Question

use the chain rule to find the indicated partial derivatives.

$w = xy + yz + zx$, $x = r\cos(\theta)$, $y = r\sin(\theta)$, $z = r\theta$;

$\frac{\partial w}{\partial r}$, $\frac{\partial w}{\partial \theta}$ when $r = 2$, $\theta = \frac{\pi}{2}$

$\frac{\partial w}{\partial r} = $

$\frac{\partial w}{\partial \theta} = $

Explanation:

Step1: Find partial derivatives of \(w\) with respect to \(x,y,z\)

By the formula for partial derivatives, \(\frac{\partial w}{\partial x}=y + z\), \(\frac{\partial w}{\partial y}=x + z\), \(\frac{\partial w}{\partial z}=y + x\)

Step2: Find partial derivatives of \(x,y,z\) with respect to \(r\) and \(\theta\)

For \(x = r\cos\theta\), \(\frac{\partial x}{\partial r}=\cos\theta\), \(\frac{\partial x}{\partial\theta}=-r\sin\theta\)
For \(y = r\sin\theta\), \(\frac{\partial y}{\partial r}=\sin\theta\), \(\frac{\partial y}{\partial\theta}=r\cos\theta\)
For \(z = r\theta\), \(\frac{\partial z}{\partial r}=\theta\), \(\frac{\partial z}{\partial\theta}=r\)

Step3: Apply the chain - rule for \(\frac{\partial w}{\partial r}\)

By the chain - rule \(\frac{\partial w}{\partial r}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial r}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial r}\)
Substitute the values: \(\frac{\partial w}{\partial r}=(y + z)\cos\theta+(x + z)\sin\theta+(y + x)\theta\)
When \(r = 2\) and \(\theta=\frac{\pi}{2}\)
\(x=r\cos\theta=2\cos\frac{\pi}{2}=0\), \(y=r\sin\theta=2\sin\frac{\pi}{2}=2\), \(z=r\theta=2\times\frac{\pi}{2}=\pi\)
\(\frac{\partial w}{\partial r}=(2+\pi)\cos\frac{\pi}{2}+(0 + \pi)\sin\frac{\pi}{2}+(2 + 0)\times\frac{\pi}{2}\)
\(=(2+\pi)\times0+\pi\times1 + 2\times\frac{\pi}{2}=\pi+\pi = 2\pi\)

Step4: Apply the chain - rule for \(\frac{\partial w}{\partial\theta}\)

By the chain - rule \(\frac{\partial w}{\partial\theta}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial\theta}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial\theta}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial\theta}\)
Substitute the values: \(\frac{\partial w}{\partial\theta}=(y + z)(-r\sin\theta)+(x + z)(r\cos\theta)+(y + x)r\)
When \(r = 2\) and \(\theta=\frac{\pi}{2}\)
\(\frac{\partial w}{\partial\theta}=(2+\pi)(-2\sin\frac{\pi}{2})+(0+\pi)(2\cos\frac{\pi}{2})+(2 + 0)\times2\)
\(=(2+\pi)(-2)+(\pi)(0)+4=-4-2\pi + 4=-2\pi\)

Answer:

\(\frac{\partial w}{\partial r}=2\pi\)
\(\frac{\partial w}{\partial\theta}=-2\pi\)