A Collection of Problems in Differential Calculus: Problems from Calculus I Final Examinations, 2000-2020 Department of Mathematics, Simon Fraser University

Section 6.15 Polar Coordinates

1. $(x^2+y^2)^3=(y^2-x^2)\text{.}$ Multiply by $r^2$ and use the fact that $\cos 2\theta =cos ^2\theta -\sin ^2\theta$

2. See Figure 6.15.1.

   

   

3. (a) $r = 2\text{,}$ (b) $r = 2\cos \theta\text{,}$ (c) $r = \sin \theta\text{.}$

4. On the given cardioid, $x = (1 + \cos \theta ) \cos \theta$ and $y = (1 + \cos \theta ) \sin \theta\text{.}$ The question is to find the maximum value of $y\text{.}$ Note that $y > 0$ is equivalent to $\sin \theta > 0\text{.}$ From $\ds \frac{dy}{d\theta } = 2\cos ^2 \theta + \cos \theta -1$ we get that the critical numbers of the function $y = y(\theta )$ are the values of $\theta$ for which $\ds \cos \theta = \frac{-1 \pm 3}{4}\text{.}$ It follows that the critical numbers are the values of $\theta$ for which $\ds \cos \theta = -1$ or $\ds \cos \theta = \frac{1}{2}\text{.}$ Since $y_{\mbox{max} } > 0$ it follows that $\ds \sin \theta = \sqrt{1-\left( \frac{1}{2}\right) ^2}=\frac{\sqrt{3}}{2}$ and the maximum height equals $\ds y = \frac{3\sqrt{3}}{4}\text{.}$ See Figure 6.15.2.

   

   

5. See Figure 6.15.3.

   

6. See the graph $r=\cos 3\theta$ (Figure 6.15.1) and apply the appropriate stretching.

7. See Figure 6.15.2.

   

8. For (a) see Figure 6.15.4 and for (b) see Figure 6.15.1. For (c) and (d) see Figure 6.15.5 and for (e) and (f) see Figure 6.15.6.

   

   

   

   

9. (a) $(x,y)=((1-\cos\theta)\cdot \cos\theta, (1-\cos\theta)\cdot \sin\theta)\text{.}$ (b) $\ds \left.\frac{dy}{dx}\right|_{\theta=\frac{\pi}{2}}=-1\text{;}$ $y=-x+1\text{.}$ (c) Horizontal tangent lines at $\ds \left(-\frac{3}{4},\frac{3\sqrt{3}}{4}\right), \ \left(-\frac{3}{4},-\frac{3\sqrt{3}}{4}\right)\text{;}$ Vertical tangent line at $\ds (0,0), \ (-2,0), \left(\frac{1}{4},\frac{\sqrt{3}}{4}\right),\ \left(\frac{1}{4},-\frac{\sqrt{3}}{4}\right)\text{.}$

10. (a) $\ds \frac{dy}{dx}=\frac{-2\sin 2\theta\sin\theta+\cos\theta\cos2\theta}{-2\sin 2\theta\cos\theta-\sin \theta\cos 2\theta}\text{.}$ (b) $\ds \left(\frac{1}{2}\sqrt{1+\frac{1}{\sqrt{2}}},\frac{1}{2}\sqrt{1-\frac{1}{\sqrt{2}}}\right)\text{.}$ (c) $y- \frac{1}{2}\sqrt{1-\frac{1}{\sqrt{2}}}= \frac{2\sqrt{2-\sqrt{2}}-\sqrt{2+\sqrt{2}}}{2\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}}\cdot\left(x-\frac{1}{2}\sqrt{1+\frac{1}{\sqrt{2}}}\right)\text{.}$ (d) See Figure 6.15.7.

    

11. (a) $(-2,-2\sqrt{3})\text{.}$ (b) See Figure 6.15.8. (c) $\ds -\frac{1}{\sqrt{3}}\text{.}$

    

    

12. (a) $(2\sqrt{3},2)\text{.}$ (b) See Figure 6.15.8. (c) $\ds \sqrt{3}\text{.}$

13. (a) $\left(\frac{5\sqrt{3}}{4},\frac{5}{4}\right)\text{.}$ (b) Observe that $r(0)=r(\pi)=4$ and $\ds r\left(\frac{\pi}{2}\right)= r\left(\frac{3\pi}{2}\right)=1\text{.}$ (c) $\ds \frac{\sqrt{3}}{23}\text{.}$

14. (a) $(0,-3)\text{.}$ (b) Solve $y=(1-2\sin \theta)\sin \theta =0\text{.}$ $(\pm 1,0)\text{.}$ (c) The middle graph corresponds to $r=1+\sin 2\theta$ and the right graph corresponds to $r=1-2\sin \theta\text{.}$

15. (a) $(1,\sqrt{3})\text{.}$ (b) $\ds \frac{1}{3\sqrt{3}}\text{.}$ (c) B.

16. (a) See Figure 6.15.9.

    

    (b) 9. (c) $\theta = 0\text{,}$ $\ds \frac{\pi }{3}\text{,}$ $\ds \frac{2\pi }{3}\text{,}$ $\pi\text{,}$ $\ds \frac{4\pi }{3}\text{,}$ $\ds \frac{5\pi }{3}\text{,}$ $2\pi\text{.}$ (d) The remaining points of intersection are obtained by solving $-1 = 1 + 2 \sin 3\theta\text{.}$

17. (a) $r(0) = 2\text{,}$ $\ds r \left( \frac{\pi }{2}\right) = 2+e\text{,}$ $\ds r \left( \frac{3\pi }{2}\right) = e^{-1}\text{.}$ (b) See Figure 6.15.10.

    

    (c) From $\ds \frac{dr}{d\theta }= \cos \theta (1 + e^{\sin \theta }) = 0$ we conclude that the critical numbers are $\ds \frac{\pi }{2}$ and $\ds \frac{3\pi }{2}\text{.}$ By the Extreme Value Theorem, the minimum distance equals $e^{-1}\text{.}$

18. (a) $\ds A=\left( r=\sqrt{2},\theta= \frac{\pi }{4}\right)\text{,}$ $\ds B=\left( 4,\frac{5\pi }{3}\right)\text{,}$ $\ds C=\left( 2,\frac{7\pi }{6}\right)\text{,}$ $\ds D=\left( 2\sqrt{2}-1,\frac{3\pi }{4}\right)\text{;}$ (b) A, B, D.

19. (a) See Figure 6.15.11. (b) Solve $\ds \sin \theta >-\frac{1}{2}$ to get $\ds \theta \in \left[ -\pi, -\frac{5\pi }{6}\right) \cup \left( -\frac{\pi }{6},\pi \right)\text{.}$ (c) To find critical numbers solve $\ds \frac{dr}{d\theta }=2\cos \theta =0$ in $[-\pi ,\pi )\text{.}$ It follows that $\ds \theta =-\frac{\pi }{2}$ and $\ds \theta =\frac{\pi }{2}$ are critical numbers. Compare $r(-\pi )=r(\pi )=1\text{,}$ $\ds r\left( -\frac{\pi }{2}\right) =-1\text{,}$ and $\ds r\left( \frac{\pi }{2}\right) =3$ to answer the question.

    

20. (a) See Figure 6.15.12. (b) The slope is given by $\ds \left. \frac{dy}{dx}\right| _{\theta =\frac{5\pi }{2}}\text{.}$ From $x=r\cos \theta =\theta \cos \theta$ and $y=\theta \sin \theta$ it follows that $\ds \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{\sin \theta +\theta \cos \theta }{\cos \theta -\theta \sin \theta}\text{.}$ Thus $\ds \left. \frac{dy}{dx}\right| _{\theta =\frac{5\pi }{2}}=-\frac{2}{5\pi }\text{.}$ (c) $\ds \sqrt{x^2+y^2}=\arctan \frac{y}{x}\text{.}$

    

21. (a) $(0,5)\text{.}$ (b) $x^2+y^2=5y\text{.}$ (c) $x=5\sin \theta \cos \theta\text{,}$ $y=5\sin ^2\theta\text{.}$ (d) $\ds \frac{dy}{dx}=\frac{2\sin \theta\cos \theta }{\cos ^2\theta - \sin ^2\theta}=\tan 2\theta\text{.}$ (e) $\ds y -\frac{5}{4}=\sqrt{3}\left( x-\frac{5\sqrt{3}}{4}\right)\text{.}$

22. Solve $2=4\cos \theta$ to get that curve intersect at $( 1, \sqrt{3})\text{.}$ To find the slope we note that the circle $r=2$ is given by parametric equations $x=2\cos \theta$ and $y=2\sin \theta\text{.}$ It follows that $\ds \frac{dy}{dx}=\frac{2\cos \theta }{-2\sin \theta }=-\cot \theta\text{.}$ The slope of the tangent line at the intersection point equals $\ds \left. \frac{dy}{dx}\right| _{\theta = \frac{\pi }{3}}=-\frac{\sqrt{3}}{3}\text{.}$

23. See Figure 6.15.13 for the graph of the case $b=1\text{,}$ $k=0.01\text{,}$ and $c=2\text{.}$ The position in $(x,y)-$plane of the bee at time $t$ is given by a vector function $\ds \vec{s}(t)=\langle be^{kt}\cos ct,be^{kt}\sin ct\rangle\text{.}$ Recall that the angle $\alpha$ between the velocity and acceleration is given by $\ds \cos\alpha=\frac{\vec{v}\cdot\vec{a}}{|\vec{v}||\vec{a}|}\text{,}$ where $\vec{v}(t)=\vec{s}^\prime(t)$ and $\vec{a}(t)=\vec{s}^{\prime\prime}(t)\text{.}$ One way to solve this problem is to consider that the bee moves in the complex plane. In that case its position is given by

    $$\begin{equation*} F(t)=be^{kt}\cos ct+i\cdot be^{kt}\sin ct=be^{kt}(\cos ct +i\sin ct)=be^{(k+ic)t}\text{,} \end{equation*}$$

    where $i$ is the imaginary unit. Observe that $\vec{v}(t)=\langle \mbox{Re} (F^\prime(t)),\mbox{Im} (F^\prime(t))\rangle$ and $\vec{a}(t)=\langle \mbox{Re} (F^{\prime\prime}(t)),\mbox{Im} (F^{\prime\prime}(t))\rangle\text{.}$ Next, observe that $F^\prime(t)=(k+ic)F(t)$ and $F^{\prime\prime}(t)=(k+ic)^2F(t)\text{.}$ From $F^{\prime\prime}(t)=(k+ic)F^\prime(t)$ it follows that $\mbox{Re} (F^{\prime\prime}(t))=k\cdot \mbox{Re} (F^\prime(t))-c\cdot \mbox{Im} (F^\prime(t))$ and $\mbox{Im} (F^{\prime\prime}(t))=k\cdot \mbox{Im} (F^\prime(t))+c\cdot \mbox{Re} (F^\prime(t))\text{.}$ Finally, $\vec{v}\cdot\vec{a}=\mbox{Re} (F^\prime(t))\cdot \mbox{Re} (F^{\prime\prime}(t))+\mbox{Im} (F^\prime(t))\cdot \mbox{Im} (F^{\prime\prime}(t))=k((\mbox{Re} (F^\prime(t)))^2+(\mbox{Im} (F^\prime(t)))^2)=k|F^\prime(t)|^2$ which immediately implies the required result.