EandH said:
When an amount of money is invested at interest rate k, compounded continuously, interest is computed every “ instant” and added to the original amount. The balance , after t years, is given by the exponential growth model. (The formula is P(t) = P subscript zero e to the superscript kt) just in case.
P(t) = P_0e^k^t
Suppose that P_0 is invested in a savings account where interest in compounded continuously at 3.1% per year.
Express p(t) in terms of p0 and 0.031
Suppose that $1000 is invested. What is the balance after 1 yr? after 2yr?
When will an investment of 1000 double itself.
P(t) = P_0e^k^t
I think the formula for continuous compounding should not have "k^t" in it, but should look like this:
P(t) = P[sub:1c3d9k1r]0[/sub:1c3d9k1r] * e[sup:1c3d9k1r]k*t[/sup:1c3d9k1r]
When the interest rate is 3.1%, k = 0.031, so
P(t) = P[sub:1c3d9k1r]0[/sub:1c3d9k1r] * e[sup:1c3d9k1r]0.031t[/sup:1c3d9k1r]
If your beginning investment is $1000, the formula becomes
P(t) = 1000*e[sup:1c3d9k1r]0.031t[/sup:1c3d9k1r]
"t" is the number of years the money is invested. For example if you want to know how much your initial $1000 has grown to in
5 years, substitute 5 for t:
P(t) = 1000 * e[sup:1c3d9k1r]0.031*5[/sup:1c3d9k1r]
P(t) = 1000 * e[sup:1c3d9k1r]0.155[/sup:1c3d9k1r]
Use your calculator to evaluate, and you should find that
P(t) = 1167.66 (rounded to the nearest cent)
You can find the amount for 1 or 2 years using the same approach.
Here's an example of a question similar to your last question. Suppose I invest $1000 at 3.1% compounded continuously. How long will it take for that investment to grow to $1500?
We know P[sub:1c3d9k1r]0[/sub:1c3d9k1r] = 1000, and that k = 0.031. And we also know that P(t) = 1500. Substitute into the formula:
1500 = 1000 * e[sup:1c3d9k1r]0.031t[/sup:1c3d9k1r]
We need to solve this for t. First, let's divide both sides of the equation by 1000 to get e[sup:1c3d9k1r]0.031t[/sup:1c3d9k1r] by itself:
1500/1000 = (1000 * e[sup:1c3d9k1r]0.031t[/sup:1c3d9k1r]) / 1000
1.5 = e[sup:1c3d9k1r]0.031t[/sup:1c3d9k1r]
Take the natural log of both sides of the equation:
ln 1.5 = ln (e[sup:1c3d9k1r]0.031t[/sup:1c3d9k1r])
ln 1.5 = 0.031t
Divide both sides by 0.031 to get just t on the right side:
(ln 1.5) / 0.031 = (0.031t)/0.031
(ln 1.5) / 0.031 = t
13.08 = t
So...it will take just a little over 13 years for the original $1000 to grow to $1500.
You can use the same approach to find out when the original $1000 has doubled...which means it grows to ????
