Chapter 11, Problem 47RE

(a)

To determine

To find: The length of the velocity vector at $t=3$ .

Answer to Problem 47RE

The velocity vector, and acceleration vector are $\langle \sqrt{3}\sec t,\sqrt{3}{\sec }^{2}t\rangle$ ,and $\langle \sqrt{3}\left(\sec t{\tan }^{2}t+{\sec }^{3}t\right),2\sqrt{3}{\sec }^{2}t\tan t\rangle$ , respectively.

Explanation of Solution

Given:

The coordinates at time $t$ are $x=3\cos \frac{\pi }{4}t$ , and $y=5\sin \frac{\pi }{4}t$ .

Calculation:

Find the derivative of the position vector with respect to time to find the velocity vector of the particle.

Find the velocity vector of the particle as follows.

  $\begin{array}{c}v\left(t\right)=\frac{d}{dt}\left(r\left(t\right)\right)\\ =\frac{d}{dt}\langle 3\cos \frac{\pi }{4}t,5\sin \frac{\pi }{4}t\rangle \\ =\langle \frac{d}{dt}\left(3\cos \frac{\pi }{4}t\right),\frac{d}{dt}\left(5\sin \frac{\pi }{4}t\right)\rangle \\ =\langle \frac{-3\pi \sin \left(\frac{\pi }{4}t\right)}{4},\frac{5\pi \cos \left(\frac{\pi }{4}t\right)}{4}\rangle \end{array}$

Find the velocity vector at $t=3$ .

  $\begin{array}{c}v{\left(t\right)}_{t=3}=\langle \frac{-3\pi \sin \left(\frac{\pi }{4}\left(3\right)\right)}{4},\frac{5\pi \cos \left(\frac{\pi }{4}\left(3\right)\right)}{4}\rangle \\ =\langle \frac{-3\pi \sin \left(\frac{3\pi }{4}\right)}{4},\frac{5\pi \cos \left(\frac{3\pi }{4}\right)}{4}\rangle \\ =\langle \frac{-3\pi }{4\sqrt{2}},\frac{-5\pi }{4\sqrt{2}}\rangle \end{array}$

Find the magnitude of the velocity vector as follows.

  $\begin{array}{c}\left|v{\left(t\right)}_{t=3}\right|=\sqrt{{\left(\frac{-3\pi }{4\sqrt{2}}\right)}^2+{\left(\frac{-5\pi }{4\sqrt{2}}\right)}^2}\\ =\sqrt{\frac{9{\pi }^2}{32}+\frac{25{\pi }^2}{32}}\\ =\sqrt{\frac{34{\pi }^2}{32}}\\ =\frac{\pi \sqrt{17}}{4}\end{array}$

Therefore, the length of the velocity vector is $\frac{\pi \sqrt{17}}{4}$ .

(b)

To determine

To find: The $x$ and $y$ components of the acceleration of the particle at $t=3$ .

Answer to Problem 47RE

The $x$ and $y$ components of the acceleration of the particle at $t=3$ is $\langle \frac{3{\pi }^2}{16\sqrt{2}},-\frac{5{\pi }^2}{16\sqrt{2}}\rangle$ .

Explanation of Solution

Given:

The velocity vector is $v\left(t\right)=\langle \frac{-3\pi \sin \left(\frac{\pi }{4}t\right)}{4},\frac{5\pi \cos \left(\frac{\pi }{4}t\right)}{4}\rangle$

Calculation:

Find the acceleration vector of the particle as follows.

  $\begin{array}{c}a\left(t\right)=\frac{d}{dt}\left(v\left(t\right)\right)\\ =\frac{d}{dt}\langle \frac{-3\pi \sin \left(\frac{\pi }{4}t\right)}{4},\frac{5\pi \cos \left(\frac{\pi }{4}t\right)}{4}\rangle \\ =\langle \frac{-3{\pi }^2\cos \left(\frac{\pi }{4}t\right)}{16},\frac{-5{\pi }^2\sin \left(\frac{\pi }{4}t\right)}{16}\rangle \end{array}$

Find the acceleration vector at $t=3$ .

  $\begin{array}{c}a{\left(t\right)}_{t=3}=\langle \frac{-3{\pi }^2\cos \left(\frac{\pi }{4}\left(3\right)\right)}{16},\frac{-5{\pi }^2\sin \left(\frac{\pi }{4}\left(3\right)\right)}{16}\rangle \\ =\langle \frac{3{\pi }^2}{16\sqrt{2}},-\frac{5{\pi }^2}{16\sqrt{2}}\rangle \end{array}$

Therefore, the $x$ and $y$ components of the acceleration of the particle at $t=3$ is $\langle \frac{3{\pi }^2}{16\sqrt{2}},-\frac{5{\pi }^2}{16\sqrt{2}}\rangle$ .

(c)

To determine

To find: The single equation in $x$ and $y$ for the path of the particle.

Answer to Problem 47RE

The required equation is $\frac{x^2}{9}+\frac{y^2}{25}=1$ .

Explanation of Solution

Given:

The coordinates at time $t$ are $x=3\cos \frac{\pi }{4}t$ , and $y=5\sin \frac{\pi }{4}t$ .

Calculation:

Rewrite the equations $x=3\cos \frac{\pi }{4}t$ , and $y=5\sin \frac{\pi }{4}t$ as follows.

  $\begin{array}{c}\cos \frac{\pi }{4}t=\frac{x}{3}\\ \sin \frac{\pi }{4}t=\frac{y}{5}\end{array}$

Use the identity ${\sin }^{2}t+{\cos }^{2}t=1$ .

  $\begin{array}{c}\frac{x^2}{9}+\frac{y^2}{25}={\cos }^2\frac{\pi }{4}t+{\sin }^2\frac{\pi }{4}t\\ \frac{x^2}{9}+\frac{y^2}{25}=1\end{array}$

Therefore, the required equation is $\frac{x^2}{9}+\frac{y^2}{25}=1$ .