EBK PRECALCULUS W/LIMITS
EBK PRECALCULUS W/LIMITS
4th Edition
ISBN: 9781337516853
Author: Larson
Publisher: CENGAGE CO
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Question
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Chapter 3.5, Problem 58E

a.

To determine

Use a graphing utility to graph each function.

a.

Expert Solution
Check Mark

Answer to Problem 58E

  EBK PRECALCULUS W/LIMITS, Chapter 3.5, Problem 58E , additional homework tip  1

Explanation of Solution

Calculation:

The amount paid toward the interest is

  u=M(MPr12)(1+r12)12t

The amount paid toward the reduction of the principal is

  v=(MPr12)(1+r12)12t

Where P the amount of the mortgage is is, r is the interest rate (in decimal form), M is the monthly payment, and  t is the time in years.

The given functions can be simplified as,

  u=839.06(839.06120000*0.07512)(1+0.07512)12tu=839.06(89.06)(1.00625)12t

And,

  v=(839.06120000*0.07512)(1+0.07512)12tv=(89.06)(1.00625)12t

Now, graph the functions as shown below:

  EBK PRECALCULUS W/LIMITS, Chapter 3.5, Problem 58E , additional homework tip  2

Hence, the graph is shown above.

b.

To determine

Find the approximate time.

b.

Expert Solution
Check Mark

Answer to Problem 58E

  t=20.7 Years.

Explanation of Solution

Calculation: The amount paid toward the interest is

  u=M(MPr12)(1+r12)12t

The amount paid toward the reduction of the principal is

  v=(MPr12)(1+r12)12t

Where P the amount of the mortgage , r is the interest rate (in decimal form), M is the monthly payment, and  t is the time in years.

Let us consider that the monthly payment is evenly divided between interest and principle then,

  u=v

  839.06(89.06)(1.00625)12t=(89.06)(1.00625)12t

On dividing both sides by 89.06

  839.0689.06(1.00625)12t=(1.00625)12t

Adding (1.00625)12t on both sides

  839.0689.06=2(1.00625)12t

Taking natural logarithm of both sides

  ln839.0689.06=ln(2(1.00625)12t)

  ln839.0689.06=ln2+ln(1.00625)12t

Subtract ln2 from both sides,

  ln839.0689.06ln2=12tln(1.00625)

Dividing both sides by 12ln(1.00625)

Hence, t=20.7 years.

c.

To determine

Conclude for a repayment period of 20 years.

c.

Expert Solution
Check Mark

Answer to Problem 58E

As the year decreases , the monthly payment increases, the graph becomes much more linear.

Explanation of Solution

Calculation:

The amount paid toward the interest is

  u=M(MPr12)(1+r12)12t

The amount paid toward the reduction of the principal is

  v=(MPr12)(1+r12)12t

Where P the amount of the mortgage, r is the interest rate (in decimal form), M is the monthly payment, and  t is the time in years.

The amount and interest paid towards the principle can be written as,

  u=966.71(839.06120000*0.07512)(1+0.07512)12tu=966.71(89.06)(1.00625)12t

  v=(966.71120000*0.07512)(1+0.07512)12tv=(216.71)(1.00625)12t

Now, graph corresponding to the function is,

  EBK PRECALCULUS W/LIMITS, Chapter 3.5, Problem 58E , additional homework tip  3

Let us consider that the monthly payment is evenly divided between interest and principle then,

  u=v

  996.71(216.71)(1.00625)12t=(216.71)(1.00625)12t

On dividing both sides by 216.71

  996.71216.71(1.00625)12t=(1.00625)12t

Adding (1.00625)12t on both sides

  996.71216.71=2(1.00625)12t

Taking natural logarithm of both sides

  ln996.71216.71=ln(2(1.00625)12t)

  ln996.71216.71=ln2+ln(1.00625)12t

Subtract ln2 from both sides,

  ln996.71216.71ln2=12tln(1.00625)

Dividing both sides by 12ln(1.00625)

  t=10.7 years.

Hence, we conclude from both the graphs,we can see that as the year decreases , the monthly payment increases, the graph becomes much more linear.

Chapter 3 Solutions

EBK PRECALCULUS W/LIMITS

Ch. 3.1 - Prob. 11ECh. 3.1 - Prob. 12ECh. 3.1 - Prob. 13ECh. 3.1 - Prob. 14ECh. 3.1 - Prob. 15ECh. 3.1 - Prob. 16ECh. 3.1 - Prob. 17ECh. 3.1 - Prob. 18ECh. 3.1 - Prob. 19ECh. 3.1 - Prob. 20ECh. 3.1 - Prob. 21ECh. 3.1 - Prob. 22ECh. 3.1 - Prob. 23ECh. 3.1 - Prob. 24ECh. 3.1 - Prob. 25ECh. 3.1 - Prob. 26ECh. 3.1 - Prob. 27ECh. 3.1 - Prob. 28ECh. 3.1 - Prob. 29ECh. 3.1 - Prob. 30ECh. 3.1 - Prob. 31ECh. 3.1 - Prob. 32ECh. 3.1 - Prob. 33ECh. 3.1 - Prob. 34ECh. 3.1 - Prob. 35ECh. 3.1 - Prob. 36ECh. 3.1 - Prob. 37ECh. 3.1 - Prob. 38ECh. 3.1 - Prob. 39ECh. 3.1 - Prob. 40ECh. 3.1 - Prob. 41ECh. 3.1 - Prob. 42ECh. 3.1 - Prob. 43ECh. 3.1 - Prob. 44ECh. 3.1 - Prob. 45ECh. 3.1 - Prob. 46ECh. 3.1 - Prob. 47ECh. 3.1 - Prob. 48ECh. 3.1 - Prob. 49ECh. 3.1 - Prob. 50ECh. 3.1 - Prob. 51ECh. 3.1 - Prob. 52ECh. 3.1 - Prob. 53ECh. 3.1 - Prob. 54ECh. 3.1 - Prob. 55ECh. 3.1 - Prob. 56ECh. 3.1 - Prob. 57ECh. 3.1 - Prob. 58ECh. 3.1 - Prob. 59ECh. 3.1 - Prob. 60ECh. 3.1 - Prob. 61ECh. 3.1 - Prob. 62ECh. 3.1 - Prob. 63ECh. 3.1 - 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