## Exercises4.2Parametric Curves

Solve the following problems.

###### 1.

An object is moving counter-clockwise along a circle with the centre at the origin. At $t=0$ the object is at point $A(0,5)$ and at $t=2\pi$ it is back to point $A$ for the first time. Determine the parametric equations $x=f(t)\text{,}$ $y=g(t)$ that describe the motion for the duration $0\leq t\leq 2\pi\text{.}$

$x=-5\sin t, \ y=5\cos t, \ t\in[0,2\pi]\text{.}$
###### 2.

The trajectory of a particle in a plane as a function of the time $t$ in seconds is given by the parametric equations

\begin{equation*} x=3t^3+2t-3, \ \ y=2t^3+2\text{.} \end{equation*}

Prove that there is exactly one time when the particle crosses the line $y=x\text{.}$

Hint

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.

Let $x=2\sin t+1$ and $y=2t^3-3$ define a parametric curve. Find $\ds \frac{d^2y}{dx^2}$ as a function of $t\text{,}$ without simplifying your answer.

$\ds \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left( \frac{dy}{dx}\right) }{\frac{dx}{dt}}=\frac{3t(2\cos t+t\sin t)}{2\cos ^3t}\text{.}$
###### 4.

Sketch the curve

\begin{equation*} x=\sin ^2\pi t, \ y=\cos ^2 \pi t, \ 0\leq t\leq 2\text{.} \end{equation*}

Clearly label the initial and terminal points and describe the motion of the point $(x(t),y(t))$ as $t$ varies in the given interval.

Hint
Note that $x+y=1$ and that $x,y\in [0,1]\text{.}$ ###### 5.

Find an equation of the tangent line to the curve $x=t-t^{-1}\text{,}$ $y=1+t^2$ at $t=1\text{.}$

$y=x+2\text{.}$
###### 6.

The parametric equations for a curve are given by

\begin{equation*} x=\theta -\sin \theta, \ \ y=1-\cos \theta\text{.} \end{equation*}
1. Find $\ds \frac{dx}{dy}$ as a function of $\theta\text{.}$

2. Find $\ds \frac{d^2x}{dy^2}$ as a function of $\theta\text{.}$

3. Find the tangent line to the curve at the point of the curve obtained by setting $\ds \theta =\frac{\pi }{3}\text{.}$

1. $\ds \frac{dx}{dy}=\frac{1-\cos \theta }{\sin \theta}\text{.}$

2. $\ds \frac{d^2x}{dy^2}=\frac{1-\cos \theta }{\sin ^3 \theta}\text{.}$

3. $\ds y=\sqrt{3}x-\frac{\pi \sqrt{3}}{3}+2\text{.}$

###### 7.

The graphs of the parametric equations $x=f(t)\text{,}$ $y=g(t)$ are shown in Figure 4.1 Identify the corresponding parametric curves in Figure 4.2. Figure 4.1. $x= f(t), y=g(t)$ Figure 4.2. Only one is right
A. Follow the curve as $t$ increases.
###### 8.

Consider the parametric curve $x(t)=-2+2\cos t\text{,}$ $y(t)=1-2\sin t\text{.}$

1. State the Cartesian equation of the curve and the sketch the curve. Determine the direction of evolution of the curve for increasing $t$ and indicate it on the graph.

2. Find the points on the curve for which the tangent line has a slope of 1.

1. 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.

2. Solve $\ds \frac{dy}{dx}=\cot t=1$ for $t\in (0,2\pi )\text{.}$

###### 9.

This question concerns the curve $x=2\cos 3t\text{,}$ $y=2\sin 2t$ for $0\leq t\leq 2\pi\text{.}$

1. Find $dy/dx$ for this curve.

2. Find the equation of the two tangent lines at the origin.

3. Identify the graph on Figure 4.3 that corresponds to the parametric curve.

1. $\ds \frac{dy}{dx}=-\frac{2\cos 2t}{3\sin 3t}\text{.}$

2. Note that for both $\ds t=\frac{\pi }{2}$ and $\ds t=\frac{3\pi }{2}$ the curve passes through the origin. Thus, $\ds y=\pm \frac{2x}{3}\text{.}$

3. C.

###### 10.

For the following parametric curve $x=t-1\text{,}$ $y=t^2-2t+2\text{:}$

1. Find the derivative $\ds \frac{dy}{dx}$ as a function of $t\text{.}$

2. Eliminate the parametric dependance to determine an expression of the form $y=f(x)$ for the given curve.

3. Find an expression for $m$ and $b$ (as functions of $x_1$) such that the equation of the line $y=mx+b$ is tangent to the curve $y=f(x)$ at $(x_1,f(x_1))\text{.}$

4. Find an expression for all tangent lines to the curve $y=f(x)$ that pass through the point $(2,0)\text{.}$

1. $\ds \frac{dy}{dx}=2t-2\text{.}$

2. $y=x^2+1\text{.}$

3. $m=2x_1\text{,}$ $b=1-x_1^2\text{.}$

4. 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.

This question concerns the parametric curve $x=t^3-4t\text{,}$ $y=2t^2-4t\text{,}$ $-\infty \lt t\lt \infty\text{.}$

1. Which of the two graphs below corresponds to the given parametric curve?  2. Find the $y$-coordinates of all points where the curve crosses the $y$-axis.

3. This curve crosses itself at exactly one point. Find equations of both tangent lines at that point.

1. Right. Note that if $t=0$ then $x=y=0\text{.}$

2. Solve $x=t^3-4t=t(t^2-4)=0\text{.}$ Next, $y(-2)=16\text{,}$ $y(0)=y(2)=0\text{.}$

3. 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.

Consider the parametric curve

\begin{equation*} x=\frac{3t}{1+t^3}, \ y=\frac{3t^2}{1+t^3}, \ t\in \mathbb{R}\backslash\{-1\}\text{.} \end{equation*}
1. Find an ordinary equation on $x$ and $y$ for this curve by eliminating the parameter $t\text{.}$

2. Find the slope of the tangent line to the curve at the point where $t=1\text{.}$

3. Find $\ds \frac{d^2y}{dx^2}$ at $t=1\text{.}$

1. $x^3+y^3=3xy\text{.}$

2. $-1\text{.}$

3. $-\frac{32}{3}\text{.}$

Solution
1. $x^3+y^3=3xy\text{.}$

2. 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 $x\text{.}$

3. 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.

Consider the parametric curve $C$ described by

\begin{equation*} x=e^{-t}, \ y=2^{2t},\ \mbox{ where } -\infty \lt t\lt \infty\text{.} \end{equation*}
1. Calculate $\ds \frac{dy}{dx}$ (as a function of $t$) directly from these equations.

2. Calculate $\ds \frac{d^2y}{dx^2}$ (as a function of $t$) directly from these equations.

3. Does the graph of $C$ have any points of inflection?

4. Eliminate $t$ from the equations of $C$ thus obtaining a description of the graph $C$ of the form $y=f(x)\text{.}$

5. Sketch $C\text{.}$ Label the axis. Plot at least three points of the curve $C\text{,}$ labeling the points with the corresponding values of $t\text{.}$ Draw an arrow on your curve indicating the direction of motion as $t$ increases toward $+\infty\text{.}$

1. $\ds \frac{dy}{dx}=-e^t\cdot 2^{2t+1}\ln 2\text{.}$

2. $\ds \frac{d^2y}{dx^2}=e^{2t}\cdot 2^{2t+1}\cdot (1+2\ln 2)\ln 2\text{.}$

3. No, the second derivative is never zero.

4. $\ds y=2^{-2\ln x}\text{.}$

5. ###### 14.

Consider the parametric curve described by

\begin{equation*} x(t)=e^{t}, \ y(t)=te^{t},\ \mbox{ where } t\in \mathbb{R}\text{.} \end{equation*}
1. Find $\ds \frac{dy}{dx}$ in terms of $t\text{.}$

2. Find the tangent line to the curve at the point where $t=0\text{.}$

3. Find the second derivative $\ds \frac{d^2y}{dx^2}$ in terms of $t\text{.}$

4. For which values of $t$ is the curve concave upward?

1. $\ds \frac{dy}{dx}=1+t\text{.}$

2. $y=x-1\text{.}$

3. $\ds \frac{d^2y}{dx^2}=e^{-t}\text{.}$

4. Observe that $\ds \frac{d^2y}{dx^2}>0$ for all $t\in\mathbb{R}\text{.}$

###### 15.

Consider the parametric curve $C$ described by

\begin{equation*} x(t)=-e^{4t}, \ y(t)=e^{1-t},\ \mbox{ where } t\in \mathbb{R}\text{.} \end{equation*}
1. Find $\ds \frac{dy}{dx}$ as a function of $t$ directly from the above equations.

2. Find $\ds \frac{d^2y}{dx^2}$ as a function of $t\text{.}$ Simplify your answer.

3. Determine if $C$ is concave up or concave down at $t=0\text{.}$

1. $\ds \frac{dy}{dx}=\frac{1}{4}e^{1-5t}\text{.}$

2. $\ds \frac{d^2y}{dx^2}=\frac{5}{16}e^{1-9t}\text{.}$

3. Note that $\ds \left. \frac{d^2y}{dx^2}\right| _{t=0}=\frac{5}{16}e>0\text{.}$

###### 16.

Consider the parametric curve $C$ described by

\begin{equation*} x(t)=1+3t, \ y(t)=2-t^3,\ \mbox{ where } t\in \mathbb{R}\text{.} \end{equation*}
1. Find $\ds \frac{dy}{dx}$ as a function of $t$ directly from the above equations.

2. Eliminate $t$ from the equations of $C$ thus obtaining a description of the graph of $C$ having the form $y=f(x)\text{.}$

1. $\ds \frac{dy}{dx}=-t^2\text{.}$

2. $\ds y=2-\frac{(x-1)^2}{3}\text{.}$

###### 17.

For the following parametric curve:

\begin{equation*} \left\{ \begin{array}{l} x=t\sin t\\ y=t\cos t \end{array} \right. \end{equation*}
1. Find the derivative $\ds \frac{dy}{dx}$ as a function of $t\text{.}$

2. Find the equation of the tangent line at $\ds t=\frac{\pi }{2}\text{.}$

3. Determine if the curve is concave upwards or downwards at $\ds t=\frac{\pi }{2}\text{.}$

1. $\ds \frac{dy}{dx}=\frac{\cos t-t\sin t}{\sin t+t\cos t}\text{.}$

2. $\ds y=-\frac{\pi }{2}\left( x-\frac{\pi }{2}\right)\text{.}$

3. $\ds \left. \frac{d^2y}{dx^2}\right| _{t=\frac{\pi }{2}}=-2-\frac{\pi ^2}{4}\lt 0\text{.}$

###### 18.

A small ball is fastened to a long rubber band and twirled around in such a way that the ball moves in a circular path given by the equation $\overrightarrow{r}=b\langle \cos \omega t, \sin \omega t\rangle\text{,}$ where $b$ and $\omega$ are constants.

1. Find the velocity and speed.

2. Find the acceleration.

1. $\overrightarrow{v}=b\omega \langle -\sin \omega t, \cos \omega t\rangle\text{;}$ $\mbox{speed } =|\overrightarrow{v}|=|b\omega|\text{.}$

2. $\overrightarrow{a}=-b\omega ^2\langle\cos \omega t, \sin \omega t\rangle\text{.}$

###### 19.

A curve is defined by the parametric equations

\begin{equation*} x=3(t^2-3), \ y=t^3-3t\text{.} \end{equation*}
1. Calculate $\ds \frac{dy}{dx}$ in terms of $t\text{.}$

2. Find the equation of the tangent line to the curve at the point corresponding to $t=2\text{.}$

1. $\ds \frac{dy}{dx}=\frac{t^2-1}{2t}\text{.}$

2. $\ds y-2=\frac{3}{4}\left( x-3\right)\text{.}$

###### 20.

Consider the parametric curve

\begin{equation*} x=3(t^2-9), \ y=t(t^2-9), \ -\infty \lt t\lt \infty\text{.} \end{equation*}
1. Find the $x$-coordinate of all points where the curve crosses the $x$-axis.

2. Find the coordinate of the point on the curve corresponding to $t=-1\text{.}$

3. Find the tangent line to the curve at the point where $t=-1\text{.}$

4. Find the second derivative at the point where $t=-1\text{.}$

1. $-27\text{,}$ $0\text{.}$

2. $(-24,8)\text{.}$

3. $y=x+32\text{.}$

4. $\ds \frac{d^2y}{dx^2}=\frac{t^2+3}{12t^3}\text{.}$

###### 21.

Consider the parametric curve: \begin{equation*} x=3\cos (2t), \ y=\sin (4t), \ 0\leq t\leq \pi\text{.} \end{equation*}
1. Find the coordinates of the point on the curve corresponding to $t=0\text{,}$ and draw this point on the curve.

2. Find the coordinates of the point on the curve corresponding to $\ds t=\frac{\pi}{4}\text{,}$ and draw this point on the curve.

3. Draw arrows indicating the direction the curve is sketched as the $t$ values increase from $0$ to $\pi\text{.}$

4. Find the slope of the tangent line to the curve at the point $\ds t=\frac{\pi}{12}\text{.}$

5. State the intervals of $t$ for which the curve is concave up, and the intervals of $t$ for which the curve is concave down.

1. $(3,0)\text{.}$

2. $(0,0)\text{.}$

3. 4. $4x+6y=9\sqrt{3}\text{.}$

5. 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.

Consider the perimetrically defined curve $x(t)=t-\sin 2t\text{,}$ $y= 4-3\cos t\text{,}$ $0\leq t\leq 10\text{.}$ Find the values for the parameter $t$ where the tangent line of the curve is vertical.

$\ds \theta \in \{ \frac{\pi }{6}, \frac{5\pi }{6}, \frac{7\pi }{6}, \frac{11\pi }{6}, \frac{13\pi }{6}, \frac{17\pi }{6}\}\text{.}$
Solution

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.

Consider the parametric curve $C$ described by

\begin{equation*} x=e^{-t}, \ y=e^{2t},\mbox{ where } -\infty \lt t\lt \infty\text{.} \end{equation*}
1. Calculate $\ds \frac{dy}{dx}$ (as a function of $t$) directly from the above equations.

2. Calculate $\ds \frac{d^2y}{dx^2}$ (as a function of $t$).

3. Find all inflection points of $C$ or otherwise show that $C$ has no inflection points.

4. Eliminate $t$ from the equations $C$ to obtain a function of the form $y=f(x)$ that describes C.

1. $\ds \frac{dy}{dx}=-2e^{3t}\text{.}$

2. $\ds \frac{d^2y}{dx^2}=6e^{4t}\text{.}$

3. $\ds \frac{d^2y}{dx^2}>0\text{.}$

4. $\ds y=x^{-2}\text{.}$

###### 24.

Given the parametric curve

\begin{equation*} x=\cos ^3t, \ y=\sin ^3 t\text{.} \end{equation*}
1. Without eliminating the parameter $t\text{,}$ show that $\ds \frac{dy}{dx}=-\tan t\text{.}$

2. Determine the concavity of this curve when $t=1\text{.}$

1. $\ds \frac{dy}{dx}=\frac{3\sin ^2t\cos t}{-3\cos ^2t\sin t}=-\tan t\text{.}$

2. concave up.

Solution
1. $\ds \frac{dy}{dx}=\frac{3\sin ^2t\cos t}{-3\cos ^2t\sin t}=-\tan t\text{.}$

2. 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.

Consider the parametric curve: \begin{equation*} x=\cos ^3t, \ y=\sin ^3t, \ 0\leq t \lt 2\pi\text{.} \end{equation*}
1. Find the $x$ and $y$ coordinates of the point on the curve corresponding to $t=0\text{,}$ and draw and clearly label this point on the graph.

2. Find the $x$ and $y$ coordinates of the point on the curve corresponding to $t=3\pi/4\text{,}$ and draw and clearly label this point on the graph.

3. On the graph draw arrows indicating the direction the curve is sketched as the $t$ value increases from $0$ to $2\pi\text{.}$

4. Find the tangent line to the curve at the point where $t=3\pi/4\text{.}$

5. State the intervals of $t$ for which the curve is concave up, and the intervals of $t$ for which the curve is concave down.

1. $(1,0)\text{.}$

2. $\ds \left( -\frac{\sqrt{2}}{4},\frac{\sqrt{2}}{4}\right)\text{.}$

3. 4. Observe that $\ds \frac{dy}{dx}=-\tan t\text{,}$ $t\not\in \{0,\pi/2, \pi,3\pi/2\}\text{.}$ $2x-2y+\sqrt{2}=0\text{.}$

###### 26.

Consider the parametric curve $x=\cos^2 t\text{,}$ $y=\cos t\text{,}$ for $0\leq t\leq \pi\text{.}$

1. Sketch the curve.

2. Write the equation of the tangent line where the slope is $\ds \frac{1}{\sqrt{3}}\text{.}$

1. 2. $4x-4\sqrt{3}y+3=0\text{.}$

Solution
1. Observe that by eliminating the parameter $t\text{,}$ the equation becomes $y^2=x\text{,}$ $(x,y)\in[0,1]\times[-1,1]\text{.}$ 2. 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.

Given the parametric curve

\begin{equation*} x=e^t, \ y=e^{-t}\text{.} \end{equation*}
1. Find $\ds \frac{dy}{dx}$ and $\ds \frac{d^2y}{dx^2}\text{.}$

2. Find the equation of the line tangent to the curve that is parallel to the line $y+x=1\text{.}$

1. $\ds \frac{dy}{dx}=-e^{-2t}, \frac{d^2y}{dx^2}=e^{-3t}\text{.}$

2. From $\ds \frac{dy}{dx}=-e^{-2t}=-1$ we get that $t=0\text{.}$ Thus the tangent line is $y=-x\text{.}$

###### 28.

Given the parametric curve

\begin{equation*} x=t(t^2-3), \ y=3(t^2-3)\text{.} \end{equation*}
1. Find the $y$-intercepts of the curve.

2. Find the points on the curve where the tangent line is horizontal or vertical.

3. Sketch the curve.

1. $(0,0)\text{,}$ $(0,-9)\text{.}$

2. 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 $(2,-6)\text{.}$

3. ###### 29.

A parametric curve $C$ is described by $\displaystyle x(t)=-e^{3t+2}$ and $y(t)=2e^{1-t}\text{.}$ where $t$ is a real number.

1. Find $\displaystyle \frac{dy}{dx}$ as a function of $t$ directly from the equation above. Simplify your answer.

2. Find $\displaystyle \frac{d^2y}{dx^2}$ as a function of $t\text{.}$ Simplify your answer.

3. Determine if $C$ is concave up or concave down at $t=0\text{.}$

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