*Seond Order Derivative*

  • Thread starter physixguru
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  • #1
physixguru
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Been doing calculus all my life , that's why i am at the Indian Institute of Technology:shy:..
but never faced a headache like this...

Prove that:

If x=cos theta and y = sin^3 theta

then [d^2y/dx^2] + [dy/dx]^2 = 3sin^2 theta [ 5 cos ^2 theta -1 ]

My side:

dx/d theta = -sin theta
dy/d theta = 3 sin^2 theta cos theta
>> {dy/dx}== -3 sin theta cos theta

now d^2y/dx^2= -3 [ cos^2 theta - sin ^2 theta ]* 1/-sin theta

moving with the give expression...

3cos^2 theta-3 sin^2 theta/sin theta + 9 sin^2 theta cos^2 theta

Spent a lot of time...cud not get anywhere...:grumpy:
Am i messing up the trigo functions?:frown:

This is not a homework question. I don't have to do homework anymore.I work in a MNC.
 
Last edited:

Answers and Replies

  • #2
Ben Niehoff
Science Advisor
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You've never faced a "headache" like that? Use trig identities.
 
  • #3
physixguru
335
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Please be a bit more clear.Even a moron knows that you got to use trigo identities.Its no rocket science!
 
  • #4
maze
661
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Ok, I'm going to assume you didn't make any algebra/differentiation mistakes so far (I didn't check)
If that is the case, then you are starting here:
3cos^2 theta-3 sin^2 theta/sin theta + 9 sin^2 theta cos^2 theta

And you want to get here:
3sin^2 theta [ 5 cos ^2 theta -1 ]

I would write each expression in terms of complex exponentials, then multiply all the exponentials out completely, working both backwards and forwards simultaneously. Then when your results meet in the middle, rearrange your steps to go from where you started to where you finished.

Here is a small (easier) example of this strategy in action: Say we want to show [itex]cos^2\left(x\right) = 1-sin^2\left(x\right)[/tex]
Working forwards:
[tex]cos^2\left(x\right)=\left(\frac{e^{i x}+e^{-i x}}{2}\right)^2=\frac{e^{2 i x} + 2 + e^{-2 i x}}{4}[/tex]
Now hold that thought while we work backwards...

Working backwards:
[tex]1-sin^2\left(x\right) = 1-\left(\frac{e^{i x}-e^{-i x}}{2 i}\right)^2=1-\frac{e^{2 i x}-2+e^{-2 i x}}{-4}=\frac{e^{2 i x} + 2 + e^{-2 i x}}{4}[/tex]

Then to write the final proof, you would just rearrange the steps:
[tex]cos^2\left(x\right)=\left(\frac{e^{i x}+e^{-i x}}{2}\right)^2=\frac{e^{2 i x} + 2 + e^{-2 i x}}{4}=1-\frac{e^{2 i x}-2+e^{-2 i x}}{-4}=1-\left(\frac{e^{i x}-e^{-i x}}{2 i}\right)^2=1-sin^2\left(x\right)[/tex]
 
Last edited:
  • #5
physixguru
335
0
Ok, I'm going to assume you didn't make any algebra/differentiation mistakes so far (I didn't check)
If that is the case, then you are starting here:
3cos^2 theta-3 sin^2 theta/sin theta + 9 sin^2 theta cos^2 theta

And you want to get here:
3sin^2 theta [ 5 cos ^2 theta -1 ]

I would write each expression in terms of complex exponentials, then multiply all the exponentials out completely, working both backwards and forwards simultaneously. Then when your results meet in the middle, rearrange your steps to go from where you started to where you finished.

Here is a small (easier) example of this strategy in action: Say we want to show [itex]cos^2\left(x\right) = 1-sin^2\left(x\right)[/tex]
Working forwards:
[tex]cos^2\left(x\right)=\left(\frac{e^{i x}+e^{-i x}}{2}\right)^2=\frac{e^{2 i x} + 2 + e^{-2 i x}}{4}[/tex]
Now hold that thought while we work backwards...

Working backwards:
[tex]1-sin^2\left(x\right) = 1-\left(\frac{e^{i x}-e^{-i x}}{2 i}\right)^2=1-\frac{e^{2 i x}-2+e^{-2 i x}}{-4}=\frac{e^{2 i x} + 2 + e^{-2 i x}}{4}[/tex]

Then to write the final proof, you would just rearrange the steps:
[tex]cos^2\left(x\right)=\left(\frac{e^{i x}+e^{-i x}}{2}\right)^2=\frac{e^{2 i x} + 2 + e^{-2 i x}}{4}=1-\frac{e^{2 i x}-2+e^{-2 i x}}{-4}=1-\left(\frac{e^{i x}-e^{-i x}}{2 i}\right)^2=1-sin^2\left(x\right)[/tex]

Dear maze, thanks a lot for your efforts and spending your precious time on the problem.
Your method is really impressive and shows your multidimensional thinking.But the problem is that the method you supplied is a lengthy one and can be extremely confusing in long trigo expressions like mine is the case.I would like a simple approach.
 

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