Single Variable Calculus: Concepts and Contexts, Enhanced Edition
Single Variable Calculus: Concepts and Contexts, Enhanced Edition
4th Edition
ISBN: 9781337687805
Author: James Stewart
Publisher: Cengage Learning
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Chapter 7.2, Problem 27E

a.

To determine

To draw: A direction field for the differential equation.

a.

Expert Solution
Check Mark

Explanation of Solution

Given information:

Differential equation: RdQdt+1CQ=E(t)

  R=5Ω , C=0.05F , E=60V

Concept used :

If dydx=F(x,y) , F(x,y) gives the slope of the solution curve at point (x,y) . If small line segments with slope F(x,y) are drawn on several points (x,y) , the results are called direction field.

Graph:

  RdQdt+1CQ=E(t)

  R=5Ω , C=0.05F , E=60V

  5dQdt+10.05Q=60⇒5dQdt+20Q=60⇒dQdt+4Q=12⇒dQdt=12−4Q

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 7.2, Problem 27E , additional homework tip 1

Interpretation: The graph is the direction field of the differential equation dQdt=12−4Q

b.

To determine

To find: The limiting value of the charge.

b.

Expert Solution
Check Mark

Answer to Problem 27E

The limiting value of the charge is 3 Coulomb.

Explanation of Solution

Given information:

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 7.2, Problem 27E , additional homework tip 2

It is clear from the direction field that all the solutions approaches to the value 3 .

i.e. limt→∞Q(t)=3

c.

To determine

To find: Whether there are any equilibrium solution or not.

c.

Expert Solution
Check Mark

Answer to Problem 27E

  Q(t)=3is an equilibrium solution.

Explanation of Solution

Given information:

  dQdt=12−4Q

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 7.2, Problem 27E , additional homework tip 3

Concept used :

If for a particular solution , dQdt be zero, then that solution will be equilibrium solution.

Calculation:

From the direction field it appears that Q=3 is an equilibrium solution of dQdt=12−4Q .

  dQdt=12−4Q=12−4×3=12−12=0

d.

To determine

To draw: The graph of the solution when Q(0)=0C using direction field.

d.

Expert Solution
Check Mark

Explanation of Solution

Given information:

  Q(0)=0 C

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 7.2, Problem 27E , additional homework tip 4

Graph :

  Q(0)=0 C

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 7.2, Problem 27E , additional homework tip 5

Interpretation:

The graph shows the particular solution when Q(0)=0 C and hence it passes through origin.

e.

To determine

To estimate: The charge after half a second using Euler’s method.

e.

Expert Solution
Check Mark

Answer to Problem 27E

The charge after half a second is 2.7672 C

Explanation of Solution

Given information:

  dQdt=12−4Q ; Q(0)=0 C

  h=0.1

Formula used :

Euler’s Method: Approximate values for the solution of the initial-value problem y′=F(x,y) , y(x0)=y0 with the step size h , at xn=xn−a+h .

  yn=yn−1+hF(xn−1,yn−1)     n=1,2,3.....

Calculation:

  h=0.1

  dQdt=12−4Q and Q(0)=0 C

Here, Q′=F(Q,t)=12−4Q

  Q0=0

  Q1=Q0+hF(Q0,t0)=0+0.1×12=1.2Q2=Q1+hF(Q1,t1)=1.2+0.1×7.2=1.92Q3=Q2+hF(Q2,t2)=1.92+0.1×4.32=2.352Q4=Q3+hF(Q3,t3)=2.352+0.1×2.592=2.612Q5=Q4+hF(Q4,t4)=2.612+0.1×1.552=2.7672

  Q(0.5)=2.7672

Therefore, the charge after half a second is 2.7672 C

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Chapter 7.2, Problem 26E
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Chapter 7.2, Problem 28E

Chapter 7 Solutions

Single Variable Calculus: Concepts and Contexts, Enhanced Edition

Ch. 7.1 - Prob. 1ECh. 7.1 - Prob. 2ECh. 7.1 - Prob. 3ECh. 7.1 - Prob. 4ECh. 7.1 - Prob. 5ECh. 7.1 - Prob. 6ECh. 7.1 - Prob. 7ECh. 7.1 - Prob. 8ECh. 7.1 - Prob. 9ECh. 7.1 - Prob. 10E
Ch. 7.1 - Prob. 11ECh. 7.1 - Prob. 12ECh. 7.1 - Prob. 13ECh. 7.1 - Prob. 14ECh. 7.1 - Prob. 15ECh. 7.2 - Prob. 1ECh. 7.2 - Prob. 2ECh. 7.2 - Prob. 3ECh. 7.2 - Prob. 4ECh. 7.2 - Prob. 5ECh. 7.2 - Prob. 6ECh. 7.2 - Prob. 7ECh. 7.2 - Prob. 8ECh. 7.2 - Prob. 9ECh. 7.2 - Prob. 10ECh. 7.2 - Prob. 11ECh. 7.2 - Prob. 12ECh. 7.2 - Prob. 13ECh. 7.2 - Prob. 14ECh. 7.2 - Prob. 15ECh. 7.2 - Prob. 16ECh. 7.2 - Prob. 17ECh. 7.2 - Prob. 18ECh. 7.2 - Prob. 19ECh. 7.2 - Prob. 20ECh. 7.2 - Prob. 21ECh. 7.2 - Prob. 22ECh. 7.2 - Prob. 23ECh. 7.2 - Prob. 24ECh. 7.2 - Prob. 25ECh. 7.2 - Prob. 26ECh. 7.2 - Prob. 27ECh. 7.2 - Prob. 28ECh. 7.3 - Prob. 1ECh. 7.3 - Prob. 2ECh. 7.3 - Prob. 3ECh. 7.3 - Prob. 4ECh. 7.3 - Prob. 5ECh. 7.3 - Prob. 6ECh. 7.3 - Prob. 7ECh. 7.3 - Prob. 8ECh. 7.3 - Prob. 9ECh. 7.3 - Prob. 10ECh. 7.3 - Prob. 11ECh. 7.3 - Prob. 12ECh. 7.3 - Prob. 13ECh. 7.3 - Prob. 14ECh. 7.3 - Prob. 15ECh. 7.3 - Prob. 16ECh. 7.3 - Prob. 17ECh. 7.3 - Prob. 18ECh. 7.3 - Prob. 19ECh. 7.3 - Prob. 20ECh. 7.3 - Prob. 21ECh. 7.3 - Prob. 22ECh. 7.3 - Prob. 23ECh. 7.3 - Prob. 24ECh. 7.3 - Prob. 25ECh. 7.3 - Prob. 26ECh. 7.3 - Prob. 27ECh. 7.3 - Prob. 28ECh. 7.3 - Prob. 29ECh. 7.3 - Prob. 30ECh. 7.3 - Prob. 31ECh. 7.3 - Prob. 32ECh. 7.3 - Prob. 33ECh. 7.3 - Prob. 34ECh. 7.3 - Prob. 35ECh. 7.3 - Prob. 36ECh. 7.3 - Prob. 37ECh. 7.3 - Prob. 38ECh. 7.3 - Prob. 39ECh. 7.3 - Prob. 40ECh. 7.3 - Prob. 41ECh. 7.3 - Prob. 42ECh. 7.3 - Prob. 43ECh. 7.3 - Prob. 44ECh. 7.3 - Prob. 45ECh. 7.3 - Prob. 46ECh. 7.3 - Prob. 47ECh. 7.3 - Prob. 48ECh. 7.3 - Prob. 49ECh. 7.3 - Prob. 50ECh. 7.3 - Prob. 51ECh. 7.3 - Prob. 52ECh. 7.3 - Prob. 53ECh. 7.3 - Prob. 54ECh. 7.4 - Prob. 1ECh. 7.4 - Prob. 2ECh. 7.4 - Prob. 3ECh. 7.4 - Prob. 4ECh. 7.4 - Prob. 5ECh. 7.4 - Prob. 6ECh. 7.4 - Prob. 7ECh. 7.4 - Prob. 8ECh. 7.4 - Prob. 9ECh. 7.4 - Prob. 10ECh. 7.4 - Prob. 11ECh. 7.4 - Prob. 12ECh. 7.4 - Prob. 13ECh. 7.4 - Prob. 14ECh. 7.4 - Prob. 15ECh. 7.4 - Prob. 16ECh. 7.4 - Prob. 17ECh. 7.4 - Prob. 18ECh. 7.4 - Prob. 19ECh. 7.4 - Prob. 20ECh. 7.4 - Prob. 21ECh. 7.4 - Prob. 22ECh. 7.5 - Prob. 1ECh. 7.5 - Prob. 2ECh. 7.5 - Prob. 3ECh. 7.5 - Prob. 4ECh. 7.5 - Prob. 5ECh. 7.5 - Prob. 6ECh. 7.5 - Prob. 7ECh. 7.5 - Prob. 8ECh. 7.5 - Prob. 9ECh. 7.5 - Prob. 10ECh. 7.5 - Prob. 11ECh. 7.5 - Prob. 12ECh. 7.5 - Prob. 13ECh. 7.5 - Prob. 14ECh. 7.5 - Prob. 15ECh. 7.5 - Prob. 16ECh. 7.5 - Prob. 17ECh. 7.5 - Prob. 18ECh. 7.5 - Prob. 19ECh. 7.5 - Prob. 20ECh. 7.6 - Prob. 1ECh. 7.6 - Prob. 2ECh. 7.6 - Prob. 3ECh. 7.6 - Prob. 4ECh. 7.6 - Prob. 5ECh. 7.6 - Prob. 6ECh. 7.6 - Prob. 7ECh. 7.6 - Prob. 8ECh. 7.6 - Prob. 9ECh. 7.6 - Prob. 10ECh. 7.6 - Prob. 11ECh. 7.6 - Prob. 12ECh. 7 - Prob. 1RCCCh. 7 - Prob. 2RCCCh. 7 - Prob. 3RCCCh. 7 - Prob. 4RCCCh. 7 - Prob. 5RCCCh. 7 - Prob. 6RCCCh. 7 - Prob. 7RCCCh. 7 - Prob. 8RCCCh. 7 - Prob. 1RQCh. 7 - Prob. 2RQCh. 7 - Prob. 3RQCh. 7 - Prob. 4RQCh. 7 - Prob. 5RQCh. 7 - Prob. 1RECh. 7 - Prob. 2RECh. 7 - Prob. 3RECh. 7 - Prob. 4RECh. 7 - Prob. 5RECh. 7 - Prob. 6RECh. 7 - Prob. 7RECh. 7 - Prob. 8RECh. 7 - Prob. 9RECh. 7 - Prob. 10RECh. 7 - Prob. 11RECh. 7 - Prob. 12RECh. 7 - Prob. 13RECh. 7 - Prob. 14RECh. 7 - Prob. 15RECh. 7 - Prob. 16RECh. 7 - Prob. 17RECh. 7 - Prob. 18RECh. 7 - Prob. 19RECh. 7 - Prob. 20RECh. 7 - Prob. 21RECh. 7 - Prob. 22RECh. 7 - Prob. 23RECh. 7 - Prob. 24RECh. 7 - Prob. 1PCh. 7 - Prob. 2PCh. 7 - Prob. 3PCh. 7 - Prob. 4PCh. 7 - Prob. 5PCh. 7 - Prob. 6PCh. 7 - Prob. 7PCh. 7 - Prob. 8PCh. 7 - Prob. 9PCh. 7 - Prob. 10PCh. 7 - Prob. 11PCh. 7 - Prob. 12PCh. 7 - Prob. 13PCh. 7 - Prob. 14P
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