1. Let $f(x,y) = x^3 - xy$. Set $\vec a = (0,1)$ and $\vec b =(1,3)$. Find a point $\vec c$ on the line segment connecting $\vec a$ and $\vec b$ for which
$$
f(\vec b) - f(\vec a) = \nabla f(\vec c) \cdot (\vec b - \vec a).
$$

> (Sol) The line segament connecting $\vec a$ and $\vec b$ can be parameterized by $(1-t)\vec a +t\vec b$ for all $t \in [0,1]$. Hence $$x(t) = t \quad \text{and}\quad y(t) = (1-t) + 3t = 1+2t$$ Define $g(t) = f(x(t),y(t)) = t^3 - t(1+2t) = t^3 - 2t^2 - t$. Then, first, $$f(\vec b) - f(\vec a) = g(1) - g(0) = -2 - 0 = -2$$ and second, by mean value theorem, there exists some $s \in [0,1]$ such that $$g(1) - g(0) = g'(s) = 3s^2 - 4s -1$$ Now solve $3s^2 - 4s - 1 = -2$, we have $s = {1\over 3},1$. Hence the point $\vec c = (x(s),y(s)) = ({1\over 3},{5\over 3})$ or $(1,3)$.

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2. Let $f$ be a smooth function on $\mathbb{R}^3$. Show that if $f(\vec a) = f(\vec b)$, then there exists a point $\vec c$ between $\vec a$ and $\vec b$ for which $\nabla f(\vec c) \perp (\vec b - \vec a)$.
> (Sol) Set $g(t) = f((1-t)\vec a + t\vec b)$. Then $$g(0) = f(\vec a) = f(\vec b) = g(1)$$By mean value theorem, there exists an $s \in [0,1]$ such that $g'(s) = 0$. Now $$0 = g'(s) = \left.{d\over dt}\right|_{t = s}f((1-t)\vec a + t\vec b) = (\nabla f)((1-s)\vec a +s\vec b) \cdot (\vec b - \vec a)$$Hence there exists a point $\vec c = (1-s)\vec a + s\vec b$ such that $f(\vec c) \perp (\vec b - \vec a)$.

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5. Let $$x = r\cos \theta \qquad \text{and} \qquad y = r \sin \theta.$$
Suppose $u=u(x,y)$ is a smooth function.

a. Show that $$\nabla u = \dfrac{\partial u}{\partial r} \vec e_r + \dfrac{1}{r} \dfrac{\partial u}{\partial \theta} \vec e_\theta,$$ where $r \neq 0$, $$\vec e_r = \cos \theta \,\vec i + \sin \theta \,\vec j \qquad \text{and} \qquad \vec e_\theta = - \sin \theta \,\vec i + \cos \theta \,\vec j.$$
> (Sol) By definition, $$\begin{aligned}{\partial\over \partial r} &= {\partial x\over \partial r}{\partial \over \partial x} + {\partial y \over \partial r}{\partial \over \partial y} = \cos\theta {\partial \over \partial x} + \sin\theta {\partial \over \partial y} \\ {\partial\over \partial \theta} &= {\partial x\over \partial \theta}{\partial \over \partial x} + {\partial y \over \partial \theta}{\partial \over \partial y} = -r\sin\theta{\partial \over \partial x} + r\cos\theta {\partial \over \partial y}\end{aligned}$$To simplified the notation, write $\partial_{x} = {\partial \over \partial x}$ and so are others. We may use a matrix notation: $$\begin{pmatrix}\partial_{r}\\ \partial_{\theta}\end{pmatrix} = \begin{pmatrix}\cos \theta & \sin \theta \\-r \sin\theta & r \cos\theta\end{pmatrix}\begin{pmatrix}\partial_{x}\\\partial_{y}\end{pmatrix}$$By solving the system $$\begin{pmatrix}\partial_{x}\\ \partial_{y}\end{pmatrix} = {1\over r}\begin{pmatrix}r\cos \theta & -\sin \theta \\r \sin\theta & \cos\theta\end{pmatrix}\begin{pmatrix}\partial_{r}\\\partial_{\theta}\end{pmatrix}$$Now $$\begin{aligned}\nabla u &= ({\partial_{x}}u) \vec i + ({\partial_{y}u})\vec j \\ &= \left(\cos\theta \partial_{r}u - {1\over r}\sin\theta \partial_{\theta}u\right)\vec i + \left(\sin \theta\partial_{r}u + {1\over r}\cos \theta \partial_{\theta}u\right)\vec j \\ &= (\partial_{r}u) \vec{e}_{r} + {1\over r}(\partial_{\theta}u) \vec{e}_{\theta}\end{aligned}$$

b. Show that $$ \dfrac{\partial^2 u}{\partial x^2}+ \dfrac{\partial^2 u}{\partial y^2} = \dfrac{\partial^2 u}{\partial r^2}+ \dfrac{1}{r^2}\dfrac{\partial^2 u}{\partial \theta^2}+\dfrac{1}{r}\dfrac{\partial u}{\partial r} .$$
> (Sol) By definition, $$\partial_{x}^{2}u + \partial_{y}^{2}u = \partial_{x}(\partial_{x}u) + \partial_{y}(\partial_{y}u)$$Use the computation in (a), $$\begin{aligned}\partial_{x}(\partial_{x}u) &= \left(\cos\theta\partial_{r} - {1\over r}\sin\theta \partial_{\theta}\right)\left(\cos\theta\partial_{r}u - {1\over r}\sin\theta \partial_{\theta}u\right)\\ &= \cos^{2}\theta\partial_{r}^{2}u+{1\over r^2}\cos\theta\sin\theta\partial_{\theta}u - {1\over r}\cos\theta\sin\theta\partial_{r}\partial_{\theta}u+{1\over r}\sin^{2}\theta\partial_{r}u \\ &\quad - {1\over r}\sin\theta\cos\theta \partial_{\theta}\partial_{r}u + {1\over r^2}\sin\theta\cos\theta\partial_{\theta}u + {1\over r^2}\sin^{2}\theta\partial_{\theta}^{2}u \\ \partial_{y}(\partial_{y}u) &= \left(\sin \theta\partial_{r} + {1\over r}\cos \theta \partial_{\theta}\right)\left(\sin \theta\partial_{r}u + {1\over r}\cos \theta \partial_{\theta}u\right) \\ &= \sin^{2}\theta\partial_{r}^{2}u - {1\over r^2}\sin\theta\cos\theta\partial_{\theta}u +{1\over r}\sin\theta\cos\theta\partial_{r}\partial_{\theta}u + {1\over r}\cos^2\theta \partial_{r}u \\ &\quad + {1\over r}\cos\theta\sin\theta\partial_{\theta}\partial_{r}u - {1\over r^2}\cos\theta\sin\theta\partial_{\theta}u + {1\over r^2}\cos^{2}\theta\partial_{\theta}^{2}u\end{aligned}$$The result follows by adding them up.

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