BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 2.1, Problem 8E

(a)

To determine

To find: The average velocity for given time periods.

Expert Solution

Answer to Problem 8E

(i) The average velocity over the time interval [1, 2] is 6 cm/s.

(ii) The average velocity over the time interval [1, 1.1] is 4.71 cm/s.

(iii) The average velocity over the time interval [1, 1.01] is 6.13 cm/s.

(iv) The average velocity over the time interval [1, 1.001] is 6.27 cm/s.

Explanation of Solution

Given:

The equation of a motion s=2sinπt+3cosπt (in centimeters), where t is measured in seconds.

Formula used:

The average velocity over the time interval [t,t+h] is,

vavg=s(t+h)s(t)(t+h)t (1)

Calculation:

Section-(i)

Obtain the average velocity over the time interval [1, 2].

Substitute t=1 and t+h=2 in equation (1),

vavg=s(t+h)s(t)(t+h)t=s(2)s(1)21=(2sinπ(2)+3cosπ(2))(2sinπ(1)+3cosπ(1))1=2sin2π+3cos2π2sinπ3cosπ

Substitute the respective trigonometric values and simplify the terms.

vavg=2(0)+3(1)2(0)3(1)=0+30+3=6

Thus, the average velocity over the time interval [1, 2] is 6 cm/s.

Section-(ii)

Obtain the average velocity over the time interval [1, 1.1].

Substitute t=1 and t+h=1.1 in equation (1),

vavg=s(t+h)s(t)(t+h)t=s(1.1)s(1)1.11=(2sinπ(1.1)+3cosπ(1.1))(2sinπ(1)+3cosπ(1))0.1=2sin(1.1π)+3cos(1.1π)2sinπ3cosπ0.1

Substitute the respective trigonometric values and simplify the terms.

vavg=2(0.30902)+3(0.9511)2(0)3(1)0.1=0.618042.8533+30.1=0.471340.14.71

Thus, the average velocity over the time interval [1, 1.1] is 4.71 cm/s.

Section-(iii)

Obtain the average velocity over the time interval [1, 1.01].

Substitute t=1 and t+h=1.01 in equation (1),

vavg=s(t+h)s(t)(t+h)t=s(1.01)s(1)1.011=(2sinπ(1.01)+3cosπ(1.01))(2sinπ(1)+3cosπ(1))0.01=2sin(1.01π)+3cos(1.01π)2sinπ3cosπ0.01

Substitute the respective trigonometric values and simplify the terms.

vavg=2(0.03141)+3(0.9995)2(0)3(1)0.01=0.062822.9985+30.01=0.061320.016.13

Thus, the average velocity over the time interval [1, 1.01] is 6.13 cm/s.

Section-(iv)

Obtain the average velocity over the time interval [1, 1.001].

Substitute t=1 and t+h=1.001 in equation (1).

vavg=s(t+h)s(t)(t+h)t=s(1.001)s(1)1.0011=(2sinπ(1.001)+3cosπ(1.001))(2sinπ(1)+3cosπ(1))0.001=2sin(1.001π)+3cos(1.001π)2sinπ3cosπ0.001

Substitute the respective trigonometric values and simplify the terms.

vavg=2(0.0031416)+3(0.999995)2(0)3(1)0.001=0.00628322.999985+30.001=0.00626820.0016.27

Thus, the average velocity over the time interval [1, 1.001] is 6.27 cm/s.

(b)

To determine

To estimate: The instantaneous velocity of the particle when t=1.

Expert Solution

Answer to Problem 8E

The estimated instantaneous velocity when t=1 is 6.3 cm/s.

Explanation of Solution

The average velocity over the time interval [1, 1+h] is,

vavg=s(1+h)s(1)(1+h)1=2sinπ(1+h)+3cosπ(1+h)(2sinπ(1)+3cosπ(1))h=2sinπ(1+h)+3cosπ(1+h)(2(0)+3(1))h=2sinπ(1+h)+3cosπ(1+h)+3h

The instantaneous velocity when t=1 is expressed as the limiting value of the average velocities over smaller time periods that start at t=1.

That is, when h approaches 0, then t is closer to 1.

v=limh0s(1+h)s(1)h=limh0(2sinπ(1+h)+3cosπ(1+h)+3h)=6.28316.3

Thus, the estimated instantaneous velocity when t=1 is 6.3 cm/s.

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