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Math Help - Compound Interest. (Kind of long..)

  1. #1
    Affinity
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    Smile Compound Interest. (Kind of long..)

    Four thousand dollars is deposited into a savings account at 3.5% interest compounded continuously.

    a) What is the formula for A(t), the balance after t years?
    b) What differential equation is satisfied by A(t), the balance after t years?
    c) How much money will be in the account after 2 years?
    d) When will the balance reach $5000?
    e) How fast is the balance growing when it reaches $5000?
    f) When will the balance triple?
    g) How fast is the balance growing at the time it triples in value?

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by Affinity View Post
    Four thousand dollars is deposited into a savings account at 3.5% interest compounded continuously.

    a) What is the formula for A(t), the balance after t years?
    b) What differential equation is satisfied by A(t), the balance after t years?
    c) How much money will be in the account after 2 years?
    d) When will the balance reach $5000?
    e) How fast is the balance growing when it reaches $5000?
    f) When will the balance triple?
    g) How fast is the balance growing at the time it triples in value?

    Thanks in advance!
    Hello,

    I don't know much about money (I don't have so much of it) but your problem seems to be about exponential growth:

    to a) A(t)=4000 \cdot e^{\ln(1+0.035)\cdot t}\approx 4000 \cdot e^{0.0344\cdot t}
    Control: Annually compounded at 3.5%: A(1) = 4000 + 4000 \cdot \frac{3.5}{100}=4140\$
    Continuously compounded: A(1)=4000 \cdot e^{0.0344} \approx 4139.99\$
    Usually the continuously compounded capital is slightly higher than the annually compounded capital. My result contradict this rule because I rounded down ln(1+0.035) to 0.0344.

    to b) I'm not sure what you are asking here, so this is only a try:
    A'(t)=0.0344 \cdot A(t)

    to c) Use the equation of a), plug in t = 2 and calculate:
    A(2)=4000 \cdot e^{0.0344\cdot 2} \approx 4284.89\$

    to d) Now A(t) = 5000 $ and you have to solve for t:
    5000=4000 \cdot e^{0.0344\cdot t}. Divide by 4000 and logarithmize both sides:
    \ln\left(\frac{5000}{4000}\right)=0.0344 \cdot t. Solve for t. I've got t\approx 6.487\text{ years}

    to e) Use the equation of b): Rate of change if A(t) = 5000 $ is A'(t) = 0.0344*5000 = 172 $ per year. (Remember: The rate of change at the beginning was 140 $/year. So now the capital is growing much faster)

    to f and g) I guess that you want to know when the account reaches 12,000 $(?) Have a look at the answers to d) and e).
    I've got the result: 31.936 years and 412.80 $/year.

    EB
    Last edited by earboth; November 23rd 2006 at 10:42 PM.
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