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Rate of change of x wrt sin(x)

by Jhenrique
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Jhenrique
#1
Apr9-14, 12:59 PM
P: 686
Is possible to compute the rate of change of x wrt sin(x)? ##\frac{dx}{d \sin(x)}##
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berkeman
#2
Apr9-14, 01:18 PM
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Quote Quote by Jhenrique View Post
Is possible to compute the rate of change of x wrt sin(x)? ##\frac{dx}{d \sin(x)}##
Have you tried just plotting it in Excel? You will get some 1/0 inflection points that you will need to deal with...
PeroK
#3
Apr9-14, 02:16 PM
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Quote Quote by Jhenrique View Post
Is possible to compute the rate of change of x wrt sin(x)? ##\frac{dx}{d \sin(x)}##
Compare with ##\frac{dsin^{-1}(x)}{dx}##

Or, turn the graph of sin(x) on its side.

mathman
#4
Apr9-14, 04:04 PM
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Rate of change of x wrt sin(x)

Quote Quote by PeroK View Post
Compare with ##\frac{dsin^{-1}(x)}{dx}##

Or, turn the graph of sin(x) on its side.
Compare to reciprocal dsin(x)/dx. Sin-1(x) is not relevant.
1MileCrash
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Apr9-14, 04:21 PM
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Quote Quote by mathman View Post
Compare to reciprocal dsin(x)/dx. Sin-1(x) is not relevant.
Why isn't it relevant?

If we have dx/d(sinx), why is it not reasonable for me to let sinx = y and therefore have that x = arcsin(y)?

Then I have d(arcsin(y))/y = 1 / sqrt(1-y²)

And thus dx/d(sinx) = 1/sqrt(1-sin²(x)) = 1/sqrt(cos²(x)) = 1/|cos(x)|

Just curious, so where does this break down? This is a very unique question so I'm sure there is an error in my logic. One may note that this comes out to be only slightly different than the reciprocal method (no absolute value). I suspect this is because in assigning these variables in the way I did, I restricted/changed domain?
micromass
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Apr9-14, 07:20 PM
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Quote Quote by mathman View Post
Compare to reciprocal dsin(x)/dx. Sin-1(x) is not relevant.
Due to the inverse function theorem, it is very relevant.
1MileCrash
#7
Apr10-14, 12:05 AM
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Quote Quote by micromass View Post
Due to the inverse function theorem, it is very relevant.
After a quick Google, it looks like I followed this theorem with f(a) = sin(x). So why does f'(b) evaluate to 1/|cos(x)| according to my work rather than 1/cos(x) as the inverse function theorem claims it should?

I mean, hopefully we can agree that the answer to the OP's question is actually 1/cos(x) and not 1/|cos(x)|.
HallsofIvy
#8
Apr10-14, 08:51 AM
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There are two ways to "differentiate x with respect to sin(x)". The first is to use the fact that
[tex]\frac{dx}{dy}= \frac{1}{\frac{dy}{dx}}[/tex].
Here y= sin(x) so dy/dx= cos(x). Then dx/dy= 1/cos(x) is a perfectly good answer.

The other way is to say that if y= sin(x) then [itex]x= sin^{-1}(y)[/itex] so that the derivative of "x with respect to sin(x)" is [itex]dx/dy= d(sin^{-1}(y))/dy= 1/\sqrt{1- y^2}[/itex].

It's not at all difficult to prove that those are the same. If y= sin(x) then [itex]1- y^2= 1- sin^2(x)= cos^2(x)[/itex] so [itex]1/\sqrt{1- y^2}= 1/cos(x)[/itex].
Jhenrique
#9
Apr10-14, 09:17 AM
P: 686
So, I can differentiate any function f(x) wrt another any function g(x).

If df = f'(x) dx and dg = g'(x) dx, thus df/dg = f'/g' ...
1MileCrash
#10
Apr10-14, 10:48 AM
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Quote Quote by HallsofIvy View Post
There are two ways to "differentiate x with respect to sin(x)". The first is to use the fact that
[tex]\frac{dx}{dy}= \frac{1}{\frac{dy}{dx}}[/tex].
Here y= sin(x) so dy/dx= cos(x). Then dx/dy= 1/cos(x) is a perfectly good answer.

The other way is to say that if y= sin(x) then [itex]x= sin^{-1}(y)[/itex] so that the derivative of "x with respect to sin(x)" is [itex]dx/dy= d(sin^{-1}(y))/dy= 1/\sqrt{1- y^2}[/itex].

It's not at all difficult to prove that those are the same. If y= sin(x) then [itex]1- y^2= 1- sin^2(x)= cos^2(x)[/itex] so [itex]1/\sqrt{1- y^2}= 1/cos(x)[/itex].
Our work is the same, but I don't understand how you conclude that sqrt(cos²(x)) = cos(x). This is |cos(x)|, don't you agree?
micromass
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Apr10-14, 10:54 AM
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Quote Quote by 1MileCrash View Post
After a quick Google, it looks like I followed this theorem with f(a) = sin(x). So why does f'(b) evaluate to 1/|cos(x)| according to my work rather than 1/cos(x) as the inverse function theorem claims it should?

I mean, hopefully we can agree that the answer to the OP's question is actually 1/cos(x) and not 1/|cos(x)|.
Well, the sine function is not invertible, since it's not injective and not surjective. This is of course no problem for the inverse function theorem, since it will take a "local inverse".
You worked with the arcsine, which is indeed a local inverse. But not all local inverses are like the arcsine. That is, if we restrict the sine to ##(-\pi/2,\pi/2)## then the arcsine is an inverse. But guess what? The cosine function is actually positive on that said, so the absolute values drop.

If you apply the inverse function theorem on some other domain, then you can't use the arcsine anymore in that form.

Does that make sense?
1MileCrash
#12
Apr10-14, 11:40 AM
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Quote Quote by micromass View Post
Well, the sine function is not invertible, since it's not injective and not surjective. This is of course no problem for the inverse function theorem, since it will take a "local inverse".
You worked with the arcsine, which is indeed a local inverse. But not all local inverses are like the arcsine. That is, if we restrict the sine to ##(-\pi/2,\pi/2)## then the arcsine is an inverse. But guess what? The cosine function is actually positive on that said, so the absolute values drop.

If you apply the inverse function theorem on some other domain, then you can't use the arcsine anymore in that form.

Does that make sense?
Yes, thank you.


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