Advance Function 12

Question

After college you invest $8500 at 7.2% a compounded quarterly. estimate the instantaneous rate of change at 10 years.

Answer 1

To calculate your return on compound interest that is not compounded anually, you use the formula:

A = P(1 + r/n)^nt

Where P is the initial principal, r is the interest rate, t is the number of years passed, and n is the number of times interest is compounded per year.

So our t = 10 and n = 4 here. To calculate the monetary return, we plug it in this value:

A = 8500(1 + (.072/4))^4*10

= 8500(1 + .018)⁴⁰

= 8500(1.018)⁴⁰

So over the course of 10 years, we are compounding interest on the principal a total of 40 times. The coordinate point (t, A) would be (10, 8500(1.018)⁴⁰). To estimate the rate of change at this time, we want to pick another coordinate point that is close to this one, and use the slope formula on the two. We can pick (10.25, 8500(1.018)⁴¹).

m = (8500(1.018)⁴¹ - 8500(1.018)⁴⁰)/(10.25 - 10)

= (8500/.25)*((1.018)⁴¹ - (1.018)⁴⁰)

= 34000*(2.078 - 2.041)

= 34000*.037

= 1,249.29 dollars per year.

Answer 2

This is more a Calculus problem than an Algebra 2 problem; but, since you list this problem under Algebra 2, I'm going to give you an answer that is more appropriate to Algebra 2.

The generic equation for quarterly interest growth, where V is the value of the investment, V₀ is the initial investment amount, t is the investment time, and r is the interest rate is V(t) = V₀(1 + r/4)^4t so in this case it becomes:

V(t) = 8500(1 + 0.072/4)^4t or

V(t) = 8500(1.018)^4t

The instantaneous rate of change (I'll call it IRC) can be estimate using:

[f(a + h) - f(a)]/h for small values of h, and using a = 10 and h is a small unit of time in this case. I'll arbitrarily pick a value of h (aka Δt) of 1 year, which is not going to give a very good answer because its not a very small value Δt, but it's a starting place. So:

IRC = [f(10 + 1) - f(10)]/1 = [f(11) - f(10)]/1 = (18634.65-17351.22)/1 = 1283.42 dollars/year

Now, lets go down by a factor of 10 to Δt = 0.1 years

IRC = [f(10 + 0.1) - f(10)]/0.1 = [f(10.1) - f(10)]/0.1 = (17475.48-17351.22)/0.1 = 1242.61 dollars/year

Now lets go down by a factor of 10 to Δt = 0.01 years

IRC = [f(10 + 0.01) - f(10)]/0.01 = [f(10.01) - f(10)]/0.01 = (17363.61-17351.22)/0.01 = 1238.62 dollars/year

Now lets go down by a factor of 10 to Δt = 0.001 years

IRC = [f(10 + 0.001) - f(10)]/0.001 = [f(10.001) - f(10)]/0.001 = (17352.46-17351.22)/0.001 = 1238.22 dollars/year

Now lets go down by a factor of 10 to Δt = 0.0001 years

IRC = [f(10 + 0.0001) - f(10)]/0.0001 = [f(10.0001) - f(10)]/0.0001 = (17351.35-17351.22)/0.0001 = 1238.18 dollars/year

Now lets go down by a factor of 10 to Δt = 0.00001 years

IRC = [f(10 + 0.00001) - f(10)]/0.00001 = [f(10.00001) - f(10)]/0.00001 = (17351.23-17351.22)/0.0001 = 1238.18 dollars/year

There was no change (in two decimal points) for this answer and the previous one so this is our final answer.

If you wanted to, you could convert this to $/day or some other more convenient unit of measure but since the problem didn't specify it, I'll just leave it in $/year