Question

When a deposit of $1000 is made into an account paying 2% interest, compounded annually, the balance, $B, in the account after t years is given by B = 1000(1.02)t. Find the average rate of change in the balance over the interval t = 0 to t = 5. Give units and interpret your answer in terms of the balance in the account.

Answer 1

Answer:

The average rate of change in the balance over the interval t = 0 to t = 5 is of $20.82 a year. This means that the balance increased by $20.82 a year over the interval t = 0 to t = 5.

Step-by-step explanation:

Given a function y, the average rate of change S of y=f(x) in an interval $(x_{s}, x_{f})$ will be given by the following equation:

$S = \frac{f(x_{f}) - f(x_{s})}{x_{f} - x_{s}}$

In this problem, we have that:

$B(t) = 1000(1.02)^{t}$

Find the average rate of change in the balance over the interval t = 0 to t = 5.

$B(0) = 1000(1.02)^{0} = 1000$

$B(5) = 1000(1.02)^{5} = 1104.08$

Then

$S = \frac{1104.08 - 1000}{5-0} = 20.82$

The average rate of change in the balance over the interval t = 0 to t = 5 is of $20.82 a year. This means that the balance increased by $20.82 a year over the interval t = 0 to t = 5.