# A investment, orginally worth $20,000, growns continously at the rate of 1000e^0.10t dollars per year, where t is the number of years since the investment was made. Find a formula for the value of...

A investment, orginally worth $20,000, growns continously at the rate of 1000e^0.10t dollars per year, where t is the number of years since the investment was made. Find a formula for the value of the investment after t years, and using this find the value of it after 7 years.

### 1 Answer | Add Yours

Let the amount of the investment be given by `A(t)` . We know that `A(0)=20000` and `A'(t)=1000e^{0.10t}` . Since the antiderivative of the exponential function `e^{kt}` is `C/ke^{kt}` where C is the constant of integration to get

`A(t)=C(1000)/(0.1)e^{0.1t}` now sub in `t=0` to solve for C

`20000=10000Ce^0=10000C`

which means that `C=2` .

**The formula for the investment is `A(t)=20000e^{0.1t}` .**

To find the value of the investment after 7 years, substitute for t to get

`A(7)=20000e^{0.1(7)}=20000e^{0.7} approx 54365.64`

**The approximate value of the investment after 7 years is $54465.64.**