Chapter 10.2, Problem 46E

$\text{(a)}$

To determine

To find: The velocity and the speed of the particle at $t=1/4$ .

Answer to Problem 46E

The velocity:

  

And the speed:

  

Explanation of Solution

Given information:

The position vector $(\sec \pi t,\tan \pi t)$ for $0\le t<1/2$ .

Calculation:

Know that,

  

Then,

  

In this exercise,

  

Then,

  $\begin{array}{c}\frac{dx}{dt}=-\frac{1}{{\cos }^2(\pi t)}\cdot (-\sin (\pi t))\cdot \pi \\ =\pi \cdot \frac{1}{\cos (\pi t)}\cdot \frac{\sin (\pi t)}{\cos (\pi t)}\\ =\pi \sec (\pi t)\tan (\pi t)\\ \frac{dy}{dt}=\frac{\cos (\pi t)\cdot \pi \cdot \cos (\pi t)-\sin (\pi t)\cdot (-\sin (\pi t))\cdot \pi }{{\cos }^2(\pi t)}\\ =\frac{\pi {\cos }^2(\pi t)+\pi {\sin }^2(\pi t)}{{\cos }^2(\pi t)}\\ =\frac{\pi \cdot \left({\sin }^2(\pi t)+{\cos }^2(\pi t)\right)}{{\cos }^2(\pi t)}\\ =\pi \cdot \frac{1}{{\cos }^2(\pi t)}\\ =\pi {\sec }^2(\pi t)\end{array}$

To use the previous result,

For $t=1/4$

Then,

  $\begin{array}{c}\frac{dx}{dt}(1/4)=\pi \sec (\pi \cdot (1/4))\tan (\pi \cdot (1/4))\\ =\pi \cdot \sqrt{2}\cdot 1\\ =\sqrt{2}\pi \\ \frac{dy}{dt}(1/4)=\pi {\sec }^2(\pi \cdot (1/4))\\ =\pi {\sec }^2(\pi /4)\\ =\pi \cdot {(}^{\sqrt{2}}\\ =2\pi \end{array}$

The previous results show that

  

Know that the magnitude of a vector.

  

To use the previous result,

  

Therefore, the required velocity of the particle is

And the speed is .

$\text{(b)}$

To determine

To find: The estimate the parameter to find an equation of hyperbola.

Answer to Problem 46E

The answer: $x^2-y^2=1$

Explanation of Solution

Given information:

The position vector $(\sec \pi t,\tan \pi t)$ for $0\le t<1/2$ .

Calculation:

  $\begin{array}{c}x=\sec \pi t\\ y=\tan \pi t\\ {\tan }^2\theta +1={\sec }^2\theta \\ y^2+1=x^2\\ x^2-y^2=1\text{ (hyperbola)}\end{array}$

Therefore, the required equation of hyperbola is $x^2-y^2=1$ .

$(c)$

To determine

To sketch: The path of the particle.

Answer to Problem 46E

The answer: The path of the particle is the parametric curve defined by

  $\begin{array}{l}x(t)=\sec (\pi t)\\ y(t)=\tan (\pi t)\end{array}$

Where,

  $0\le t<1/2$

Explanation of Solution

Given information:

The position vector $(\sec \pi t,\tan \pi t)$ for $0\le t<1/2$ .

Calculation:

Know that,

  

In this exercise,

  

The following graph is the graph of the path of the particle.

  

Therefore, the required The path of the particle is the parametric curve defined by

  $\begin{array}{l}x(t)=\sec (\pi t)\\ y(t)=\tan (\pi t)\end{array}$

Where,

  $0\le t<1/2$