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Compound interest The balance A (in dollars) in a saving account is given by A = 5000e0.08t,Where t is measured in years.Find the rates at which the balance is changing when (a) t = 1 year,and (c) t = 50 years.

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Answer:

a)
(dA)/(dt)=400 e^(0.08(1)) =433.315

b)
(dA)/(dt)=400 e^(0.08(10)) =890.216

c)
(dA)/(dt)=400 e^(0.08(50)) =21839.260

Explanation:

For this case we have the following expression for the balance:


A = 5000 e^(0.08t)

And we are interested on the rate at which the balance is changing for some values of t. The correct way to answer this is finding the derivate respect to t of A like this:


(dA)/(dt)= (d)/(dt) (5000 e^(0.08t))


(dA)/(dt)= 5000 (d)/(dt)(e^(0.08t))


(dA)/(dt)= 5000*0.08 e^(0.08t)


(dA)/(dt)=400 e^(0.08t)

So then we can solve for the different years like this

Part a

t= 1 years


(dA)/(dt)=400 e^(0.08(1)) =433.315

So on this case would be $433.315 dollars per year

Part b

t=10 years


(dA)/(dt)=400 e^(0.08(10)) =890.216

So on this case would be $890.216 dollars per year

Part c

t=50 years


(dA)/(dt)=400 e^(0.08(50)) =21839.260

So on this case would be $21839.260 dollars per year

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