MATH113 Calculus and Geometry

Test 2016: Solutions

• i) $𝐚\cdot 𝐛=(1,0,2)\cdot (-2,2,1)=0$. [2]

  $$ $𝐚\times 𝐛=(1,0,2)\times (-2,2,1)=(-4,-5,2)$. [2]

  ii) The point $C$ is $\overrightarrow{OA}+\overrightarrow{OB}=(2,1,0)+(3,1,1)=(5,2,1)$. [2]

  iii) The area of $OACB$ is $|\overrightarrow{OA}\times \overrightarrow{OB}|$. [1] We have $\overrightarrow{OA}\times \overrightarrow{OB}=(1,-2,-1)$. [2] The length of this vector is $\sqrt{6}$. [1]

  iv) We have $\cos \theta =\frac{𝐮\cdot 𝐯}{|𝐮||𝐯|}=\frac{\sqrt{3}}{|𝐮||𝐯|}$ [1] $$ and $|𝐮\times 𝐯|=|𝐮||𝐯|\sin \theta$. [1]

  Further, $|𝐮\times 𝐯|=|(1,2,2)|=3$. [1]

  Thus, $\cos \theta =\frac{\sqrt{3}\sin \theta }{|𝐮\times 𝐯|}=\frac{\sqrt{3}}{3}\sin \theta$, and hence $\tan \theta =\frac{3}{\sqrt{3}}=\sqrt{3}$. [1]

  
  

• i) To calculate the arc length, we recall the formula: $L={\int }_0^3|{\gamma }^{\prime }(t)|𝑑t$. [1]

  Here ${\gamma }^{\prime }(t)=(e^t-e^{-t},-2)$ [1], hence $|{\gamma }^{\prime }(t)|=\sqrt{{(e^t-e^{-t})}^2+4}=\sqrt{e^{2t}+e^{-2t}+2}$. [1] This equals $e^t+e^{-t}$. [1] It follows that the arc length is ${\int }_0^3(e^t+e^{-t})𝑑t=e^3-\frac{1}{e^3}$. [1]

  ii) We have $\gamma (0)=(2,5)$ [1] and ${\gamma }^{\prime }(0)=(0,-2)$ [1]. So an equation of the tangent line to the image of $\gamma$ at the point $\gamma (0)$ is $(x,y)=\gamma (0)+\lambda {\gamma }^{\prime }(0)=(2,5)+\lambda (0,-2)$. [1]

  
  

• The surface is $f(x,y,z)=0$, where $f(x,y,z)=x^4-y^2{z}^2$. [1]

  We have $\nabla f=(4x^3,-2y{z}^2,-2z{y}^2)$ [1] and $\nabla f(2,2,2)=(32,-16,-16)$. [1]

  So the normal line is $\{(2,2,2)+\lambda (2,-1,-1)|\lambda \in ℝ\}$ [1]

  and the tangent plane is $(2,-1,-1)\cdot (x,y,z)=(2,-1,-1)\cdot (2,2,2)=0$. [2]

  
  

• i) We check the partial derivatives ([1] for knowing the right condition). We have: ${(f_2)}_z=2z$ and ${(f_3)}_y=2y$ [1], so the partial derivatives do not match and hence $𝐟$ cannot be expressed as $\nabla \varphi$. [1]

  ii) Here ${(g_1)}_y=2x-z$, ${(g_2)}_x=2x-z$. [1] Similarly ${(g_1)}_z=-y-4z$, ${(g_3)}_x=-4z-y$. [1] Finally, ${(g_2)}_z=6y-x$, ${(g_3)}_y=6y-x$ [1], so all of the conditions are satisfied and hence $𝐠$ can be expressed as $\nabla \varphi$ for some $\varphi$. To find $\varphi$, since ${\varphi }_x=2xy-2z^2-yz$, we obtain: $\varphi =x^{2}y-2x{z}^2-xyz+h(y,z)$. [1] Now, checking against ${\varphi }_y=g_2$, ${\varphi }_z=g_3$, we obtain:

  $${\varphi }_y=x^2-xz+h_y=x^2+6yz-xz,{\varphi }_z=-4xz-xy+h_z=3y^2-4xz-xy[\mathrm{𝟏}]$$

  Hence we have: $h_y=6yz$ and $h_z=3y^2$, so we deduce: $h(y,z)=3y^{2}z$ ($+c$). It follows that $\varphi (x,y,z)=x^{2}y-2x{z}^2+3y^{2}z-xyz$. [1]

  
  

• Let $f(x,y)=x^2+2y^2$. We form the auxiliary function $\mathrm{\Lambda }(x,y,\lambda )=x^2+2y^2-\lambda (x^2+x^2-1)$. [2] Then we look for stationary points:

  $${\mathrm{\Lambda }}_x=2x-2\lambda x=0,{\mathrm{\Lambda }}_y=4y-2\lambda y=0[\mathrm{𝟐}]$$

  By the first equation, we have $x=0$ or $\lambda =1$. [1]

  If $x=0$, then $y^2=1$ (since ${\mathrm{\Lambda }}_{\lambda }=0$) and hence $y=\pm 1$.

  If $\lambda =1$, then $y=0$ (by the second equation) and hence $x=\pm 1$.

  So the possible extreme points are $(0,1),(0,-1),(1,0),(-1,0)$. [1]

  We have $f(0,1)=2$, $f(0,-1)=2$, $f(1,0)=1$, $f(-1,0)=1$.

  We deduce that the greatest value of $f$ is 2 [1], and the least value is $1$. [1]

  
  

• The region $D$ is given by: $0\le r\le 3,\mathrm{\hspace{0.33em}0}\le \theta \le \pi /2$ [2], so

  $$I={\int }_0^{\pi /2}{\int }_0^3\frac{1}{r^2+1}rdrd\theta .\mathit{\hspace{1em}\hspace{1em}}[\mathrm{𝟐}]$$

  Now

  $${\int }_0^3\frac{r}{r^2+1}𝑑r={\left[\frac{1}{2}\log (r^2+1)\right]}_0^3=\frac{1}{2}\log 10,[\mathrm{𝟏}]$$

  so   $I=\pi /4\log 10$. [1]