Parametric Curves

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Exercises 4.2 Parametric 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 \(y=x\text{.}\)

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.

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*}

Find \(\ds \frac{dx}{dy}\) as a function of \(\theta\text{.}\)
Find \(\ds \frac{d^2x}{dy^2}\) as a function of \(\theta\text{.}\)
Find the tangent line to the curve at the point of the curve obtained by setting \(\ds \theta =\frac{\pi }{3}\text{.}\)

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

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 \(x(t)=-2+2\cos t\text{,}\) \(y(t)=1-2\sin t\text{.}\)

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.
Find the points on the curve for which the tangent line has a slope of 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.
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 \(0\leq t\leq 2\pi\text{.}\)

Find \(dy/dx\) for this curve.
Find the equation of the two tangent lines at the origin.
Identify the graph on Figure 4.3 that corresponds to the parametric curve.

Figure 4.3. Which graph is right?

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

10.

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

Find the derivative \(\ds \frac{dy}{dx}\) as a function of \(t\text{.}\)
Eliminate the parametric dependance to determine an expression of the form \(y=f(x)\) for the given curve.
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 \((x_1,f(x_1))\text{.}\)
Find an expression for all tangent lines to the curve \(y=f(x)\) that pass through the point \((2,0)\text{.}\)

\(\ds \frac{dy}{dx}=2t-2\text{.}\)
\(y=x^2+1\text{.}\)
\(m=2x_1\text{,}\) \(b=1-x_1^2\text{.}\)
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 \(x=t^3-4t\text{,}\) \(y=2t^2-4t\text{,}\) \(-\infty \lt t\lt \infty\text{.}\)

Which of the two graphs below corresponds to the given parametric curve?
Find the \(y\)-coordinates of all points where the curve crosses the \(y\)-axis.
This curve crosses itself at exactly one point. Find equations of both tangent lines at that point.

Right. Note that if \(t=0\) then \(x=y=0\text{.}\)
Solve \(x=t^3-4t=t(t^2-4)=0\text{.}\) Next, \(y(-2)=16\text{,}\) \(y(0)=y(2)=0\text{.}\)
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*}

Find an ordinary equation on \(x\) and \(y\) for this curve by eliminating the parameter \(t\text{.}\)
Find the slope of the tangent line to the curve at the point where \(t=1\text{.}\)
Find \(\ds \frac{d^2y}{dx^2}\) at \(t=1\text{.}\)

\(x^3+y^3=3xy\text{.}\)
\(-1\text{.}\)
\(-\frac{32}{3}\text{.}\)

\(x^3+y^3=3xy\text{.}\)
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 \(x\text{.}\)
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*}

Calculate \(\ds \frac{dy}{dx}\) (as a function of \(t\)) directly from these equations.
Calculate \(\ds \frac{d^2y}{dx^2}\) (as a function of \(t\)) directly from these equations.
Does the graph of \(C\) have any points of inflection?
Eliminate \(t\) from the equations of \(C\) thus obtaining a description of the graph \(C\) of the form \(y=f(x)\text{.}\)
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 \(+\infty\text{.}\)

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

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*}

Find \(\ds \frac{dy}{dx}\) in terms of \(t\text{.}\)
Find the tangent line to the curve at the point where \(t=0\text{.}\)
Find the second derivative \(\ds \frac{d^2y}{dx^2}\) in terms of \(t\text{.}\)
For which values of \(t\) is the curve concave upward?

\(\ds \frac{dy}{dx}=1+t\text{.}\)
\(y=x-1\text{.}\)
\(\ds \frac{d^2y}{dx^2}=e^{-t}\text{.}\)
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*}

Find \(\ds \frac{dy}{dx}\) as a function of \(t\) directly from the above equations.
Find \(\ds \frac{d^2y}{dx^2}\) as a function of \(t\text{.}\) Simplify your answer.
Determine if \(C\) is concave up or concave down at \(t=0\text{.}\)

\(\ds \frac{dy}{dx}=\frac{1}{4}e^{1-5t}\text{.}\)
\(\ds \frac{d^2y}{dx^2}=\frac{5}{16}e^{1-9t}\text{.}\)
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*}

Find \(\ds \frac{dy}{dx}\) as a function of \(t\) directly from the above equations.
Eliminate \(t\) from the equations of \(C\) thus obtaining a description of the graph of \(C\) having the form \(y=f(x)\text{.}\)

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

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*}

Find the derivative \(\ds \frac{dy}{dx}\) as a function of \(t\text{.}\)
Find the equation of the tangent line at \(\ds t=\frac{\pi }{2}\text{.}\)
Determine if the curve is concave upwards or downwards at \(\ds t=\frac{\pi }{2}\text{.}\)

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

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.

Find the velocity and speed.
Find the acceleration.

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

19.

A curve is defined by the parametric equations

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

Calculate \(\ds \frac{dy}{dx}\) in terms of \(t\text{.}\)
Find the equation of the tangent line to the curve at the point corresponding to \(t=2\text{.}\)

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

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*}

Find the \(x\)-coordinate of all points where the curve crosses the \(x\)-axis.
Find the coordinate of the point on the curve corresponding to \(t=-1\text{.}\)
Find the tangent line to the curve at the point where \(t=-1\text{.}\)
Find the second derivative at the point where \(t=-1\text{.}\)

\(-27\text{,}\) \(0\text{.}\)
\((-24,8)\text{.}\)
\(y=x+32\text{.}\)
\(\ds \frac{d^2y}{dx^2}=\frac{t^2+3}{12t^3}\text{.}\)

21.

Consider the parametric curve:

\begin{equation*} x=3\cos (2t), \ y=\sin (4t), \ 0\leq t\leq \pi\text{.} \end{equation*}

Find the coordinates of the point on the curve corresponding to \(t=0\text{,}\) and draw this point on the curve.
Find the coordinates of the point on the curve corresponding to \(\ds t=\frac{\pi}{4}\text{,}\) and draw this point on the curve.
Draw arrows indicating the direction the curve is sketched as the \(t\) values increase from \(0\) to \(\pi\text{.}\)
Find the slope of the tangent line to the curve at the point \(\ds t=\frac{\pi}{12}\text{.}\)
State the intervals of \(t\) for which the curve is concave up, and the intervals of \(t\) for which the curve is concave down.

\((3,0)\text{.}\)
\((0,0)\text{.}\)
\(4x+6y=9\sqrt{3}\text{.}\)
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.

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

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*}

Calculate \(\ds \frac{dy}{dx}\) (as a function of \(t\)) directly from the above equations.
Calculate \(\ds \frac{d^2y}{dx^2}\) (as a function of \(t\)).
Find all inflection points of \(C\) or otherwise show that \(C\) has no inflection points.
Eliminate \(t\) from the equations \(C\) to obtain a function of the form \(y=f(x)\) that describes C.

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

24.

Given the parametric curve

\begin{equation*} x=\cos ^3t, \ y=\sin ^3 t\text{.} \end{equation*}

Without eliminating the parameter \(t\text{,}\) show that \(\ds \frac{dy}{dx}=-\tan t\text{.}\)
Determine the concavity of this curve when \(t=1\text{.}\)

\(\ds \frac{dy}{dx}=\frac{3\sin ^2t\cos t}{-3\cos ^2t\sin t}=-\tan t\text{.}\)
concave up.

\(\ds \frac{dy}{dx}=\frac{3\sin ^2t\cos t}{-3\cos ^2t\sin t}=-\tan t\text{.}\)
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*}

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.
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.
On the graph draw arrows indicating the direction the curve is sketched as the \(t\) value increases from \(0\) to \(2\pi\text{.}\)
Find the tangent line to the curve at the point where \(t=3\pi/4\text{.}\)
State the intervals of \(t\) for which the curve is concave up, and the intervals of \(t\) for which the curve is concave down.

\((1,0)\text{.}\)
\(\ds \left( -\frac{\sqrt{2}}{4},\frac{\sqrt{2}}{4}\right)\text{.}\)
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 \(0\leq t\leq \pi\text{.}\)

Sketch the curve.
Write the equation of the tangent line where the slope is \(\ds \frac{1}{\sqrt{3}}\text{.}\)

\(4x-4\sqrt{3}y+3=0\text{.}\)

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

Find \(\ds \frac{dy}{dx}\) and \(\ds \frac{d^2y}{dx^2}\text{.}\)
Find the equation of the line tangent to the curve that is parallel to the line \(y+x=1\text{.}\)

\(\ds \frac{dy}{dx}=-e^{-2t}, \frac{d^2y}{dx^2}=e^{-3t}\text{.}\)
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*}

Find the \(y\)-intercepts of the curve.
Find the points on the curve where the tangent line is horizontal or vertical.
Sketch the curve.

\((0,0)\text{,}\) \((0,-9)\text{.}\)
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 \((2,-6)\text{.}\)

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.

Find \(\displaystyle \frac{dy}{dx}\) as a function of \(t\) directly from the equation above. Simplify your answer.
Find \(\displaystyle \frac{d^2y}{dx^2}\) as a function of \(t\text{.}\) Simplify your answer.
Determine if \(C\) is concave up or concave down at \(t=0\text{.}\)

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