Chapter 11.2, Problem 46E

(a)

To determine

To calculate: The velocity and speed of the point at $t=\frac{1}{4}$ .

Answer to Problem 46E

The required velocity and speed of the point are $\langle \sqrt{2}\pi ,2\pi \rangle$ and $\sqrt{6}\pi$ respectively.

Explanation of Solution

Given information:

The position of the particle: $\langle \sec \pi t,\tan \pi t\rangle$

Calculation:

The given position of the particle is $\overrightarrow{r}\left(t\right)=\langle \sec \pi t,\tan \pi t\rangle$ .

It is known that for a position vector $\overrightarrow{r}\left(t\right)$ of a particle, the first derivative of $\overrightarrow{r}\left(t\right)$ is the velocity vector of the particle.

So, $\overrightarrow{v}\left(t\right)=\frac{d\left(\overrightarrow{r}\left(t\right)\right)}{dt}$

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

At $t=\frac{1}{4}$ ; $\begin{array}{c}\overrightarrow{v}\left(t\right)=\langle \pi \sec \pi \left(\frac{1}{4}\right)\tan \pi \left(\frac{1}{4}\right),\pi {\sec }^2\pi \left(\frac{1}{4}\right)\rangle \\ =\langle \sqrt{2}\pi ,2\pi \rangle \end{array}$

Also, it is known that the magnitude of the velocity vector of a particle is known as the speed of the particle.

The magnitude of a vector $\langle \overrightarrow{a},\overrightarrow{b}\rangle$ is $\left|\langle \overrightarrow{a},\overrightarrow{b}\rangle \right|=\sqrt{{\overrightarrow{a}}^2+{\overrightarrow{b}}^2}$ .

At $t=\frac{1}{4}$ ; substitute $\sqrt{2}\pi$ for $\overrightarrow{a}$ and $2\pi$ for $\overrightarrow{b}$ in the above formula.

  $\begin{array}{c}\sqrt{{\overrightarrow{a}}^2+{\overrightarrow{b}}^2}=\sqrt{{\left(\sqrt{2}\pi \right)}^2+{\left(2\pi \right)}^2}\\ =\sqrt{6}\pi \end{array}$

Hence, the required velocity and speed of the point are $\langle \sqrt{2}\pi ,2\pi \rangle$ and $\sqrt{6}\pi$ respectively.

(b)

To determine

To determine: The equation of the hyperbola in x and y by eliminating the parameter.

Answer to Problem 46E

The required equation is $x^2-y^2=1$ .

Explanation of Solution

Given information:

The position of the point: $\langle \sec \pi t,\tan \pi t\rangle$

Calculation:

The given position of the point is $\langle \sec \pi t,\tan \pi t\rangle$ .

Use the trigonometric identity ${\sec }^{2}t-{\tan }^{2}t=1$ .

Write the expressions $\overrightarrow{x}=\sec \pi t$ and $\overrightarrow{y}=\tan \pi t$ as:

  $\begin{array}{l}{x}^2={\left(\sec \pi t\right)}^2\\ x^2={\sec }^2\pi t\end{array}$

Also,

  $\begin{array}{l}{y}^2={\left(\tan \pi t\right)}^2\\ y^2={\tan }^2\pi t\end{array}$

Use subtraction on the above two equations:

  $\begin{array}{c}{x}^2-y^2={\sec }^2\pi t-{\tan }^2\pi t\\ x^2-y^2=1\end{array}$

Hence, the required equation is $x^2-y^2=1$ .

(c)

To determine

To graph: The path of the particle over the time interval

Explanation of Solution

Given information:

The position of the point: $\langle \sec \pi t,\tan \pi t\rangle$

Calculation:

The given position of the point is $\langle \sec \pi t,\tan \pi t\rangle$ .

The required graph for the path of the particle is shown below: