# Compound Interest. (Kind of long..)

• Nov 23rd 2006, 10:33 AM
Affinity
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?

• Nov 23rd 2006, 11:31 PM
earboth
Quote:

Originally Posted by Affinity
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?

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