Question

An investment is initially worth $11,000. Write an equation representing the value of this investment V after t years in each of the following situations.

Answer 1

We have a present value of the investment of PV = 11,000.

We have to express the future value V in function of different situations.

a) The value increases by 9% per year.

In this case, the value at year t will be 1.09 times the value at year (t-1).

We can then derive the value at year t as:

`\begin{gathered} V(0)=11000 \\ V(1)=V(0)\cdot1.09=11000\cdot1.09 \\ V(2)=V(1)\cdot1.09=(11000\cdot1.09)\cdot1.09=11000\cdot1.09^2 \\ \Rightarrow V(t)=11000\cdot1.09^t \end{gathered}`

b) The value decreases by $852 per year.

In this case, the rate of change per year is constant (m = -852), so the value in function of time is a linear relation.

We can then write:

`V(t)=11000-852\cdot t`

c) The value increases by $816 per year.

This situation is similar to point b, but with a positive slope (m = 816) instead of a negative slope.

We can express V(t) as:

`V(t)=11000+816\cdot t`

d) The value decreases by 11% per year.

This situation is similar to point a, but with and exponential decay instead of an exponential growth: the ratio is smaller than 1.

Each year, the value at year t is 1-0.11 = 0.89 times the value at year t-1, so we can find the value V(t) as:

`\begin{gathered} V(0)=11000 \\ V(1)=V(0)\cdot0.89=11000\cdot0.89 \\ V(2)=V(1)\cdot0.89=(11000\cdot0.89)\cdot0.89=11000\cdot0.89^2 \\ \Rightarrow V(t)=11000\cdot0.89^t \end{gathered}`

Answer:

a) V = 11000*1.09^t

b) V = 11000 - 852t

c) V = 11000 + 816t

d) V = 11000*0.89^t