wolfram|Alpha and a CRC handbook, skip the class. This business about limits and continuity, on the other hand, and the exploration of the "infinitesimal", this sly attempt to touch the face of infinity; if that does not make one blush, I humbly submit one has no soul.
#### ThePerfectHacker
##### Well-known member
Let me see if I have this straight: you differentiate a function without any justification that it is differentiable. Why do I think this is not a good idea?
I only read your first paragraph. I did not read anything else you wrote because I think your teaching standards are abysmal if you have a problem with writing down down $b^x$ in the way that makes sense to most people when they actually think about it.
Historically speaking all the calculus developed in a non-precise way, but it was motivated, and my presentation of it most closely parallels it. Go ahead why not spend the the first two months of class defining Dedekind cuts and what it means to even add two real numbers together. But you do not do that. You say that is too much. So if you are willing to accept that it is okay to teach calculus without defining what addition means, rather appealing to students intuitive feeling about what it is, then why not define exponentials in the most intuitive way possible?
I am willing to bet if you explained to a student why $e$ is natural using my explanation they will have a much better chance of repeating it back to you than your explanation.
#### Deveno
##### Well-known member
MHB Math Scholar
I'm not sure what it is you think I am getting at. It might be a bit more illuminating to read my entire post, if only to provide more context, but I cannot force you to do this.
Not every function is differentiable. Students should not, even if they do not get all the details think this is "mostly true". Continuity is *special* and differentiability even more so. | {
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But again, I do not hear an answer to my original question: assuming, for the time being that $b^x$ *is* differentiable (and we certainly HOPE it would be so), how do you actually differentiate it?
What I imagine is something invoking the chain rule, and writing:
$b^x = e^{x\log(b)} = \exp \circ (x \cdot \log(b))$
and using the product (or multiplying by a constant) rule for the $x \log(b)$ part, and then re-writing:
$(b^x)' = (\exp)'(x \log(b))\cdot \log(b) = \log(b)\cdot b^x$.
To get to that conclusion, we need to know at least these two facts:
1. $(e^x)' = e^x$
2. what the function $\log$ is (in particular that it is a functional *inverse* of $\exp$).
Perhaps you show (or merely indicate the "plausibility" of) the fact that $e^0 = 1$. I would be *interested* in how you derive fact 1. There are some different ways to do this, and you haven't indicated which one you prefer. You also have not indicated how the number $e$ (rather than "some exponential function $\exp$") enters into this. I think that this is critical, since the whole POINT of your initial post is that $e$ represents a "special base".
Fact 2 also needs prior establishment of at least SOME facts about $e^x$ (like, that it is 1-1, so it *has* an inverse, contrast this with the contortions we must undergo for inverse trig functions).
I know that you feel that it is somehow more complicated to start with a way of describing $\log$ and deriving $\exp$ as ITS inverse function, but I really don't see this. The crucial property of $\log$:
$\log(xy) = \log(x) + \log(y)$
is an EASY consequence of the definition:
$$\displaystyle \log(x) = \int_1^x \frac{1}{t}\ dt$$, by taking the interval:
$[1,xy]$ and splitting it into the two sub-intervals of $[1,x]$ and $[x,xy]$, and using the linearity of the integral (presumably you find this property not so challenging as to withhold it from your students). | {
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At least you answered my original question of "what is $b^x$?" by answering it via some kind of "rational approximation". As I indicated to Evgeny, in his reply, I am perfectly OK with that. The real number system was designed to let us do things like that, so that's fair.
How you derive the fact:
$b^x = e^{x\log(b)}$ is a more subtle question. If in fact, you already have the inverse-pair $\exp$ and $\log$ previously established (you don't say if this is true or not), one could use THIS property of (rational, and in approximation, real) exponents:
$(a^b)^c = a^{bc}$.
Perhaps "most" (or even possibly "all") of your students do not care. I cannot change that, nor do I have the singular pleasure of addressing them. I feel you should care, especially since the tone of this thread indicates a pedagogical approach of choosing a "simple case" as desirable.
Historically, the logical foundations of calculus were seen as something of a "crisis", and provided the impetus for some truly worth-while mathematics by Cauchy, Weierstrass, Riemann and Dedekind (to name a few note-worthy examples). The "soundness" of Newton's (and Liebniz's) original justifications were viewed with some suspicion well into the 20th century, although with the advent of "non-standard analysis" these fears have been mostly allayed. Not of all mathematical progress is made at "the outer reaches" sometimes thinking about "the very beginnings" proves fruitful, as well.
If this is not so, then tell me: why do we even "bother" with "epsilon-delta" proofs? I bet even a beginning calculus text makes some (perhaps brief) mention of these strange animals (or so it seems to me with the wealth of questions asked about them on sites such as this one). | {
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Personally, I think talking about Dedekind cuts or Cauchy sequences (or even the oft-maligined "infinite decimal" approach) is a worth-while undertaking, if only to show that these "real numbers" have some "concrete" realization as something we can use rational numbers to approximate (because, underneath it all, the rationals is what we "really" use). I understand if the syllabus you follow does not allow time for this, but it certainly would make "some theorems" less confusing (anything that uses glb's or lub's to establish the existence of some desired real number, for example).
*******************
I do get it, you know, that a "simple" explanation is easier to remember, and that perhaps as much as 80% of what a first-year calculus student learns will be forgotten in 3 months time (unless they *immediately* take a refresher course, or a "building-upon-it" continuation).
If I were a student of yours, asking these same questions, would I get the very same answers?
*******************
I have talked at perhaps too much length about this particular subject. It's hard for me to judge whether this becomes a purely "ad hominem" discussion, so let us abandon it. What I have neglected to say in all of this, is that the SECOND part of your post, is something I WHOLE-HEARTEDLY agree with, and that thinking in terms of "turns" is the "natural" way to measure angle (and very profitable for physicists, and engineers).
#### ThePerfectHacker
##### Well-known member
I'm not sure what it is you think I am getting at. It might be a bit more illuminating to read my entire post, if only to provide more context, but I cannot force you to do this.
Not every function is differentiable. Students should not, even if they do not get all the details think this is "mostly true". Continuity is *special* and differentiability even more so. | {
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But again, I do not hear an answer to my original question: assuming, for the time being that $b^x$ *is* differentiable (and we certainly HOPE it would be so), how do you actually differentiate it?
How do you "prove" that $\sin x$ and $\cos x$ are differenciable? You draw a picture illustrating important limits. That aint a proof at all. However, the picture is more illuminating to understanding the analysis of trigonometric functions than a boring non-inspired unmotivated, but rigorous, power series definition.
How do you "define" $\sqrt{x}$? You have no problem saying that it is that positive number when squared results in $x$. This definition is not exactly rigorous as it assumes the reals has positive square roots. The proof of which requires the completeness property. Something which is never even mentioned in a calculus course at all. You have no problem using that definition at all.
But the moment I write $b^x$ and I say that means raise $b$ to the power of $x$, you suddenly have a problem. You ask how do I even define. I "define" it in the most natural way most people would expect, $b^q$, were $q$ is a rational number approaching $x$. There is nothing natural about writing $\exp(x\log b)$, it is confusing and entirely unmotivated for anyone who sees it for their first time. Now if you applied this standard consistently you will have a problem with how one "defines" $\sin x$ and $\sqrt{x}$. But you do not. You are okay with how it is presented in a calculus class and assume the necessarily properties, including being differenciable, when needed.
In the end you apply selective criticisms and make the concept of exponential functions exponentially more difficult for first-time students because you think it justifies it. Calculus is not supposed to be about justifying analytic concepts. It is supposed to introduce one to them and try to motivate the reasons for why it works and why it is true.
#### Deveno | {
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#### Deveno
##### Well-known member
MHB Math Scholar
How do you "prove" that $\sin x$ and $\cos x$ are differenciable? You draw a picture illustrating important limits. That aint a proof at all. However, the picture is more illuminating to understanding the analysis of trigonometric functions than a boring non-inspired unmotivated, but rigorous, power series definition.
How do you "define" $\sqrt{x}$? You have no problem saying that it is that positive number when squared results in $x$. This definition is not exactly rigorous as it assumes the reals has positive square roots. The proof of which requires the completeness property. Something which is never even mentioned in a calculus course at all. You have no problem using that definition at all.
But the moment I write $b^x$ and I say that means raise $b$ to the power of $x$, you suddenly have a problem. You ask how do I even define. I "define" it in the most natural way most people would expect, $b^q$, were $q$ is a rational number approaching $x$. Now if you applied this standard consistently you will have a problem with how one "defines" $\sin x$ and $\sqrt{x}$. But you do not. You are okay with how it is presented in a calculus class and assume the necessarily properties, including being differenciable, when needed.
In the end you apply selective criticisms and make the concept of exponential functions exponentially more difficult for first-time students because you think it justifies it. Calculus is not supposed to be about justifying analytic concepts. It is supposed to introduce one to them and try to motivate the reasons for why it works and why it is true.
I think differentiating trig functions is likewise somewhat thorny. Similar glib use of $\pi$ is made in this case, usually without any justification. Again, a "geometric" limit argument is the usual "first look" approach taken, and this is not out of character with the geometric flavor that much of calculus has. | {
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I think that transcendental functions are "subtler" than many people realize...indeed, that even the completeness property of the reals is a subtle concept. These things are usually "glossed over" in favor of expediency. I remain unconvinced that this is a NECESSARY evil. It's convenient, yes, and perhaps even "motivational" (whatever that means), but the truth is kept "out of view".
Many calculus texts start with the assertion that "they will assume the reader is familiar with the basic properties of real numbers" and dispense with 100 years of deep (and HARD) mathematics in a few words.
Square roots are a bit thorny, as well. Many students know (or have at least seen a proof) that $\sqrt{2}$ is irrational, but they have no idea what a big can of worms that really is. It is easy to write $\sqrt{x}$ and think you know what it means. I'm sure it comes as a shock to some math majors when they realize that actually, they do NOT (and maybe non-math-majors who never take anything more demanding than calculus ever realize there is "more to it than that").
I believe in DEMONSTRATING that a function is differentiable, by showing the limit that defines its derivative EXISTS. I think this is good "practice" for later when one encounters a function whose differentiability status is unknown. Not all "real problems" have answer keys in the back of some book.
Of course, you have me at a disadvantage: you know your students, and I do not. I ask you for an explanation of what you mean in your original post, and you reply with remarks about "selective criticisms". | {
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I cannot recall the particular thread, but there was a similar discussion some time back about defining the area of a circle, and arriving at a value for $\pi$. Some argued that a "proper definition" of trig functions would have to be delayed until after power series, as that was logically more satisfiying. I made the point that it could be done with the tools of integral calculus (although this might not be very "direct").
One often sees T-shirts with:
$e^{i\pi} + 1 = 0$
printed on them. The simplicity of this formula belies the depth of information it summarizes: how the circle ties together exponentials, logarithms, trigonometry, and an intrinsic notion of distance in 7 scant symbols. I'm not even confident I understand entirely everything it says...but the small portion I DO get, fills me with a kind of awe.
If, in the final analysis, you feel that I am looking "too deeply" at what you think should be taken more lightly, I have no ready answer. I have my reasons for thinking the way I do, but if my explanations are not sufficient, I suppose that is my failing. I *try* to be clear, but I have no assurances I ever am, for I only see the world with my own perceptions, and never that of another human being.
Indeed, I often fear I might be embarrassing my comrades here at MHB, for being too outspoken, or having views well outside of the conventional wisdom, or even worse, perhaps just plain being wrong. I have only a vague notion of INTERNAL consistency to guide me, and I find it enormously difficult to communicate this to other people. Surely, I reason to myself, Evgeny.Makarov or Ackbach or Jameson or MarkFL or chisigma (or some other soul who I have in my ignorance failed to mention) will point out to me what is wrong with my views, so I may improve my ability to express math to other people...but alas, I hear only silence. | {
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# dsolve
Solve system of differential equations
Support for character vector or string inputs will be removed in a future release. Instead, use syms to declare variables and replace inputs such as dsolve('Dy = -3*y') with syms y(t); dsolve(diff(y,t) == -3*y).
## Description
example
S = dsolve(eqn) solves the differential equation eqn, where eqn is a symbolic equation. Use diff and == to represent differential equations. For example, diff(y,x) == y represents the equation dy/dx = y. Solve a system of differential equations by specifying eqn as a vector of those equations.
example
S = dsolve(eqn,cond) solves eqn with the initial or boundary condition cond.
example
S = dsolve(___,Name,Value) uses additional options specified by one or more Name,Value pair arguments.
example
[y1,...,yN] = dsolve(___) assigns the solutions to the variables y1,...,yN.
## Examples
collapse all
Solve the first-order differential equation $\frac{\mathit{dy}}{\mathit{dt}}=\mathit{ay}$.
Specify the first-order derivative by using diff and the equation by using ==. Then, solve the equation by using dsolve.
syms y(t) a
eqn = diff(y,t) == a*y;
S = dsolve(eqn)
S = ${C}_{1} {\mathrm{e}}^{a t}$
The solution includes a constant. To eliminate constants, see Solve Differential Equations with Conditions. For a full workflow, see Solving Partial Differential Equations.
Solve the second-order differential equation $\frac{{\mathit{d}}^{2}\mathit{y}}{{\mathit{dt}}^{2}}=\mathit{ay}$.
Specify the second-order derivative of y by using diff(y,t,2) and the equation by using ==. Then, solve the equation by using dsolve.
syms y(t) a
eqn = diff(y,t,2) == a*y;
ySol(t) = dsolve(eqn)
ySol(t) = ${C}_{1} {\mathrm{e}}^{-\sqrt{a} t}+{C}_{2} {\mathrm{e}}^{\sqrt{a} t}$
Solve the first-order differential equation $\frac{\mathit{dy}}{\mathit{dt}}=\mathit{ay}$ with the initial condition $y\left(0\right)=5$. | {
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Specify the initial condition as the second input to dsolve by using the == operator. Specifying condition eliminates arbitrary constants, such as C1, C2, ..., from the solution.
syms y(t) a
eqn = diff(y,t) == a*y;
cond = y(0) == 5;
ySol(t) = dsolve(eqn,cond)
ySol(t) = $5 {\mathrm{e}}^{a t}$
Next, solve the second-order differential equation $\frac{{\mathit{d}}^{2}\mathit{y}}{{\mathit{dt}}^{2}}={\mathit{a}}^{2}\mathit{y}$ with the initial conditions $y\left(0\right)=b$ and ${y}^{\prime }\left(0\right)=1$.
Specify the second initial condition by assigning diff(y,t) to Dy and then using Dy(0) == 1.
syms y(t) a b
eqn = diff(y,t,2) == a^2*y;
Dy = diff(y,t);
cond = [y(0)==b, Dy(0)==1];
ySol(t) = dsolve(eqn,cond)
ySol(t) =
$\frac{{\mathrm{e}}^{a t} \left(a b+1\right)}{2 a}+\frac{{\mathrm{e}}^{-a t} \left(a b-1\right)}{2 a}$
This second-order differential equation has two specified conditions, so constants are eliminated from the solution. In general, to eliminate constants from the solution, the number of conditions must equal the order of the equation.
Solve the system of differential equations
$\begin{array}{l}\frac{\mathit{dy}}{\mathit{dt}}=\mathit{z}\\ \frac{\mathit{dz}}{\mathit{dt}}=-\mathit{y}.\end{array}$
Specify the system of equations as a vector. dsolve returns a structure containing the solutions.
syms y(t) z(t)
eqns = [diff(y,t) == z, diff(z,t) == -y];
S = dsolve(eqns)
S = struct with fields:
z: C2*cos(t) - C1*sin(t)
y: C1*cos(t) + C2*sin(t)
Access the solutions by addressing the elements of the structure.
ySol(t) = S.y
ySol(t) = ${C}_{1} \mathrm{cos}\left(t\right)+{C}_{2} \mathrm{sin}\left(t\right)$
zSol(t) = S.z
zSol(t) = ${C}_{2} \mathrm{cos}\left(t\right)-{C}_{1} \mathrm{sin}\left(t\right)$ | {
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When solving for multiple functions, dsolve returns a structure by default. Alternatively, you can assign solutions to functions or variables directly by explicitly specifying the outputs as a vector. dsolve sorts the outputs in alphabetical order using symvar.
Solve a system of differential equations and assign the outputs to functions.
syms y(t) z(t)
eqns = [diff(y,t)==z, diff(z,t)==-y];
[ySol(t),zSol(t)] = dsolve(eqns)
ySol(t) = ${C}_{1} \mathrm{cos}\left(t\right)+{C}_{2} \mathrm{sin}\left(t\right)$
zSol(t) = ${C}_{2} \mathrm{cos}\left(t\right)-{C}_{1} \mathrm{sin}\left(t\right)$
Solve the differential equation $\frac{\partial }{\partial t}y\left(t\right)={e}^{-y\left(t\right)}+y\left(t\right)$. dsolve returns an explicit solution in terms of a Lambert W function that has a constant value.
syms y(t)
eqn = diff(y) == y+exp(-y)
eqn(t) =
sol = dsolve(eqn)
sol = ${\mathrm{W}\text{lambertw}}_{0}\left(-1\right)$
To return implicit solutions of the differential equation, set the 'Implicit' option to true. An implicit solution has the form $F\left(y\left(t\right)\right)=g\left(t\right)$.
sol = dsolve(eqn,'Implicit',true)
sol =
$\left(\begin{array}{c}\left({\int \frac{{\mathrm{e}}^{y}}{y {\mathrm{e}}^{y}+1}\mathrm{d}y|}_{y=y\left(t\right)}\right)={C}_{1}+t\\ {\mathrm{e}}^{-y\left(t\right)} \left({\mathrm{e}}^{y\left(t\right)} y\left(t\right)+1\right)=0\end{array}\right)$
If dsolve cannot find an explicit solution of a differential equation analytically, then it returns an empty symbolic array. You can solve the differential equation by using MATLAB® numerical solver, such as ode45. For more information, see Solve a Second-Order Differential Equation Numerically.
syms y(x)
eqn = diff(y) == (x-exp(-x))/(y(x)+exp(y(x)));
S = dsolve(eqn)
Warning: Unable to find symbolic solution.
S =
[ empty sym ] | {
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S =
[ empty sym ]
Alternatively, you can try finding an implicit solution of the differential equation by specifying the 'Implicit' option to true. An implicit solution has the form $F\left(y\left(x\right)\right)=g\left(x\right)$.
S = dsolve(eqn,'Implicit',true)
S =
${\mathrm{e}}^{y\left(x\right)}+\frac{{y\left(x\right)}^{2}}{2}={C}_{1}+{\mathrm{e}}^{-x}+\frac{{x}^{2}}{2}$
Solve the differential equation $\frac{\mathit{dy}}{\mathit{dt}}=\frac{\mathit{a}}{\sqrt{\mathit{y}}}+\mathit{y}$ with condition $y\left(a\right)=1$. By default, dsolve applies simplifications that are not generally correct, but produce simpler solutions. For more details, see Algorithms.
syms a y(t)
eqn = diff(y) == a/sqrt(y) + y;
cond = y(a) == 1;
ySimplified = dsolve(eqn, cond)
ySimplified =
${\left({\mathrm{e}}^{\frac{3 t}{2}-\frac{3 a}{2}+\mathrm{log}\left(a+1\right)}-a\right)}^{2/3}$
To return the solutions that include all possible values of the parameter $a$, turn off simplifications by setting 'IgnoreAnalyticConstraints' to false.
yNotSimplified = dsolve(eqn,cond,'IgnoreAnalyticConstraints',false)
yNotSimplified =
Solve the second-order differential equation $\left({x}^{2}-1{\right)}^{2}\frac{{\partial }^{2}}{\partial {x}^{2}}y\left(x\right)+\left(x+1\right)\frac{\partial }{\partial x}y\left(x\right)-y\left(x\right)=0$. dsolve returns a solution that contains a term with unevaluated integral.
syms y(x)
eqn = (x^2-1)^2*diff(y,2) + (x+1)*diff(y) - y == 0;
S = dsolve(eqn)
S =
${C}_{2} \left(x+1\right)+{C}_{1} \left(x+1\right) \int \frac{{\mathrm{e}}^{\frac{1}{2 \left(x-1\right)}} {\left(1-x\right)}^{1/4}}{{\left(x+1\right)}^{9/4}}\mathrm{d}x$
To return series solutions of the differential equation around $x=-1$, set the 'ExpansionPoint' to -1. dsolve returns two linearly independent solutions in terms of a Puiseux series expansion.
S = dsolve(eqn,'ExpansionPoint',-1)
S = | {
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S = dsolve(eqn,'ExpansionPoint',-1)
S =
$\left(\begin{array}{c}x+1\\ \frac{1}{{\left(x+1\right)}^{1/4}}-\frac{5 {\left(x+1\right)}^{3/4}}{4}+\frac{5 {\left(x+1\right)}^{7/4}}{48}+\frac{5 {\left(x+1\right)}^{11/4}}{336}+\frac{115 {\left(x+1\right)}^{15/4}}{33792}+\frac{169 {\left(x+1\right)}^{19/4}}{184320}\end{array}\right)$
Find other series solutions around the expansion point $\infty$ by setting 'ExpansionPoint' to Inf.
S = dsolve(eqn,'ExpansionPoint',Inf)
S =
$\left(\begin{array}{c}x-\frac{1}{6 {x}^{2}}-\frac{1}{8 {x}^{4}}\\ \frac{1}{6 {x}^{2}}+\frac{1}{8 {x}^{4}}+\frac{1}{90 {x}^{5}}+1\end{array}\right)$
The default truncation order of the series expansion is 6. To obtain more terms in the Puiseux series solutions, set 'Order' to 8.
S = dsolve(eqn,'ExpansionPoint',Inf,'Order',8)
S =
$\left(\begin{array}{c}x-\frac{1}{6 {x}^{2}}-\frac{1}{8 {x}^{4}}-\frac{1}{90 {x}^{5}}-\frac{37}{336 {x}^{6}}\\ \frac{1}{6 {x}^{2}}+\frac{1}{8 {x}^{4}}+\frac{1}{90 {x}^{5}}+\frac{37}{336 {x}^{6}}+\frac{37}{1680 {x}^{7}}+1\end{array}\right)$
Solve the differential equation $\frac{\mathit{dy}}{\mathit{dx}}=\frac{1}{{\mathit{x}}^{2}}{\mathit{e}}^{-\frac{1}{\mathit{x}}}$ without specifying the initial condition.
syms y(x)
eqn = diff(y) == exp(-1/x)/x^2;
ySol(x) = dsolve(eqn)
ySol(x) =
${C}_{1}+{\mathrm{e}}^{-\frac{1}{x}}$
To eliminate constants from the solution, specify the initial condition $\mathit{y}\left(0\right)=1$.
cond = y(0) == 1;
S = dsolve(eqn,cond)
S =
${\mathrm{e}}^{-\frac{1}{x}}+1$
The function ${\mathit{e}}^{-\frac{1}{\mathit{x}}}$ in the solution ySol(x) has different one-sided limits at $x=0$. The function has a right-side limit, $\underset{\mathit{x}\to {0}^{+}}{\mathrm{lim}}\text{\hspace{0.17em}}{\mathit{e}}^{-\frac{1}{\mathit{x}}}=0$, but it has undefined left-side limit, $\underset{\mathit{x}\to {0}^{-}}{\mathrm{lim}}\text{\hspace{0.17em}}{\mathit{e}}^{-\frac{1}{\mathit{x}}}=\infty$. | {
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When you specify the condition y(x0) for a function with different one-sided limits at x0, dsolve treats the condition as a limit from the right, $\mathrm{lim}\text{\hspace{0.17em}}\mathit{x}\to {\mathit{x}}_{0}^{+}$.
## Input Arguments
collapse all
Differential equation or system of equations, specified as a symbolic equation or a vector of symbolic equations. Specify a differential equation by using the == operator. If eqn is a symbolic expression (without the right side), the solver assumes that the right side is 0, and solves the equation eqn == 0.
In the equation, represent differentiation by using diff. For example, diff(y,x) differentiates the symbolic function y(x) with respect to x. Create the symbolic function y(x) by using syms and solve the equation d2y(x)/dx2 = x*y(x) using dsolve.
syms y(x)
S = dsolve(diff(y,x,2) == x*y)
Specify a system of differential equations by using a vector of equations, as in syms y(t) z(t); S = dsolve([diff(y,t) == z, diff(z,t) == -y]). Here, y and z must be symbolic functions that depend on symbolic variables, which are t in this case. The right side must be symbolic expressions that depend on t, y and z. Note that Symbolic Math Toolbox™ currently does not support composite symbolic functions, that is, symbolic functions that depend on another symbolic functions.
Initial or boundary condition, specified as a symbolic equation or vector of symbolic equations.
When a condition contains a derivative, represent the derivative with diff. Assign the diff call to a variable and use the variable to specify the condition. For example, see Solve Differential Equations with Conditions.
Specify multiple conditions by using a vector of equations. If the number of conditions is less than the number of dependent variables, the solutions contain the arbitrary constants C1, C2,....
### Name-Value Arguments | {
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### Name-Value Arguments
Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.
Example: 'IgnoreAnalyticConstraints',false does not apply internal simplifications.
Expansion point of a Puiseux series solution, specified as a number, or a symbolic number, variable, function, or expression. Specifying this option returns the solution of a differential equation in terms of a Puiseux series (a power series that allows negative and fractional exponents). The expansion point cannot depend on the series variable. For example, see Find Series Solution of Differential Equation.
Data Types: single | double | sym
Complex Number Support: Yes
Option to use internal simplifications, specified as true or false.
By default, the solver applies simplifications while solving the differential equation, which could lead to results not generally valid. In other words, this option applies mathematical identities that are convenient, but the results might not hold for all possible values of the variables. Therefore, by default, the solver does not guarantee the completeness of results. If 'IgnoreAnalyticConstraints' is true, always verify results returned by the dsolve function. For more details, see Algorithms.
To solve ordinary differential equations without these simplifications, set 'IgnoreAnalyticConstraints' to false. Results obtained with 'IgnoreAnalyticConstraints' set to false are correct for all values of the arguments. For certain equations, dsolve might not return an explicit solution if you set 'IgnoreAnalyticConstraints' to false.
Option to return an implicit solution, specified as false or true. For a differential equation with variables x and y(x), an implicit solution has the form F(y(x)) = g(x). | {
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By default, the solver tries to find an explicit solution y(x) = f(x) analytically when solving a differential equation. If dsolve cannot find an explicit solution, then you can try finding a solution in implicit form by specifying the 'Implicit' option to true.
Maximum degree of polynomial equations for which the solver uses explicit formulas, specified as a positive integer smaller than 5. dsolve does not use explicit formulas when solving polynomial equations of degrees larger than 'MaxDegree'.
Truncation order of a Puiseux series solution, specified as a positive integer or a symbolic positive integer. Specifying this option returns the solution of a differential equation in terms of a Puiseux series (a power series that allow negative and fractional exponents). The truncation order n is the exponent in the O-term: $O\left({\mathrm{var}}^{n}\right)$ or $O\left({\mathrm{var}}^{-n}\right)$.
## Output Arguments
collapse all
Solutions of differential equation, returned as a symbolic expression or a vector of symbolic expressions. The size of S is the number of solutions.
Variables storing solutions of differential equations, returned as a vector of symbolic variables. The number of output variables must equal the number of dependent variables in a system of equations. dsolve sorts the dependent variables alphabetically, and then assigns the solutions for the variables to output variables or symbolic arrays.
## Tips
• If dsolve cannot find an explicit or implicit solution, then it issues a warning and returns the empty sym. In this case, try to find a numeric solution using the MATLAB® ode23 or ode45 function. Sometimes, the output is an equivalent lower-order differential equation or an integral.
• dsolve does not always return complete solutions even if 'IgnoreAnalyticConstraints' is false. | {
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• dsolve does not always return complete solutions even if 'IgnoreAnalyticConstraints' is false.
• If dsolve returns a function that has different one-sided limits at x0 and you specify the condition y(x0), then dsolve treats the condition as a limit from the right, .
## Algorithms
If you do not set 'IgnoreAnalyticConstraints' to false, then dsolve applies these rules while solving the equation:
• log(a) + log(b) = log(a·b) for all values of a and b. In particular, the following equality is applied for all values of a, b, and c:
(a·b)c = ac·bc.
• log(ab) = b·log(a) for all values of a and b. In particular, the following equality is applied for all values of a, b, and c:
(ab)c = ab·c.
• If f and g are standard mathematical functions and f(g(x)) = x for all small positive numbers, f(g(x)) = x is assumed to be valid for all complex x. In particular:
• log(ex) = x
• asin(sin(x)) = x, acos(cos(x)) = x, atan(tan(x)) = x
• asinh(sinh(x)) = x, acosh(cosh(x)) = x, atanh(tanh(x)) = x
• Wk(x·ex) = x for all branch indices k of the Lambert W function.
• The solver can multiply both sides of an equation by any expression except 0.
• The solutions of polynomial equations must be complete.
## Compatibility Considerations
expand all
Warns starting in R2019b | {
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# Proving right angle using vectors
Let $$ABCD$$ be a rectangle, $$E$$ midpoint of $$\overline{DC}$$ and $$G$$ point on $$\overline{AC}$$ such that $$\vec{BG}$$ is perpendicular to $$\vec{AC}$$. Also, let $$F$$ be a midpoint of $$\overline{AG}$$. Prove that angle $$\angle BFE =\pi/2$$.
So, I need to prove that $$\vec{EF}\cdot \vec{FB}=0$$. I tried to just express these vectors as sums of vectors of rectangle but haven't find any "elegant" way to prove it.
I did found one "ugly" way to prove it:
Let $$|AB|=a$$, $$|AD|=b$$ and $$\vec{AB}=\vec{a}$$, $$\vec{AD}=\vec{b}$$, so $$\vec{AC}=\vec{a}+\vec{b}$$ and since $$\vec{a}\cdot\vec{b}=0$$ we have $$|AC|^2=a^2+b^2$$. We can see that $$\triangle ABC$$ and $$\triangle BCG$$ are similar so $$|CG|=|CB|^2/|AC|$$ and since $$\vec{CG}=\lambda \cdot \vec{CA}$$ we get $$\lambda=\frac{b^2}{a^2+b^2}.$$
Now, we can express $$\vec{EF}$$ and $$\vec{FB}$$ in the terms of $$\vec{a}$$, $$\vec{b}$$, more precisely: $$\vec{EF}=\frac{1}{2}\left(\frac{-b^2}{a^2+b^2}\right)\vec{a}+\frac{1}{2}\left(\frac{a^2}{a^2+b^2}-2\right)\vec{b}$$ and $$\vec{FB}=\frac{1}{2}\left(2-\frac{a^2}{a^2+b^2}\right)\vec{a}+\frac{1}{2}\left(\frac{-a^2}{a^2+b^2}\right)\vec{b}$$ and if we multiply we get that dot product is $$0$$.
But as you can see, this $$\lambda$$ is "weird" and I am wondering if someone sees a more elegant way to prove this ?
• Actually, if you write those final coefficients in terms of $\lambda$, it makes the proof of orthogonality much prettier. – J.G. Oct 9 '18 at 19:26
You can simplify and generalize at the same time.
Instead of defining $$E$$ and $$F$$ as midpoints, define them as weighted averages:
$$E=kC+(1-k)D$$ $$F=kG+(1-k)A$$
(In your question $$k=1/2.)$$ As $$k$$ goes from $$0$$ to $$1$$, $$\triangle{BFE}$$ interpolates the similar triangles $$\triangle{BAD}$$ and $$\triangle{BGC}$$. And it is also similar to these two triangles. | {
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Given that $$(B-G)\cdot(C-G)=0$$ and $$(B-A)\cdot(D-A)=0$$, you can show that $$(B-F)\cdot(E-F)=0$$ (the simplification is a bit tedious, and I used the fact that $$(Rx)\cdot y+(Ry)\cdot x=0$$ for perpendicular vectors $$x,y$$ and rotation $$R$$).
More about the phenomenon of linearly interpolating two directly similar figures can be found by Googling the terms 'spiral similarity' and 'fundamental theorem of directly similar figures'. Also, instead of using vectors you can use complex numbers, as shown in $$\textbf{2.2}$$ of Zachary Abel's Mean Geometry paper.
Let $$H$$ be a midpoint of $$BG$$. Then $$HF$$ is a middle line in triangle $$ABG$$, so $$FH||AB||CD$$ and $$FH = {1\over 2}AB = CE$$ so $$FHCE$$ is a parallelogram, so $$FE||CH$$.
Now $$HF\bot BC$$ so $$HF$$ is (second) altitude in $$\triangle BCF$$ so $$H$$ is orthocenter in this triangle, since $$BG$$ is also altitude in this triangle. So line $$CH\bot FB$$ and thus $$\angle BFE = 90^{\circ}$$.
Here is a vector solution:
Let $$G$$ be an origin of position vecotors. Then $$G=0$$, $$F = A/2$$, $$D = A-B+C$$, $$A\cdot B = C\cdot B=0$$ and $$E = {1\over 2}(C+D) = {1\over 2}(A-B+2C)$$
$$\begin{eqnarray} \vec{FB}\cdot \vec{FE} &=& (E-F)(B-F)\\ &=& {1\over 4}(-B+2C)(2B-A)\\ &=& -{1\over 2}(B^2+CA)\\ &=&0 \end{eqnarray}$$
A solution with a help of the coordinate system.
Let $$B = (0,0)$$, $$C= (2c,0)$$ and $$A= (0,2a)$$. Then $$D(2c,2a)$$ and $$E(2c,a)$$. A perpendicular to $$AC:\;\;{x\over 2c}+{y\over 2a}=1$$ through $$B$$ is $$y={c\over a}x$$ which cuts $$AC$$ at $$G= \big({2a^2c\over a^2+c^2},{2c^2a\over a^2+c^2} \big)$$
so $$F = \big({a^2c\over a^2+c^2},{2c^2a+a^3\over a^2+c^2} \big)$$
Now it is not difficult to see that $$\vec{FE}\cdot \vec{FB} =0$$.
Another solution: | {
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Now it is not difficult to see that $$\vec{FE}\cdot \vec{FB} =0$$.
Another solution:
Say $$M$$ is midpoint of $$AB$$, then $$FM$$ is a middle line in triangle $$ABG$$ so $$FM||BG$$ so $$\angle MFC = 90^{\circ}$$ and thus $$F$$ is on a circle through $$B,C$$ and $$M$$. But on this circle is also $$E$$ since $$\angle MEC = 90^{\circ}$$. So $$\angle BFE = \angle BME = 90^{\circ}$$
Alternatively Say $$AB = 2a$$ and $$BC= b$$, then $$AF = {1\over 2}AG = {1\over 2} {4a^2\over \sqrt{4a^2+b^2}} = {2a^2\over \sqrt{4a^2+b^2}}$$ Then we have $$AF\cdot AC = {2a^2\over \sqrt{4a^2+b^2}}\cdot \sqrt{4a^2+b^2} = 2a^2 = AM\cdot AB$$ and so $$B,C,F,M$$ are concylic. | {
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## Solve Algebraic Equation
Symbolic Math Toolbox™ offers both symbolic and numeric equation solvers. This topic shows you how to solve an equation symbolically using the symbolic solver solve. To compare symbolic and numeric solvers, see Select Numeric or Symbolic Solver.
### Solve an Equation
If eqn is an equation, solve(eqn, x) solves eqn for the symbolic variable x.
Use the == operator to specify the familiar quadratic equation and solve it using solve.
syms a b c x
eqn = a*x^2 + b*x + c == 0;
solx = solve(eqn, x)
solx =
-(b + (b^2 - 4*a*c)^(1/2))/(2*a)
-(b - (b^2 - 4*a*c)^(1/2))/(2*a)
solx is a symbolic vector containing the two solutions of the quadratic equation. If the input eqn is an expression and not an equation, solve solves the equation eqn == 0.
To solve for a variable other than x, specify that variable instead. For example, solve eqn for b.
solb = solve(eqn, b)
solb =
-(a*x^2 + c)/x
If you do not specify a variable, solve uses symvar to select the variable to solve for. For example, solve(eqn) solves eqn for x.
### Return the Full Solution to an Equation
solve does not automatically return all solutions of an equation. Solve the equation cos(x) == -sin(x). The solve function returns one of many solutions.
syms x
solx = solve(cos(x) == -sin(x), x)
solx =
-pi/4
To return all solutions along with the parameters in the solution and the conditions on the solution, set the ReturnConditions option to true. Solve the same equation for the full solution. Provide three output variables: for the solution to x, for the parameters in the solution, and for the conditions on the solution.
syms x
[solx, param, cond] = solve(cos(x) == -sin(x), x, 'ReturnConditions', true)
solx =
pi*k - pi/4
param =
k
cond =
in(k, 'integer') | {
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solx contains the solution for x, which is pi*k - pi/4. The param variable specifies the parameter in the solution, which is k. The cond variable specifies the condition in(k, 'integer') on the solution, which means k must be an integer. Thus, solve returns a periodic solution starting at pi/4 which repeats at intervals of pi*k, where k is an integer.
### Work with the Full Solution, Parameters, and Conditions Returned by solve
You can use the solutions, parameters, and conditions returned by solve to find solutions within an interval or under additional conditions.
To find values of x in the interval -2*pi<x<2*pi, solve solx for k within that interval under the condition cond. Assume the condition cond using assume.
assume(cond)
solk = solve(-2*pi<solx, solx<2*pi, param)
solk =
-1
0
1
2
To find values of x corresponding to these values of k, use subs to substitute for k in solx.
xvalues = subs(solx, solk)
xvalues =
-(5*pi)/4
-pi/4
(3*pi)/4
(7*pi)/4
To convert these symbolic values into numeric values for use in numeric calculations, use vpa.
xvalues = vpa(xvalues)
xvalues =
-3.9269908169872415480783042290994
-0.78539816339744830961566084581988
2.3561944901923449288469825374596
5.4977871437821381673096259207391
### Visualize and Plot Solutions Returned by solve
The previous sections used solve to solve the equation cos(x) == -sin(x). The solution to this equation can be visualized using plotting functions such as fplot and scatter.
Plot both sides of equation cos(x) == -sin(x).
fplot(cos(x))
hold on
grid on
fplot(-sin(x))
title('Both sides of equation cos(x) = -sin(x)')
legend('cos(x)','-sin(x)','Location','best','AutoUpdate','off')
Calculate the values of the functions at the values of x, and superimpose the solutions as points using scatter.
yvalues = cos(xvalues)
yvalues = | {
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yvalues = cos(xvalues)
yvalues =
$\left(\begin{array}{c}-0.70710678118654752440084436210485\\ 0.70710678118654752440084436210485\\ -0.70710678118654752440084436210485\\ 0.70710678118654752440084436210485\end{array}\right)$
scatter(xvalues, yvalues)
As expected, the solutions appear at the intersection of the two plots.
### Simplify Complicated Results and Improve Performance
If results look complicated, solve is stuck, or if you want to improve performance, see, Troubleshoot Equation Solutions from solve Function.
## Support
#### Mathematical Modeling with Symbolic Math Toolbox
Get examples and videos | {
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1. ## Find the limit
Limit as n approaches infinity [(3^n + 5^n)/(3^n+1 + 5^n+1)]
I tried to divide the numerator and denominator by 3^n. Was not successful.
What should I do next?
2. Originally Posted by Nichelle14
Limit as n approaches infinity [(3^n + 5^n)/(3^n+1 + 5^n+1)]
I tried to divide the numerator and denominator by 3^n. Was not successful.
What should I do next?
Did you consider to divide by $3^n+5^n$
Thus,
$\frac{1}{1+\frac{2}{3^n+5^n}}\to 1$ as $n\to\infty$
3. Also,
$1-\frac{1}{n}\leq \frac{3^n+5^n}{3^n+1+5^n+1} \leq 1+\frac{1}{n}$
Since,
$\lim_{n\to\infty} 1-\frac{1}{n}=\lim_{n\to\infty}1+\frac{1}{n}=1$
Thus,
$\lim_{n\to\infty}\frac{3^n+5^n}{3^n+1+5^n+1}=1$ by the squeeze theorem.
4. Hello, Nichelle14!
$\lim_{n\to\infty}\frac{3^n + 5^n}{3^{n+1} + 5^{n+1}}$
I tried to divide the numerator and denominator by $3^n$ . . .Was not successful.
What should I do next?
Divide top and bottom by $5^{n+1}.$
The numerator is: . $\frac{3^n}{5^{n+1}} + \frac{5^n}{5^{n+1}} \;= \;\frac{1}{5}\cdot\frac{3^n}{5^n} + \frac{1}{5}\;=$ $\frac{1}{5}\left(\frac{3}{5}\right)^n + \frac{1}{5}$
The denominator is: . $\frac{3^{n+1}}{5^{n+1}} + \frac{5^{n+1}}{5^{n+1}} \;=\;\left(\frac{3}{5}\right)^{n+1} + 1$
Recall that: if $|a| < 1$, then $\lim_{n\to\infty} a^n\:=\:0$
Therefore, the limit is: . $\lim_{n\to\infty}\,\frac{\frac{1}{5}\left(\frac{3} {5}\right)^n + \frac{1}{5}} {\left(\frac{3}{5}\right)^n + 1} \;=\;\frac{\frac{1}{5}\cdot0 + \frac{1}{5}}{0 + 1}\;=\;\frac{1}{5}
$
5. Originally Posted by Soroban
Hello, Nichelle14!
Divide top and bottom by $5^{n+1}.$
The numerator is: . $\frac{3^n}{5^{n+1}} + \frac{5^n}{5^{n+1}} \;= \;\frac{1}{5}\cdot\frac{3^n}{5^n} + \frac{1}{5}\;=$ $\frac{1}{5}\left(\frac{3}{5}\right)^n + \frac{1}{5}$
The denominator is: . $\frac{3^{n+1}}{5^{n+1}} + \frac{5^{n+1}}{5^{n+1}} \;=\;\left(\frac{3}{5}\right)^{n+1} + 1$
Recall that: if $|a| < 1$, then $\lim_{n\to\infty} a^n\:=\:0$ | {
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Recall that: if $|a| < 1$, then $\lim_{n\to\infty} a^n\:=\:0$
Therefore, the limit is: . $\lim_{n\to\infty}\,\frac{\frac{1}{5}\left(\frac{3} {5}\right)^n + \frac{1}{5}} {\left(\frac{3}{5}\right)^n + 1} \;=\;\frac{\frac{1}{5}\cdot0 + \frac{1}{5}}{0 + 1}\;=\;\frac{1}{5}
$
I think the trick here is that you will divide all the terms by the largest term in the expression(it is useful when the limit tends to infinity)
Keep Smiling
Malay | {
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# Spatial Co-ordinate of Maximum Value of Variable
Say I have the following code:
int nn = 50;
mesh Th = square(nn, nn);
fespace Vh(Th, P1);
Vh u;
u = sin(x);
I can get the maximum value of ‘u’ by:
real maxValueU = u[].max;
How do I get the values of x and y coordinates where maximum value of ‘u’ occurs?
yes but be careful because the max could be not unique.unc
the method imax give the dof number
int dofmaxValueU = u[].imax;
Now for P1 finite element the dof are vertices with same numbering so
Th(dofmaxValueU).x
Th(dofmaxValueU).y
given the coordinate. | {
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"url": "https://community.freefem.org/t/spatial-co-ordinate-of-maximum-value-of-variable/1364"
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# Fixed Cost & Cost Per Person
1. Jul 10, 2017
### zak100 | {
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"lm_q2_score": 0.888758798648752,
"openwebmath_perplexity": 5311.896602003922,
"openwebmath_score": 0.2869255542755127,
"tags": null,
"url": "https://www.physicsforums.com/threads/fixed-cost-cost-per-person.919820/"
} |
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