Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 3.1, Problem 29E

(a)

To determine

The representation for derivative of a function.

(a)

Expert Solution

Answer to Problem 29E

The derivative represent the speed .

Explanation of Solution

Given information:

The given table skiing distances

Time t (seconds)	Distance Traveled (feet)
0	0
1	3.3
2	13.3
3	29.9
4	53.2
5	83.2
6	119.8
7	163.0
8	212.9
9	269.5
10	332.7

Formula Used:

Speed formula of an object:

speed=distancetime

Calculation:

The derivative represent the distance covered per unit of time.

Since, speed=distancetime

Hence, the derivatives represent speed.

(b)

To determine

The units of the derivatives to be measured.

(b)

Expert Solution

Answer to Problem 29E

The unit is in feet per second.

Explanation of Solution

Given information:

The given table skiing distances

Time t (seconds)	Distance Traveled (feet)
0	0
1	3.3
2	13.3
3	29.9
4	53.2
5	83.2
6	119.8
7	163.0
8	212.9
9	269.5
10	332.7

Formula Used:

Speed formula of an object:

speed=distancetime

In the table the distance are given in feet and the time are given in second.

Since, speed=distancetime=feetsecond

Hence, the unit is in feet per second.

(c)

To determine

To calculate:The equation of the derivative.

(c)

Expert Solution

Answer to Problem 29E

The expression of the straight line is d=6.7t−0.05

Explanation of Solution

Given information:

The given table skiing distances

Time t (seconds)	Distance Traveled (feet)
0	0
1	3.3
2	13.3
3	29.9
4	53.2
5	83.2
6	119.8
7	163.0
8	212.9
9	269.5
10	332.7

Formula Used:

Slop of a straight line:

m=y2−y1x2−x1

Point slop form of a straight line:

y2−y1=m(x2−x1)

Here m is the slop and (x1,y1) and (x2,y2) are the points on the line.

Draw the graph of the derivatives. Use the data’s of the given table

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 3.1, Problem 29E

The derivatives represents a linear function. Thus it must contain a slop.

Consider any two points on the line.

Let two points be (0.5,3.3) and (1.5,10.0) .

Now, calculate the slop. Use the slop formula m=y2−y1x2−x1 .

m=10.0−3.31.5−0.5=6.7

Now, find the equation of the linear equation. Use the point slop formula y2−y1=m(x2−x1) .

d−3.3=6.7(t−0.5)d=6.7t−3.35+3.3d=6.7t−0.05

Hence, the expression of the straight line is d=6.7t−0.05

Chapter 3.1, Problem 28E

Chapter 3.1, Problem 30E

Chapter 3 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

Ch. 3.1 - Prob. 1QR Ch. 3.1 - Prob. 2QR Ch. 3.1 - Prob. 3QR Ch. 3.1 - Prob. 4QR Ch. 3.1 - Prob. 5QR Ch. 3.1 - Prob. 6QR Ch. 3.1 - Prob. 7QR Ch. 3.1 - Prob. 8QR Ch. 3.1 - Prob. 9QR Ch. 3.1 - Prob. 10QR

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Additional Math Textbook Solutions

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In Problems 35-46, find the distance d( P 1 , P 2 ) between the points P 1 and P 2 . 37.

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