Chapter 0.4, Problem 43E

(a.)

To determine

The graphs of the given parametric equations in the parametric interval $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ , for $a=1,2,3$ and $b=1,2,3$ ; and the role of $a$ and $b$ in these parametric equations.

Answer to Problem 43E

The graphs of the given parametric equations in the parametric interval $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ , for $a=1,2,3$ and $b=1,2,3$ have been provided; and it has been determined that $a$ determines the $x$ -coordinate of the vertex and $b$ determines the curvature of the hyperbola formed by the given parametric equations.

Explanation of Solution

Given:

The parametric equations; $x=a\sec t$ , $y=b\tan t$ .

Concept used:

The parametric equations are graphed for each of the given values of $a$ , $b$ .

Calculation:

The given parametric equations are $x=a\sec t$ and $y=b\tan t$ .

The graph of these parametric equations in the parametric interval $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$

are as follows:

  

It can be seen that $a$ determines the $x$ -coordinate of the vertex and $b$ determines the curvature of the hyperbola formed by the given parametric equations.

Conclusion:

The graphs of the given parametric equations in the parametric interval $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ , for $a=1,2,3$ and $b=1,2,3$ have been provided; and it has been determined that $a$ determines the $x$ -coordinate of the vertex and $b$ determines the curvature of the hyperbola formed by the given parametric equations.

(b.)

To determine

The graph of the given parametric equations in the parametric interval $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$ for $a=2$ , $b=3$ .

Answer to Problem 43E

The graph of the given parametric equations in the parametric interval $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$ for $a=2$ , $b=3$ is as follows:

  

Explanation of Solution

Given:

The parametric equations; $x=a\sec t$ , $y=b\tan t$ .

Concept used:

The parametric equations are graphed for the given values of $a$ , $b$ .

Calculation:

The given parametric equations are $x=a\sec t$ and $y=b\tan t$ .

Put $a=2$ , $b=3$ in these equations to get,

  $x=2\sec t$ and $y=3\tan t$

The graph of these parametric equations in the parametric interval $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$

are as follows:

  

Conclusion:

The graph of the given parametric equations in the parametric interval $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$ for $a=2$ , $b=3$ is as follows:

  

(c.)

To determine

The graph of the given parametric equations in the parametric interval $\left(-\frac{\pi }{2},\frac{3\pi }{2}\right)$ for $a=2$ , $b=3$ and an explanation for the appearance of the lines based on the parameter $t$ .

Answer to Problem 43E

The graph of the given parametric equations in the parametric interval $\left(-\frac{\pi }{2},\frac{3\pi }{2}\right)$ for $a=2$ , $b=3$ is as follows:

  

An explanation for the appearance of the lines based on the parameter $t$ has also been provided.

Explanation of Solution

Given:

The parametric equations; $x=a\sec t$ , $y=b\tan t$ .

Concept used:

The parametric equations are graphed for the given values of $a$ , $b$ .

Calculation:

The given parametric equations are $x=a\sec t$ and $y=b\tan t$ .

Put $a=2$ , $b=3$ in these equations to get,

  $x=2\sec t$ and $y=3\tan t$

The graph of these parametric equations in the parametric interval $\left(-\frac{\pi }{2},\frac{3\pi }{2}\right)$

are as follows:

  

Note that both $\sec t$ and $\tan t$ are undefined for $t=-\frac{\pi }{2},\frac{\pi }{2},\frac{3\pi }{2}$ .

So, both $x=2\sec t$ and $y=3\tan t$ are undefined for $t=-\frac{\pi }{2},\frac{\pi }{2},\frac{3\pi }{2}$ .

Hence, the graph of these parametric equations is discontinuous at these points.

Since the graph is in connected mode, the dotted lines show up on the graph at these discontinuities.

Conclusion:

The graph of the given parametric equations in the parametric interval $\left(-\frac{\pi }{2},\frac{3\pi }{2}\right)$ for $a=2$ , $b=3$ is as follows:

  

An explanation for the appearance of the lines based on the parameter $t$ has also been provided.

(d.)

To determine

To Explain: ${\left(\frac{x}{a}\right)}^2-{\left(\frac{y}{b}\right)}^2=1$ , using algebra.

Answer to Problem 43E

It has been explained using algebra why ${\left(\frac{x}{a}\right)}^2-{\left(\frac{y}{b}\right)}^2=1$ .

Explanation of Solution

Given:

The parametric equations; $x=a\sec t$ , $y=b\tan t$ .

Concept used:

According to trigonometric identity, ${\sec }^{2}t=1+{\tan }^{2}t$ for all $t$ .

Calculation:

The given parametric equations are; $x=a\sec t$ , $y=b\tan t$ .

Put these values in ${\left(\frac{x}{a}\right)}^2-{\left(\frac{y}{b}\right)}^2$ to get,

  ${\left(\frac{x}{a}\right)}^2-{\left(\frac{y}{b}\right)}^2={\left(\frac{a\sec t}{a}\right)}^2-{\left(\frac{b\tan t}{b}\right)}^2$

Simplifying,

  ${\left(\frac{x}{a}\right)}^2-{\left(\frac{y}{b}\right)}^2={\left(\sec t\right)}^2-{\left(\tan t\right)}^2$

On further simplification,

  ${\left(\frac{x}{a}\right)}^2-{\left(\frac{y}{b}\right)}^2={\sec }^{2}t-{\tan }^{2}t$

Put ${\sec }^{2}t=1+{\tan }^{2}t$ in the above equation to get,

  ${\left(\frac{x}{a}\right)}^2-{\left(\frac{y}{b}\right)}^2=1+{\tan }^{2}t-{\tan }^{2}t$

Simplifying,

  ${\left(\frac{x}{a}\right)}^2-{\left(\frac{y}{b}\right)}^2=1$

This is the required expression.

Conclusion:

It has been explained using algebra why ${\left(\frac{x}{a}\right)}^2-{\left(\frac{y}{b}\right)}^2=1$ .

(e.)

To determine

To Repeat: The solutions of part (a), (b) and (d) using an appropriate version of (d) for the given parametric equations.

Answer to Problem 43E

The solutions for part (a), (b) and (d) using an appropriate version of (d) have been repeated for the given parametric equations.

Explanation of Solution

Given:

The parametric equations; $x=a\tan t$ , $y=b\sec t$ .

Concept used:

The parametric equations are graphed for each of the given values of $a$ , $b$ .

Calculation:

The given parametric equations are $x=a\tan t$ and $y=b\sec t$ .

The graph of these parametric equations in the parametric interval $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$

are as follows:

  

It can be seen that $b$ determines the $y$ -coordinate of the vertex and $a$ determines the curvature of the hyperbola formed by the given parametric equations.

The graph of the given parametric equations in the parametric interval $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$ for $a=2$ , $b=3$ is as follows:

  

Now, for the given parametric equations, it follows that ${\left(\frac{y}{b}\right)}^2-{\left(\frac{x}{a}\right)}^2=1$ .

This can be shown as follows:

Put the given equations; $x=a\tan t$ and $y=b\sec t$ in ${\left(\frac{y}{b}\right)}^2-{\left(\frac{x}{a}\right)}^2$ to get,

  ${\left(\frac{y}{b}\right)}^2-{\left(\frac{x}{a}\right)}^2={\left(\frac{b\sec t}{b}\right)}^2-{\left(\frac{a\tan t}{a}\right)}^2$

Simplifying,

  ${\left(\frac{y}{b}\right)}^2-{\left(\frac{x}{a}\right)}^2={\left(\sec t\right)}^2-{\left(\tan t\right)}^2$

On further simplification,

  ${\left(\frac{y}{b}\right)}^2-{\left(\frac{x}{a}\right)}^2={\sec }^{2}t-{\tan }^{2}t$

Put ${\sec }^{2}t=1+{\tan }^{2}t$ , which is a trigonometric identity, in the above equation to get,

  ${\left(\frac{y}{b}\right)}^2-{\left(\frac{x}{a}\right)}^2=1+{\tan }^{2}t-{\tan }^{2}t$

Simplifying,

  ${\left(\frac{y}{b}\right)}^2-{\left(\frac{x}{a}\right)}^2=1$

This is the required proof.

Conclusion:

The solutions for part (a), (b) and (d) using an appropriate version of (d) have been repeated for the given parametric equations.