# Thread: Compound Interest. (Kind of long..)

1. ## 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!

2. 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?

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