Calculus B Midterm

--- $\usepackage{mathtools}$ $\usepackage[utf8]{inputenc}$ $\usepackage{xcolor}$ --- $\newcommand{\dfdx}{\frac{\partial f}{\partial x}} \newcommand{\defeq}{\mathrel{\colon}=} \newcommand{\gradf}{\nabla f} \newcommand{\fsub}[1]{f \vert_{\partial U_{#1}}} \newcommand{\ddx}{\frac{d}{dx}} \newcommand{\ddxx}{\frac{d^2}{dx^2}} \newcommand{\DDx}{\dfrac{d}{dx}} \newcommand{\vecbracket}[1]{\left\langle #1 \right\rangle}$ # Calculus B Midterm :::success 4. Let $f \, \colon \Bbb R^2 \to \Bbb R$, $f(x,y)=\sin(xy) + \cos(xy)$. **($\text{a}$)** [5pts] Find the directional derivative of $f$ at the point $(0,1)$ in the direction in the direction $\vec{v} = (3,4)$. **($\text{b}$)** [5pts] What is the direction at $(0,1)$ that $f$ increases most rapidly? Express your answer by a unit vector. **($\text{c}$)** [5pts] Find the tangent plane to the graph of $z=f(x,y)$ through $(0,1,1)$. ::: **($\text{a}$)** The partial derivatives of $f$ are $$ \begin{cases} f_x = y \cos(xy) - y \sin(xy) \quad &\color{red}{ \text{[1pt]} }\\ f_y = x \cos(xy) - x \sin(xy) \quad &\color{red}{ \text{[1pt]} } \end{cases} $$ then the gradient of $f$ at $(0,1)$ is $\gradf \vert_{(0,1)} = \vecbracket{1,0} \, .$ [1pt] Thus the directional derivative is \begin{split} D_{\vec{v}}f\,(0,1) &= \gradf \vert_{(0,1)} \cdot \frac{\vec{v}}{|\vec{v}|} \quad \color{red}{ \text{[1pt]} } \\ &= \vecbracket{1,0} \cdot \vecbracket{\frac{3}{5}, \frac{4}{5}} = \frac{3}{5} \, . \quad \color{red}{ \text{[1pt]} } \end{split} --- **($\text{b}$)** The gradient vector is exactly the direction with max rate of changing. [3pts] Thus the answer is $\vecbracket{1,0}$. [2pts] --- **($\text{c}$)** The equation to the tangent plane is $$z = f_x(0,1) \cdot (x-0) + f_y(0,1) \cdot (y-1) + f(0,1) \, . \quad \color{red}{ \text{[3pt]} }$$ Thus the answer is $$z = x+1 \, . \quad \color{red}{ \text{[2pt]} }$$ :::success 5. [20pts] Use the Lagrangian multiplier to find the area of the largest rectangle inscribed in the ellipse $\frac{x^2}{16} + \frac{y^2}{9}=1$ with sides parallel to the coordinate axes. ::: W.L.O.G., we can only find the vertex of the rectangle $(x,y)$ at the first quadrant, i.e. $x>0$ and $y>0$. The object function is the area of the rectangle $f(x,y)=4xy$ or $f(x,y)=xy\,$. [2pts] The constraint is $\displaystyle g(x,y)=\frac{x^2}{16}+\frac{y^2}{9} \ (-1)\,$. [2pts] Now compute the gradient of functions $f$ and $g$, $$ \begin{cases} \nabla f = \vecbracket{y, x} \quad \color{red}{ \text{[2pts]} } \\ \nabla g = \vecbracket{\dfrac{x}{8}, \dfrac{2y}{9}} \quad \color{red}{ \text{[2pts]} } \end{cases} $$ For the Lagrangian multiplier $\nabla f = \lambda \nabla g$, we need to solve the following set of equations $$ \begin{cases} y = \lambda \dfrac{x}{8} \quad \color{red}{ \text{[2pts]} } & (1)\\ x = \lambda \dfrac{2y}{9} \quad \color{red}{ \text{[2pts]} } & (2) \\ \dfrac{x^2}{16} + \dfrac{y^2}{9}=1 \quad \color{red}{ \text{[1pt]} } & (3) \end{cases} $$ Since it's meaningless if $x=0$ or $y=0$, thus we know $x \neq 0$, $y \neq 0$ and $\lambda \neq 0$. Now multiply the equation $(1)$ and $(2)$, we can get $\displaystyle xy = \lambda^2 \frac{2xy}{72}$, then $\lambda = \pm 6$ (choose positive) [2pts]. Hence we have $\displaystyle y=\frac{3}{4}x$ [2pts], with the equation $(3)$, we will get $\displaystyle (x,y)=(2\sqrt{2}, 3 / \sqrt{2})$. [2pts] Thus the largest area is $2\sqrt{2} \cdot 3 / \sqrt{2} \cdot 4 = 24$. [1pt] ## QR Code <img style="float: left;" src="https://i.imgur.com/4QmFo4y.png">