- #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?

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

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?

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

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