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Section 6.14 Parametric Curves

These answers correspond to the problems in Section 4.2.

  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.

    Figure 6.14.1. \(x=\sin ^2\pi t\text{,}\) \(y=\cos ^2 \pi t\)

  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.

    Figure 6.14.2. \(x=e^{-t}\text{,}\) \(y=2^{2t}\)

  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.

    Figure 6.14.3. \(x=3\cos(2t)\text{,}\) \(y=\sin(4t)\) and \(4x+6y=9\sqrt{3}\)

  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?

    Figure 6.14.4. \(x=\cos^3t\text{,}\) \(y=\sin^3t\)

  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 \(t=\ds \frac{\pi}{6}\text{.}\) \(4x-4\sqrt{3}y+3=0\text{.}\)

    Figure 6.14.5. \(x=\cos^2t\text{,}\) \(y=\cos t\)

  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.

    Figure 6.14.6. \(x=t(t^2-3)\text{,}\) \(y=3(t^2-3)\)

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