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 t i + j + t k

Question

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Answer 1

Textbook Question

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.

$r ( t ) = tan t i + j + t k$

(a)

Expert Solution

To determine

To calculate: The domain of the function $r(t)=tanti+j+tk$ .

Answer to Problem 1RE

Solution:

The domain is $t≠π2+nπ$ , n is integer.

Explanation of Solution

Given:

The vector-valued function is $r(t)=tanti+j+tk$ .

Calculation:

Consider the function:

$r(t)=tanti+j+tk$

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

$tant≠0t≠π2+nπ$

Where n is integer.

Thus, the required domain is $t≠π2+nπ$ , n is integer.

(b)

Expert Solution

To determine

To calculate: The interval on which the function $r(t)=tanti+j+tk$ is continuous.

Answer to Problem 1RE

Solution:

The function is continuous for all $t≠π2+nπ$ , n is integer.

Explanation of Solution

Given:

The function $r(t)=tanti+j+tk$ .

Calculation:

Consider the function:

$r(t)=tanti+j+tk$

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:

$f(t)=tant,g(t)=1,h(t)=t$

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

However, the function f is continuous for

$tant≠0t≠π2+nπ$

Where n is an integer.

Therefore, the function is continuous for $t≠π2+nπ$ , where n is an integer.

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Answer 2

Textbook Question

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.

$r ( t ) = tan t i + j + t k$

(a)

Expert Solution

To determine

To calculate: The domain of the function $r(t)=tanti+j+tk$ .

Answer to Problem 1RE

Solution:

The domain is $t≠π2+nπ$ , n is integer.

Explanation of Solution

Given:

The vector-valued function is $r(t)=tanti+j+tk$ .

Calculation:

Consider the function:

$r(t)=tanti+j+tk$

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

$tant≠0t≠π2+nπ$

Where n is integer.

Thus, the required domain is $t≠π2+nπ$ , n is integer.

(b)

Expert Solution

To determine

To calculate: The interval on which the function $r(t)=tanti+j+tk$ is continuous.

Answer to Problem 1RE

Solution:

The function is continuous for all $t≠π2+nπ$ , n is integer.

Explanation of Solution

Given:

The function $r(t)=tanti+j+tk$ .

Calculation:

Consider the function:

$r(t)=tanti+j+tk$

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:

$f(t)=tant,g(t)=1,h(t)=t$

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

However, the function f is continuous for

$tant≠0t≠π2+nπ$

Where n is an integer.

Therefore, the function is continuous for $t≠π2+nπ$ , where n is an integer.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

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.

$r ( t ) = tan t i + j + t k$

(a)

Expert Solution

To determine

To calculate: The domain of the function $r(t)=tanti+j+tk$ .

Answer to Problem 1RE

Solution:

The domain is $t≠π2+nπ$ , n is integer.

Explanation of Solution

Given:

The vector-valued function is $r(t)=tanti+j+tk$ .

Calculation:

Consider the function:

$r(t)=tanti+j+tk$

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

$tant≠0t≠π2+nπ$

Where n is integer.

Thus, the required domain is $t≠π2+nπ$ , n is integer.

(b)

Expert Solution

To determine

To calculate: The interval on which the function $r(t)=tanti+j+tk$ is continuous.

Answer to Problem 1RE

Solution:

The function is continuous for all $t≠π2+nπ$ , n is integer.

Explanation of Solution

Given:

The function $r(t)=tanti+j+tk$ .

Calculation:

Consider the function:

$r(t)=tanti+j+tk$

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:

$f(t)=tant,g(t)=1,h(t)=t$

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

However, the function f is continuous for

$tant≠0t≠π2+nπ$

Where n is an integer.

Therefore, the function is continuous for $t≠π2+nπ$ , where n is an integer.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

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.

$r ( t ) = tan t i + j + t k$

(a)

Expert Solution

To determine

To calculate: The domain of the function $r(t)=tanti+j+tk$ .

Answer to Problem 1RE

Solution:

The domain is $t≠π2+nπ$ , n is integer.

Explanation of Solution

Given:

The vector-valued function is $r(t)=tanti+j+tk$ .

Calculation:

Consider the function:

$r(t)=tanti+j+tk$

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

$tant≠0t≠π2+nπ$

Where n is integer.

Thus, the required domain is $t≠π2+nπ$ , n is integer.

(b)

Expert Solution

To determine

To calculate: The interval on which the function $r(t)=tanti+j+tk$ is continuous.

Answer to Problem 1RE

Solution:

The function is continuous for all $t≠π2+nπ$ , n is integer.

Explanation of Solution

Given:

The function $r(t)=tanti+j+tk$ .

Calculation:

Consider the function:

$r(t)=tanti+j+tk$

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:

$f(t)=tant,g(t)=1,h(t)=t$

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

However, the function f is continuous for

$tant≠0t≠π2+nπ$

Where n is an integer.

Therefore, the function is continuous for $t≠π2+nπ$ , where n is an integer.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

To calculate: The domain of the function $r(t)=tanti+j+tk$ .

Answer to Problem 1RE

Solution:

The domain is $t≠π2+nπ$ , n is integer.

Explanation of Solution

Given:

The vector-valued function is $r(t)=tanti+j+tk$ .

Calculation:

Consider the function:

$r(t)=tanti+j+tk$

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

$tant≠0t≠π2+nπ$

Where n is integer.

Thus, the required domain is $t≠π2+nπ$ , n is integer.

(b)

Expert Solution

To determine

To calculate: The interval on which the function $r(t)=tanti+j+tk$ is continuous.

Answer to Problem 1RE

Solution:

The function is continuous for all $t≠π2+nπ$ , n is integer.

Explanation of Solution

Given:

The function $r(t)=tanti+j+tk$ .

Calculation:

Consider the function:

$r(t)=tanti+j+tk$

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:

$f(t)=tant,g(t)=1,h(t)=t$

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

However, the function f is continuous for

$tant≠0t≠π2+nπ$

Where n is an integer.

Therefore, the function is continuous for $t≠π2+nπ$ , where n is an integer.

Answer 8

(a)

Expert Solution

To determine

To calculate: The domain of the function $r(t)=tanti+j+tk$ .

Answer to Problem 1RE

Solution:

The domain is $t≠π2+nπ$ , n is integer.

Explanation of Solution

Given:

The vector-valued function is $r(t)=tanti+j+tk$ .

Calculation:

Consider the function:

$r(t)=tanti+j+tk$

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

$tant≠0t≠π2+nπ$

Where n is integer.

Thus, the required domain is $t≠π2+nπ$ , n is integer.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

To calculate: The domain of the function $r(t)=tanti+j+tk$ .

Answer 13

Answer to Problem 1RE

Solution:

The domain is $t≠π2+nπ$ , n is integer.

Answer 14

Explanation of Solution

Given:

The vector-valued function is $r(t)=tanti+j+tk$ .

Calculation:

Consider the function:

$r(t)=tanti+j+tk$

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

$tant≠0t≠π2+nπ$

Where n is integer.

Thus, the required domain is $t≠π2+nπ$ , n is integer.

Answer 15

(b)

Expert Solution

To determine

To calculate: The interval on which the function $r(t)=tanti+j+tk$ is continuous.

Answer to Problem 1RE

Solution:

The function is continuous for all $t≠π2+nπ$ , n is integer.

Explanation of Solution

Given:

The function $r(t)=tanti+j+tk$ .

Calculation:

Consider the function:

$r(t)=tanti+j+tk$

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:

$f(t)=tant,g(t)=1,h(t)=t$

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

However, the function f is continuous for

$tant≠0t≠π2+nπ$

Where n is an integer.

Therefore, the function is continuous for $t≠π2+nπ$ , where n is an integer.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

To calculate: The interval on which the function $r(t)=tanti+j+tk$ is continuous.

Answer 20

Answer to Problem 1RE

Solution:

The function is continuous for all $t≠π2+nπ$ , n is integer.

Answer 21

Explanation of Solution

Given:

The function $r(t)=tanti+j+tk$ .

Calculation:

Consider the function:

$r(t)=tanti+j+tk$

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:

$f(t)=tant,g(t)=1,h(t)=t$

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

However, the function f is continuous for

$tant≠0t≠π2+nπ$

Where n is an integer.

Therefore, the function is continuous for $t≠π2+nπ$ , where n is an integer.

Answer 22

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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 t i + j + t k

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