Calculus: Early Transcendentals
Calculus: Early Transcendentals
3rd Edition
ISBN: 9781464114885
Author: Jon Rogawski, Colin Adams
Publisher: W. H. Freeman
Question
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Chapter 13, Problem 1CRE
To determine

(a)

Domain of the given vector valued function.

Expert Solution
Check Mark

Answer to Problem 1CRE

Solution:

Domain of the given vector-valued function is (1,0)(0,1].

Explanation of Solution

Given:

We have been given a vector valued function:

r1(t)=t1,(t+1)1,sin1t

Key concepts used:

Domain of a rational function is all real numbers except when denominator is zero. Domain of inverse sine function is [1,1]

Calculation:

In order to find the domain of a vector valued function, we find the domains of all the components of the vector function and then we consider the intersection of all the domains.

r1(t)=t1,(t+1)1,sin1t

Domain of the first component t1=1t is all real numbers except 0. Thus, domain of the first component is (,0)(0,).

Domain of the second component (t+1)1=1t+1 is all real numbers except 1. Thus, domain of the second component is (,1)(1,).

Domain of the third component sin1t is [1,1].

For writing the domain of the entire vector valued function, we consider the intersection of all three domains. Thus, the domain of the given vector valued function is (1,0)(0,1].

Conclusion:

The domain of the given vector valued function has been found by first finding the domains of each of the components and then considering the intersection.

To determine

(b)

Domain of the given vector valued function.

Expert Solution
Check Mark

Answer to Problem 1CRE

Solution:

Domain of the given vector-valued function is (0,2].

Explanation of Solution

Given:

We have been given a vector valued function:

r2(t)=8t3,lnt,et

Key concepts used:

Domain of a rational function is all real numbers except when denominator is zero. Domain of inverse sine function is [1,1]

Calculation:

In order to find the domain of a vector valued function, we find the domains of all the components of the vector function and then we consider the intersection of all the domains.

r2(t)=8t3,lnt,et

Domain of the first component 8t3 is (,2].

Domain of the second component lnt is (0,).

Domain of the third component et is [0,).

For writing the domain of the entire vector valued function, we consider the intersection of all three domains. Thus, the domain of the given vector valued function is (0,2].

Conclusion:

The domain of the given vector valued function has been found by first finding the domains of each of the components and then considering the intersection.

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Chapter 13 Solutions

Calculus: Early Transcendentals

Ch. 13.1 - Prob. 5ECh. 13.1 - Prob. 6ECh. 13.1 - Prob. 7ECh. 13.1 - Prob. 8ECh. 13.1 - Prob. 9ECh. 13.1 - Prob. 10ECh. 13.1 - Prob. 11ECh. 13.1 - Prob. 12ECh. 13.1 - Prob. 13ECh. 13.1 - Prob. 14ECh. 13.1 - Prob. 15ECh. 13.1 - Prob. 16ECh. 13.1 - Prob. 17ECh. 13.1 - Prob. 18ECh. 13.1 - Prob. 19ECh. 13.1 - Prob. 20ECh. 13.1 - Prob. 21ECh. 13.1 - Prob. 22ECh. 13.1 - Prob. 23ECh. 13.1 - Prob. 24ECh. 13.1 - Prob. 25ECh. 13.1 - Prob. 26ECh. 13.1 - Prob. 27ECh. 13.1 - Prob. 28ECh. 13.1 - Prob. 29ECh. 13.1 - Prob. 30ECh. 13.1 - Prob. 31ECh. 13.1 - Prob. 32ECh. 13.1 - Prob. 33ECh. 13.1 - Prob. 34ECh. 13.1 - Prob. 35ECh. 13.1 - Prob. 36ECh. 13.1 - Prob. 37ECh. 13.1 - Prob. 38ECh. 13.1 - Prob. 39ECh. 13.1 - Prob. 40ECh. 13.1 - Prob. 41ECh. 13.1 - Prob. 42ECh. 13.1 - Prob. 43ECh. 13.1 - Prob. 44ECh. 13.1 - Prob. 45ECh. 13.1 - Prob. 46ECh. 13.1 - Prob. 47ECh. 13.1 - Prob. 48ECh. 13.1 - Prob. 49ECh. 13.2 - Prob. 1PQCh. 13.2 - Prob. 2PQCh. 13.2 - Prob. 3PQCh. 13.2 - Prob. 4PQCh. 13.2 - Prob. 5PQCh. 13.2 - 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Prob. 29ECh. 13.5 - Prob. 30ECh. 13.5 - Prob. 31ECh. 13.5 - Prob. 32ECh. 13.5 - Prob. 33ECh. 13.5 - Prob. 34ECh. 13.5 - Prob. 35ECh. 13.5 - Prob. 36ECh. 13.5 - Prob. 37ECh. 13.5 - Prob. 38ECh. 13.5 - Prob. 39ECh. 13.5 - Prob. 40ECh. 13.5 - Prob. 41ECh. 13.5 - Prob. 42ECh. 13.5 - Prob. 43ECh. 13.5 - Prob. 44ECh. 13.5 - Prob. 45ECh. 13.5 - Prob. 46ECh. 13.5 - Prob. 47ECh. 13.5 - Prob. 48ECh. 13.5 - Prob. 49ECh. 13.5 - Prob. 50ECh. 13.5 - Prob. 51ECh. 13.5 - Prob. 52ECh. 13.5 - Prob. 53ECh. 13.5 - Prob. 54ECh. 13.5 - Prob. 55ECh. 13.5 - Prob. 56ECh. 13.5 - Prob. 57ECh. 13.5 - Prob. 58ECh. 13.5 - Prob. 59ECh. 13.5 - Prob. 60ECh. 13.5 - Prob. 61ECh. 13.6 - Prob. 1PQCh. 13.6 - Prob. 2PQCh. 13.6 - Prob. 3PQCh. 13.6 - Prob. 1ECh. 13.6 - Prob. 2ECh. 13.6 - Prob. 3ECh. 13.6 - Prob. 4ECh. 13.6 - Prob. 5ECh. 13.6 - Prob. 6ECh. 13.6 - Prob. 7ECh. 13.6 - Prob. 8ECh. 13.6 - Prob. 9ECh. 13.6 - Prob. 10ECh. 13.6 - Prob. 11ECh. 13.6 - Prob. 12ECh. 13.6 - Prob. 13ECh. 13.6 - Prob. 14ECh. 13.6 - Prob. 15ECh. 13.6 - Prob. 16ECh. 13.6 - Prob. 17ECh. 13.6 - Prob. 18ECh. 13.6 - Prob. 19ECh. 13.6 - Prob. 20ECh. 13.6 - Prob. 21ECh. 13.6 - Prob. 22ECh. 13.6 - Prob. 23ECh. 13.6 - Prob. 24ECh. 13.6 - Prob. 25ECh. 13 - Prob. 1CRECh. 13 - Prob. 2CRECh. 13 - Prob. 3CRECh. 13 - Prob. 4CRECh. 13 - Prob. 5CRECh. 13 - Prob. 6CRECh. 13 - Prob. 7CRECh. 13 - Prob. 8CRECh. 13 - Prob. 9CRECh. 13 - Prob. 10CRECh. 13 - Prob. 11CRECh. 13 - Prob. 12CRECh. 13 - Prob. 13CRECh. 13 - Prob. 14CRECh. 13 - Prob. 15CRECh. 13 - Prob. 16CRECh. 13 - Prob. 17CRECh. 13 - Prob. 18CRECh. 13 - Prob. 19CRECh. 13 - Prob. 20CRECh. 13 - Prob. 21CRECh. 13 - Prob. 22CRECh. 13 - Prob. 23CRECh. 13 - Prob. 24CRECh. 13 - Prob. 25CRECh. 13 - Prob. 26CRECh. 13 - Prob. 27CRECh. 13 - Prob. 28CRECh. 13 - Prob. 29CRECh. 13 - Prob. 30CRECh. 13 - Prob. 31CRECh. 13 - Prob. 32CRECh. 13 - Prob. 33CRECh. 13 - Prob. 34CRECh. 13 - Prob. 35CRECh. 13 - Prob. 36CRECh. 13 - Prob. 37CRECh. 13 - Prob. 38CRECh. 13 - Prob. 39CRECh. 13 - Prob. 40CRECh. 13 - Prob. 41CRECh. 13 - 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