Objective [1.8] An investment of $8,100 which earns 11.5% per year is growing continuously How fast will it be growing at year 8?

Question

Objective [1.8]

An investment of $8,100 which earns 11.5% per year is growing continuously

How fast will it be growing at year 8?

Answer 1

Each year, we add 11.5% to the value of the investment.

More easily, each year, the value of the investment is multiplied by 1.115

(Here's why it's easier to do with multiplication. The amount we add on is *different* each year: it

gets bigger and bigger. But the amount by which we multiply is the *same* each year: it's always 11.5%.)

Let f(y) be the value when y years have passed since the start of the investment. Therefore

f(0) = 8100 (because no time has passed since the start of the investment)

f(1) = 8100 · 1.115 (because 1 year has passed since the start of the investment)

f(2) = 8100 · 1.115 · 1.115 (because 2 years have passed since the start of the investment)

f(3) = 8100 · 1.115 · 1.115 · 1.115 (because 3 years have passed since the start of the investment)

etc.

In general, f(y) will be

f(y) = 8100 · 1.115^y

Now it's time to use Calculus. To find how fast it's growing, we will take the derivative of f(y) with respect to y.

f'(y) = 8100 · 1.115^y · ln 1.115

When y=8, we have

f'(8) = 8100 · 1.115⁸ · ln 1.115

= 8100 · 2.38891 · 0.108854 (rounding off a little bit)

= 2106.34 dollars per year

That's how fast the investment is growing at year 8.

Answer 2

same rate of growth: 11.5% every year, including the 8th year

A = 8100e^.115(8) in the 8th year

A= 8100e^.92

= $20,325.25 rounded off to nearest cent, in 8 years, compounded continuously