compounded continuously

[killasnake]

HI I am having trouble with these two problems.

A bank pays interest at a rate of 10.6%, compounded continuously. How much should be invested so as to have 11 thousand dollars in 6 years?

Formula: \(\displaystyle Ae^r^t\)

So I have:

\(\displaystyle 11000=A_0e^0^.^1^0^6*^6\)

I would need to solve for \(\displaystyle A_0\) correct?

I would have \(\displaystyle A_0 = 11000/.696\) but it's the wrong answer.

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The rat population in a major metropolitan city is given by the formula \(\displaystyle n(t)=63e^0^.^0^3^5^t\)where \(\displaystyle t\)is measured in years since 1993 and \(\displaystyle n(t)\)is measured in millions.

What was the rat population in 1993 ?______

What is the rat population going to be in the year 2008 ? ________

I don't have a clue where to start this problem.

 

[o_O]

1. You should check your calculations again.
\(\displaystyle e^{0.106 \cdot 6} \neq .696\)

2. For the function n(t), t represents the time that has elapsed since 1993. Since you're finding the rat population in 1993, what would t be equal to? Just simply plug this into your equation and solve for n(t). Similarly, what would t be equal to if you're talking about the year 2008?

 

[jwpaine]

Are you now looking for an answer using calculus? Surly we can provide you with a limit or definite integral!

 

[killasnake]

ahh.. Thank you