Chapter 12, Problem 1RE

Domain and Continuity In Exercises 1-4, (a) find the domain of r, and (b) determine the interval(s) on which the function is continuous.

$r(t)=\tan ti+j+tk$

To determine

To calculate: The domain of the function $r\left(t\right)=\tan ti+j+tk$.

Answer to Problem 1RE

Solution:

The domain is $t\ne \frac{\pi }{2}+n\pi$, n is integer.

Explanation of Solution

Given:

The vector-valued function is $r\left(t\right)=\tan ti+j+tk$.

Calculation:

Consider the function:

$r\left(t\right)=\tan ti+j+tk$

The x-coordinate of the function cannot be zero. Thus,

$\begin{array}{c}\tan t\ne 0\\ t\ne \frac{\pi }{2}+n\pi \end{array}$

Where n is integer.

Thus, the required domain is $t\ne \frac{\pi }{2}+n\pi$, n is integer.

To determine

To calculate: The interval on which the function $r\left(t\right)=\tan ti+j+tk$ is continuous.

Answer to Problem 1RE

Solution:

The function is continuous for all $t\ne \frac{\pi }{2}+n\pi$, n is integer.

Explanation of Solution

Given:

The function $r\left(t\right)=\tan ti+j+tk$.

Calculation:

Consider the function:

$r\left(t\right)=\tan ti+j+tk$

Evaluate the continuity of the vector valued function by evaluating the continuity of the component functions and then taking the intersection of the two sets.

The component functions of the vector valued function are:

$f\left(t\right)=\tan t,g\left(t\right)=1,h\left(t\right)=t$

Both the functions g and h are continuous for all real values of t.

However, the function f is continuous for

$\begin{array}{c}\tan t\ne 0\\ t\ne \frac{\pi }{2}+n\pi \end{array}$

Where n is an integer.

Therefore, the function is continuous for $t\ne \frac{\pi }{2}+n\pi$, where n is an integer.