Find the indicated partial derivatives
$$
\begin{array}{l}
w=x y+y z+z x, \quad x=r \cos \theta, \quad y=r \sin \theta, \quad z=r \theta \
\frac{\partial w}{\partial r}, \frac{\partial w}{\partial \theta} \quad \text { when } r=2, \theta=\pi / 2
\end{array}
$$

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Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Given the function $w = xy + yz + zx$ with $x = r \cos \theta$, $y = r \sin \theta$, and $z = r \theta$, we need to find the partial derivatives $\frac{\partial w}{\partial r}$ and $\frac{\partial w}{\partial \theta}$ at $r = 2$ and $\theta = \frac{\pi}{2}$. ‖ ‖ step 2 ⋮ First, we differentiate $w$ with respect to $r$: $$ \frac{\partial w}{\partial r} = \frac{\partial}{\partial r} (r \cos \theta \cdot r \sin \theta + r \sin \theta \cdot r \theta + r \theta \cdot r \cos \theta) $$ $$ = \frac{\partial}{\partial r} (r^2 \cos \theta \sin \theta + r^2 \sin \theta \theta + r^2 \theta \cos \theta) $$ $$ = 2r \cos \theta \sin \theta + 2r \sin \theta \theta + 2r \theta \cos \theta $$ $$ = 2r (\cos \theta \sin \theta + \sin \theta \theta + \theta \cos \theta) $$ \] ‖ ‖ step 3 ⋮ Next, we differentiate $w$ with respect to $\theta$: $$ \frac{\partial w}{\partial \theta} = \frac{\partial}{\partial \theta} (r \cos \theta \cdot r \sin \theta + r \sin \theta \cdot r \theta + r \theta \cdot r \cos \theta) $$ $$ = \frac{\partial}{\partial \theta} (r^2 \cos \theta \sin \theta + r^2 \sin \theta \theta + r^2 \theta \cos \theta) $$ $$ = r^2 (\cos \theta \cos \theta - \sin \theta \sin \theta + \sin \theta + \theta \cos \theta - \theta \sin \theta) $$ $$ = r^2 (\cos^2 \theta - \sin^2 \theta + \sin \theta + \theta \cos \theta - \theta \sin \theta) $$ \] ‖ ‖ step 4 ⋮ Now, we evaluate $\frac{\partial w}{\partial r}$ at $r = 2$ and $\theta = \frac{\pi}{2}$: $$ \frac{\partial w}{\partial r} \bigg|_{r=2, \theta=\frac{\pi}{2}} = 2 \cdot 2 (\cos \frac{\pi}{2} \sin \frac{\pi}{2} + \sin \frac{\pi}{2} \cdot \frac{\pi}{2} + \frac{\pi}{2} \cos \frac{\pi}{2}) $$ $$ = 4 (0 \cdot 1 + 1 \cdot \frac{\pi}{2} + \frac{\pi}{2} \cdot 0) $$ $$ = 4 \cdot \frac{\pi}{2} $$ $$ = 2\pi $$ \] ‖ ‖ step 5 ⋮ Finally, we evaluate $\frac{\partial w}{\partial \theta}$ at $r = 2$ and $\theta = \frac{\pi}{2}$: $$ \frac{\partial w}{\partial \theta} \bigg|_{r=2, \theta=\frac{\pi}{2}} = 2^2 (\cos^2 \frac{\pi}{2} - \sin^2 \frac{\pi}{2} + \sin \frac{\pi}{2} + \frac{\pi}{2} \cos \frac{\pi}{2} - \frac{\pi}{2} \sin \frac{\pi}{2}) $$ $$ = 4 (0 - 1 + 1 + 0 - \frac{\pi}{2}) $$ $$ = 4 (0 - \frac{\pi}{2}) $$ $$ = -2\pi $$ \] ‖ ∻Answer∻ ⚹ $\frac{\partial w}{\partial r} = 2\pi$ and $\frac{\partial w}{\partial \theta} = -2\pi$ ⚹ ∻Key Concept∻ ⚹Partial Derivatives⚹ ∻Explanation∻ ⚹Partial derivatives measure how a function changes as its input variables change. Here, we computed the partial derivatives of $w$ with respect to $r$ and $\theta$ and evaluated them at specific values.⚹◊What is the gradient of $w$ with respect to $r$ and $\theta$ at $r=2$ and $\theta=\frac{\pi}{2}$?⍭ Generate me a similar question◊

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