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

Section 6.14 Parametric Curves

1. $\displaystyle x=-5\sin t, \ y=5\cos t, \ t\in[0,2\pi]$

2. From $3t^3+2t-3=2t^3+2$ obtain $t^3+2t-5=0\text{.}$ What can you tell about the function $f(t)=t^3+2t-5\text{?}$

3. $\ds \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left( \frac{dy}{dx}\right) }{\frac{dx}{dt}}=\frac{3t(2\cos t+t\sin t)}{2\cos ^3t}\text{.}$

4. Note that $x+y=1$ and that $x,y\in [0,1]\text{.}$ See Figure 6.14.1.

   

5. $y=x+2\text{.}$

6. (a) $\ds \frac{dx}{dy}=\frac{1-\cos \theta }{\sin \theta}\text{.}$ (b) $\ds \frac{d^2x}{dy^2}=\frac{1-\cos \theta }{\sin ^3 \theta}\text{.}$ (c) $\ds y=\sqrt{3}x-\frac{\pi \sqrt{3}}{3}+2\text{.}$

7. A. Follow the curve as $t$ increases.

8. (a) Express $\cos t$ and $\sin t$ in terms of $x$ and $y$ to get the circle $(x+2)^2+(y-1)^2=4\text{.}$ Check which points correspond to $t=0$ and $\ds t=\frac{\pi }{2}$ to get the orientation. (b) Solve $\ds \frac{dy}{dx}=\cot t=1$ for $t\in (0,2\pi )\text{.}$

9. (a) $\ds \frac{dy}{dx}=-\frac{2\cos 2t}{3\sin 3t}\text{;}$ (b) Note that for both $\ds t=\frac{\pi }{2}$ and $\ds t=\frac{3\pi }{2}$ the curve passes through the origin. Thus, $\ds y=\pm \frac{2x}{3}\text{.}$ (c) C.

10. (a) $\ds \frac{dy}{dx}=2t-2\text{.}$ (b) $y=x^2+1\text{.}$ (c) $m=2x_1\text{,}$ $b=1-x_1^2\text{.}$ (d) Note that the point $(2,0)$ does not belong to the curve. Since all tangent lines to the curve are given by $y=2x_1x+1-x_1^2\text{,}$ $x_1\in \mathbb{R}\text{,}$ we need to solve $0=4x_1+1-x_1^2$ for $x_1\text{.}$ Hence $x_1=2\pm \sqrt{5}\text{.}$ The tangent lines are given by $y=2(2\pm \sqrt{5})x-8\mp 4\sqrt{5}\text{.}$

11. (a) Note that if $t=0$ then $x=y=0\text{.}$ (b) Solve $x=t^3-4t=t(t^2-4)=0\text{.}$ Next, $y(-2)=16\text{,}$ $y(0)=y(2)=0\text{.}$ (c) From $\ds \frac{dy}{dx}=\frac{4(t-1)}{3t^2-4}$ it follows that $\ds \left. \frac{dy}{dx}\right| _{t=0}=1$ and $\ds \left. \frac{dy}{dx}\right| _{t=2}=\frac{1}{2}\text{.}$ The tangent lines are $\ds y=x$ and $\ds y=\frac{x}{2}\text{.}$

12. Observe that $\ds t=\frac{y}{x}\text{.}$ (a) $x^3+y^3=3xy\text{.}$ (b) From $\ds \left.\frac{dx}{dt}\right|_{t=1}=-\frac{3}{4}$ and $\ds \left.\frac{dy}{dt}\right|_{t=1}=\frac{3}{4}$ it follows that the slope equals to $\ds \left.\frac{dy}{dx}\right|_{t=1}=-1\text{.}$ Alternatively, differentiate the expression in (a) with respect to $x\text{.}$ (c) Use the fact that $\ds \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{1}{\frac{dx}{dt}}\cdot \frac{d}{dt}\left(\frac{dy}{dx}\right)$ to obtain $\ds \left.\frac{d^2y}{dx^2}\right|_{t=1}=-\frac{32}{3}\text{.}$ Alternatively, differentiate the expression in (a) with respect to $x$ twice.

13. (a) $\ds \frac{dy}{dx}=-e^t\cdot 2^{2t+1}\ln 2\text{.}$ (b) $\ds \frac{d^2y}{dx^2}=e^{2t}\cdot 2^{2t+1}\cdot (1+2\ln 2)\ln 2\text{.}$ (c) No, the second derivative is never zero. (d) $\ds y=2^{-2\ln x}\text{.}$ (e) See Figure 6.14.2.

    

14. (a) $\ds \frac{dy}{dx}=1+t\text{.}$ (b) $y=x-1\text{.}$ (c) $\ds \frac{d^2y}{dx^2}=e^{-t}\text{.}$ (d) Observe that $\ds \frac{d^2y}{dx^2}>0$ for all $t\in\mathbb{R}\text{.}$

15. a) $\ds \frac{dy}{dx}=\frac{1}{4}e^{1-5t}\text{.}$ (b) $\ds \frac{d^2y}{dx^2}=\frac{5}{16}e^{1-9t}\text{.}$ (c) Note that $\ds \left. \frac{d^2y}{dx^2}\right| _{t=0}=\frac{5}{16}e>0\text{.}$

16. a) $\ds \frac{dy}{dx}=-t^2\text{.}$ (b) $\ds y=2-\frac{(x-1)^2}{3}\text{.}$

17. (a) $\ds \frac{dy}{dx}=\frac{\cos t-t\sin t}{\sin t+t\cos t}\text{.}$ (b) $\ds y=-\frac{\pi }{2}\left( x-\frac{\pi }{2}\right)\text{.}$ (c) $\ds \left. \frac{d^2y}{dx^2}\right| _{t=\frac{\pi }{2}}=-2-\frac{\pi ^2}{4}\lt 0\text{.}$

18. (a) $\overrightarrow{v}=b\omega \langle -\sin \omega t, \cos \omega t\rangle\text{;}$ $\mbox{speed } =|\overrightarrow{v}|=|b\omega|\text{.}$ (b) $\overrightarrow{a}=-b\omega ^2\langle\cos \omega t, \sin \omega t\rangle\text{.}$

19. (a) $\ds \frac{dy}{dx}=\frac{t^2-1}{2t}\text{.}$ (b) $\ds y-2=\frac{3}{4}\left( x-3\right)\text{.}$

20. (a) $-27\text{,}$ $0\text{.}$ Solve $y=0$ for $t\text{.}$ (b) $(-24,8)\text{.}$ (c) $y=x+32\text{.}$ (d) $\ds \frac{d^2y}{dx^2}=\frac{t^2+3}{12t^3}\text{.}$

21. (a) $(3,0)\text{.}$ (b) $(0,0)\text{.}$ (c) See Figure 6.14.3. (d) $4x+6y=9\sqrt{3}\text{.}$ (e) Conclude from the graph and from what you have observed in (a) — (c) that the curve is concave up if $\ds t\in \left(\frac{\pi}{4},\frac{\pi}{2}\right)\cup \left(\frac{3\pi}{4},\pi\right)\text{.}$ Alternatively you may discuss the sign of the second derivative.

    

22. From $\ds \frac{dy}{dx}=\frac{3\sin t}{1-2\cos 2t}$ we conclude that $\ds \frac{dy}{dx}$ is not defined if $1-2\cos 2t=0\text{,}$ $0\leq t\leq 10\text{.}$ Thus, $\ds \theta \in \{ \frac{\pi }{6}, \frac{5\pi }{6}, \frac{7\pi }{6}, \frac{11\pi }{6}, \frac{13\pi }{6}, \frac{17\pi }{6}\}\text{.}$

23. (a) $\ds \frac{dy}{dx}=-2e^{3t}\text{.}$ (b) $\ds \frac{d^2y}{dx^2}=6e^{4t}\text{.}$ (c) $\ds \frac{d^2y}{dx^2}>0\text{.}$ (d) $\ds y=x^{-2}\text{.}$

24. (a) $\ds \frac{dy}{dx}=\frac{3\sin ^2t\cos t}{-3\cos ^2t\sin t}=-\tan t\text{.}$ (b) From $\ds \frac{d^2y}{dx^2}=-\frac{\sec ^2 t}{-3\cos ^2t\sin t}=\frac{1}{3\cos ^4t \sin t}$ we get $\ds \left. \frac{d^2y}{dx^2}\right| _{t=1}=\frac{1}{3\cos ^41 \sin 1}>0\text{.}$

25. (a) $(1,0)\text{.}$ (b) $\ds \left( -\frac{\sqrt{2}}{4},\frac{\sqrt{2}}{4}\right)\text{.}$ (c) See Figure 6.14.4. (d) Observe that $\ds \frac{dy}{dx}=-\tan t\text{,}$ $t\not\in \{0,\pi/2, \pi,3\pi/2\}\text{.}$ $2x-2y+\sqrt{2}=0\text{.;}$ (e) Concave down. See the graph or discuss the sign of the second derivative. What is the domain of the second derivative?

    

26. (a) Observe that by eliminating the parameter $t\text{,}$ the equation becomes $y^2=x\text{,}$ $(x,y)\in[0,1]\times[-1,1]\text{.}$ See Figure 6.14.5. (b) Observe that solving $\ds \frac{dy}{dx}=\frac{1}{2\cos t}=\frac{1}{\sqrt{3}}$ gives $t=\ds \frac{\pi}{6}\text{.}$ $4x-4\sqrt{3}y+3=0\text{.}$

    

27. (a) $\ds \frac{dy}{dx}=-e^{-2t}, \frac{d^2y}{dx^2}=e^{-3t}\text{.}$ (b) From $\ds \frac{dy}{dx}=-e^{-2t}=-1$ we get that $t=0\text{.}$ Thus the tangent line is $y=-x\text{.}$

28. (a) $(0,0)\text{,}$ $(0,-9)\text{.}$ (b) From $\ds \frac{dy}{dx}=\frac{2t}{t^2-1}$ we get that the tangent line is horizontal at the point $(0,-9)$ and vertical at the points $(-2,-6)$ and $(2,-6)\text{.}$ (c) See Figure 6.14.6.

    

29. (a) $\ds \frac{dy}{dx}=\frac{2}{3}e^{-4t-1}\text{.}$ (b) $\ds \frac{d^2y}{dx^2}=\frac{8}{9}e^{-7t-3}\text{.}$