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A investment, orginally worth $20,000, growns continously at the rate of 1000e^0.10t...

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jjmgingrich | Student, Undergraduate | (Level 1) Salutatorian

Posted July 13, 2013 at 5:16 PM via web

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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. 

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lfryerda | High School Teacher | (Level 2) Educator

Posted July 14, 2013 at 12:23 AM (Answer #1)

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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.

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