Question

Use the Chain Rule to find the indicated partial derivatives. u = x2 + yz, x = pr cos(θ), y = pr sin(θ), z = p + r; (partial u)/(partial p), (partial u)/(partial r), (partial u)/(partial theta) when p = 2, r = 2, θ = 0

Answer 1

$u(x,y,z)=x^2+yz$

$\begin{cases}x(p,r,\theta)=pr\cos\theta\\y(p,r,\theta)=pr\sin\theta\\z(p,r,\theta)=p+r\end{cases}$

At the point $(p,r,\theta)=(2,2,0)$, we have

$\begin{cases}x(2,2,0)=4\\y(2,2,0)=0\\z(2,2,0)=4\end{cases}$

Denote by $f_x:=\dfrac{\partial f}{\partial x}$ the partial derivative of a function $f$ with respect to the variable $x$. We have

$\begin{cases}u_x=2x\\u_y=z\\u_z=y\end{cases}$

The Jacobian is

$\begin{bmatrix}x_p&x_r&x_\theta\\y_p&y_r&y_\theta\\z_p&z_r&z_\theta\end{bmatrix}=\begin{bmatrix}r\cos\theta&p\cos\theta&-pr\sin\theta\\r\sin\theta&p\sin\theta&pr\cos\theta\\1&1&0\end{bmatrix}$

By the chain rule,

$u_p=u_xx_p+u_yy_p+u_zz_p=2xr\cos\theta+zr\sin\theta+y$

$u_p(2,2,0)=2\cdot4\cdot2\cos0+4\cdot2\sin0+0\implies\boxed{u_p(2,2,0)=16}$

$u_r=u_xx_r+u_yy_r+u_zz_r=2xp\cos\theta+zp\sin\theta+y$

$u_r(2,2,0)=2\cdot4\cdot2\cos0+4\cdot2\sin0+0\implies\boxed{u_r(2,2,0)=16}$

$u_\theta=u_xx_\theta+u_yy_\theta+u_zz_\theta=-2xpr\sin\theta+zpr\cos\theta$

$u_\theta(2,2,0)=-2\cdot4\cdot2\cdot2\sin0+4\cdot2\cdot2\cos0\implies\boxed{u_\theta(2,2,0)=16}$