Chapter 11, Problem 40RE

(a)

To determine

To find: The velocity and acceleration vectors.

Answer to Problem 40RE

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 position vector is $r\left(t\right)=\langle \sqrt{3}\sec t,\sqrt{3}\tan t\rangle$ at $t=0$ .

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.

  $\begin{array}{c}v\left(t\right)=\frac{d}{dt}\left(r\left(t\right)\right)\\ =\frac{d}{dt}\langle \sqrt{3}\sec t,\sqrt{3}\tan t\rangle \\ =\langle \frac{d}{dt}\left(\sqrt{3}\sec t\right),\frac{d}{dt}\left(\sqrt{3}\tan t\right)\rangle \\ =\langle \sqrt{3}\sec t\tan t,\sqrt{3}{\sec }^{2}t\rangle \end{array}$

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

Find the acceleration vector of the particle as follows.

  $\begin{array}{c}a\left(t\right)=\frac{d}{dt}\left(v\left(t\right)\right)\\ =\frac{d}{dt}\langle \sqrt{3}\sec t\tan t,\sqrt{3}{\sec }^{2}t\rangle \\ =\langle \frac{d}{dt}\left(\sqrt{3}\sec t\right),\frac{d}{dt}\left(\sqrt{3}{\sec }^{2}t\right)\rangle \\ =\langle \sqrt{3}\left(\sec t{\tan }^{2}t+{\sec }^{3}t\right),2\sqrt{3}{\sec }^{2}t\tan t\rangle \end{array}$

Therefore, 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.

(b)

To determine

To find: The speed at the given value of $t$ .

Answer to Problem 40RE

The speed at $t=0$ is $\sqrt{3}$ units.

Explanation of Solution

Given:

The position vector is $r\left(t\right)=\langle \sqrt{3}\sec t,\sqrt{3}\tan t\rangle$ at $t=0$ .

Calculation:

Find the velocity vector of the particle at $t=0$ as follows.

  $\begin{array}{c}v\left(t\right)=\langle \sqrt{3}\sec \left(0\right)\tan \left(0\right),\sqrt{3}{\sec }^2\left(0\right)\rangle \\ =\langle 0,\sqrt{3}\rangle \end{array}$

Find the magnitude of the velocity vector as follows.

  $\begin{array}{c}{\left|v\left(t\right)\right|}_{t=0}=\sqrt{{\left(0\right)}^2+{\left(\sqrt{3}\right)}^2}\\ =\sqrt{0+3}\\ =\sqrt{3}\end{array}$

Therefore, the speed at $t=0$ is $\sqrt{3}$ units.