If z = x^2 + 2y^2 , x = r cos θ , y = r sin θ , find the partial derivative (∂z/∂θ)_y ?

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Answer 1 · 2017-12-16T10:45:52

The answer is $=-r^2sin2theta$

Reminder

$sin2theta=2sinthetacostheta$

We are given

$z=x^2+2y^2$

$x=rcostheta$

$y=rsintheta$

$(delz)/(deltheta)=(delz)/(delx).(delx)/(deltheta)+(delz)/(dely).(dely)/(deltheta)$

$(delz)/(delx)=2x$

$(delz)/(dely)=4y$

$(delx)/(deltheta)=-rsintheta$

$(dely)/(deltheta)=rcostheta$

$(delz)/(deltheta)=-2xrsintheta+4yrcostheta$

$=-2rsinthetarcostheta+4rsinthetarcostheta$

$=2r^2sinthetacostheta$

If $y$ is constant then

$(delz)/(dely)=0$

Therefore,

$((delz)/(deltheta))_y=-2r^2sinthetacostheta=-r^2sin2theta$