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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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 $= - r^{2} sin 2 θ$ $=-r^2sin2theta$

Explanation:

Reminder

$sin 2 θ = 2 sin θ cos θ$ $sin2theta=2sinthetacostheta$

We are given

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

$x = r cos θ$ $x=rcostheta$

$y = r sin θ$ $y=rsintheta$

$\frac{\partial z}{\partial θ} = \frac{\partial z}{\partial x} . \frac{\partial x}{\partial θ} + \frac{\partial z}{\partial y} . \frac{\partial y}{\partial θ}$ $(delz)/(deltheta)=(delz)/(delx).(delx)/(deltheta)+(delz)/(dely).(dely)/(deltheta)$

$\frac{\partial z}{\partial x} = 2 x$ $(delz)/(delx)=2x$

$\frac{\partial z}{\partial y} = 4 y$ $(delz)/(dely)=4y$

$\frac{\partial x}{\partial θ} = - r sin θ$ $(delx)/(deltheta)=-rsintheta$

$\frac{\partial y}{\partial θ} = r cos θ$ $(dely)/(deltheta)=rcostheta$

$\frac{\partial z}{\partial θ} = - 2 x r sin θ + 4 y r cos θ$ $(delz)/(deltheta)=-2xrsintheta+4yrcostheta$

$= - 2 r sin θ r cos θ + 4 r sin θ r cos θ$ $=-2rsinthetarcostheta+4rsinthetarcostheta$

$= 2 r^{2} sin θ cos θ$ $=2r^2sinthetacostheta$

If $y$ $y$ is constant then

$\frac{\partial z}{\partial y} = 0$ $(delz)/(dely)=0$

Therefore,

${(\frac{\partial z}{\partial θ})}_{y} = - 2 r^{2} sin θ cos θ = - r^{2} sin 2 θ$ $((delz)/(deltheta))_y=-2r^2sinthetacostheta=-r^2sin2theta$

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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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