Remark

  1. In Question 3, we use simply compare values to determine the maximum or minimum on the boundary. While we acknowledge that this method is not entirely rigorous, we have not covered a more rigorous method in this course.
  2. It is recommended that you review the double angle formula before the midterm.
  3. We did not explicitly introduce the second method used in Question 6. We suggest that you review the method.

Let $f(x,y)=\sqrt{1+x+2y}$.

  1. Find Gradient of $f$, and explain what the meaning of gradient.
  2. Find vectors $\vec{v}$ such that the directional derivative of $f$ along the vector $\vec{v}$ at $(2,3)$ is $0$.
  3. Find an equation of the tangent plain to the graph $z=f(x,y)$ when $x=2$ and $y=3$.
  4. Use the linearization of $f$ at $(2,3)$ to estimate the value of $f(2.01,2.99)$.
Answer
  1. $f_x = \frac{1}{2\sqrt{1+x+2y}}$, $f_y = \frac{1}{\sqrt{1+x+2y}}$. The gradient of a function $f$ at a point $(a,b)$ is a vector pointing to the direction where has largest directional derivative at the point, i.e. \(\frac{\partial f}{\partial\nabla f} = \max_\vec{v}\left\\{\frac{\partial f}{\partial \vec{v}}\right\\}.\)
  2. We notice that $\vec{v}$ being perpendicular to $\nabla f$ follows from the following equation \(\frac{\partial f}{\partial\vec{v}} = \vec{v}\cdot\nabla f =0.\) Therefore, $\vec{v}= \left(\frac{\partial f}{\partial y}(2,3),-\frac{\partial f}{\partial x}(2,3)\right) = \left(-\frac{1}{3},\frac{1}{6}\right)$
  3. $\frac{1}{6}(x-2)+\frac{1}{3}(y-3)=z-3$
  4. Let $L(x,y)=3+\frac{1}{6}(x-2)+\frac{1}{3}(y-3)$. We have $f(2.01,2.99)\approx 3-\frac{0.01}{3}$.

Let $f(x,y)=x^2y-xy^2$. Suppose both $x$ and $y$ are functions of $t$. Say $x(t)=\ln(t)$ and $y(t)=\tan(t)$. Find $\frac{\partial F}{\partial t}$.

Answer

$\frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t} = \frac{2\ln(t)-\tan^2(t)}{t} + (2\ln(t)\tan(t)-\tan^2(t))\sec^2(t)$.

Suppose the function $f(x,y)=ax^2+bxy+cy^2+3x$ has a saddle point when $x = -\frac{6}{5}$ and $y = \frac{3}{5}$, and suppose $f(0,1)=1$.

  1. Find $a$, $b$, and $c$.
  2. Find the local maximal and local minimal on the region $x^2+y^2\leq 9$.
Answer
  1. Since $\left(-\frac{6}{5},\frac{3}{5}\right)$ is a critical point of $f$, so we may set the following system of equations \(\nabla f\left(-\frac{6}{5},\frac{3}{5}\right) = (0,0)\Rightarrow \begin{cases} 4a-b=3\\\\ b-c=0 \end{cases}.\) Moreover, $c=1$ follows $f(0,1)=1$. Hence, we have $a=1$ and $b=1$.
  2. We already know $\left(-\frac{6}{5},\frac{3}{5}\right)$ is a critical point in the region $x^2+y^2\leq 9$. It remains to identify critical points on the boundary. By Lagrange method, we have to solve, for $g(x,y)=x^2+y^2$, \(\begin{cases} \nabla f =\lambda\nabla g\\\\ g = 9 \end{cases}\Rightarrow \begin{cases} 2x + y + 3 = \lambda 2x\\\\ x + 2y = \lambda 2y\\\\ x^2 + y^2 = 9 \end{cases}.\) We devide the first equation by the second, and derive \(\begin{cases} \frac{2x+y+3}{x+2y}=\frac{x}{y}\\\\ x^2 + y^2 = 9 \end{cases}\Rightarrow \begin{cases} x^2 - y^2 = 3y\\\\ x^2 + y^2 = 9 \end{cases}\Rightarrow 2y^2 + 3y - 9 = 0.\) Therefore, we have $y=\frac{3}{2}$ or $-3$, and $x=\pm\frac{3\sqrt{3}}{2}$ or $0$ respectively.
    For $(x,y) = (\frac{3\sqrt{3}}{2},\frac{3}{2})$, $f(x,y) = \frac{27(1+\sqrt{3})}{4}$ (max).
    For $(x,y) = (0,3)$, $f(x,y) = 9$ (min).

Evaluate $\int_0^4\int_{\frac{\sqrt{y}}{2}}^1\sqrt{x^3+3}dxdy$.

Answer

To interchange the order of integration, we have to rewrite the limits of $\Omega: 0<y<4,\ \frac{\sqrt{y}}{2}<x<1$. It is \(\Omega: 0<x<1,\ 0<y<4x^2.\) Thus, the integral is \(\int_0^1\int_0^4x^2 \sqrt{x^3+3}dydx = \int_0^1 4x^2\sqrt{x^3+3}dx = \left.\frac{8}{9}(x^3+3)^{3/2}\right|^1_0.\)

Calculate the area of the region inside a half cardioid \(r=1-\sin(\theta)\)

polar_plot(1-sin(x), (x,-pi/2,pi/2), color='blue')
Answer

The area is the following double integral: \(\begin{aligned} \int_{-\pi/2}^{\pi/2}\int_0^{1-\sin(\theta)} r drd\theta &= \frac{1}{2}\int_{-\pi/2}^{\pi/2} (1-\sin\theta)^2 d\theta\\\\ &= -\int_{-\pi/2}^{\pi/2} \sin\theta d\theta + \frac{1}{2}\int_{-\pi/2}^{\pi/2} 1 + \sin^2\theta d\theta\\\\ &= \left. \cos\theta\right\vert_{-\pi/2}^{\pi/2} + \frac{1}{2}\int_{-\pi/2}^{\pi/2}\frac{3}{2} - \frac{1}{2} + \sin^2\theta d\theta\\\\ &= 0 + \left. \frac{3}{4}\theta \right\vert_{-\pi/2}^{\pi/2} - \frac{1}{2}\int_{-\pi/2}^{\pi/2} \cos2\theta d\theta\\\\ &= \frac{3}{4}\pi + \left. \frac{1}{4}\sin 2\theta \right\vert_{-\pi/2}^{\pi/2} = \frac{3}{4}\pi. \end{aligned}\)

Evaluate $\int\int_\Omega (x^2-y^2)dxdy$ where $\Omega$ is enclosed by $y-x=1$, $y-x=0$, $xy=2$ and $xy=1$.

Answer

Consider a change of variable $(u,\ v) = \Phi(x,y) = (y-x,\ xy)$. Then, we have \(\Omega' = \Phi(\Omega): 0\leq u\leq 1,\ 1\leq v\leq 2.\)

  • Method 1: If we desire to change variables from $x$ and $y$ to $u$ and $v$, we have to find “inverse” function of $\Phi$. We notice that \(y+x = \sqrt{u^2+4v}.\) Thus, we can consider $\Psi(u,v) = \left(\frac{\sqrt{u^2+4v}-u}{2},\frac{u+\sqrt{u^2+4v}}{2}\right)$, and note that $\Phi\circ\Psi = Id(u,v)$, the identity map on $uv$-coordinate. We compute Jacobian for the change of variable \(J_{\Psi}(u,v) = \frac{1}{\sqrt{u^2+4v}}.\) Then, the double integral is \(\int_1^2\int_0^1 u(\sqrt{u^2+4v})\frac{1}{\sqrt{u^2+4v}}dudv = \int_1^2\int_0^1 u dudv.\) It follows that the value of the double integral is $1$.
  • Method 2: Observing $dxdy = J_{\Psi}(u,v)dudv$ and $dudv = J_{\Phi}dxdy$, we notice that $J_{\Psi}(u,v) = J_{\Phi}(x,y)^{-1}$. Therefore, we have \(\begin{aligned} \int\int_\Omega (x^2-y^2)dxdy &= \int\int_{\Omega'}(x^2-y^2)J_{\Psi}(u,v)dudv\\\\ &= \int\int_{\Omega'}(x^2-y^2)J_{\Phi}(x,y)^{-1}dudv. \end{aligned}\) where \(J_{\Phi}(x,y) = -x-y.\) Thus, it follows that \(\int\int_{\Omega'}(x^2-y^2)J_{\Phi}(x,y)^{-1}dudv = \int\int_{\Omega'}-(x-y)dudv = \int_1^2\int_0^1 ududv = 1.\)