If $\$9000$ is invested in a savings account offering $3\%$ per year, compounded continuously, how fast (in dollars/yr) is the balance growing after $4$ years? (Round your answer to the nearest cent)
My Answer :
Using the formula for continuous compound interest we get the final amount, $$P_f = P_0 e^{rt} \\
= 9000 \cdot e^{0.03 \times 4} \\
= 10147.47$$
Therefore, the rate if growth is $= \frac{P_f – P_0}{t} = 286.86$.
My question is that I'm a bit confused about the term "how fast (in dollars/yr) is the balance growing?". Did I get the meaning right in my answer?
Best Answer
We have a function $$f(t)=9000e^{0.03t}\implies f'(t)=270e^{0.03t}.$$ The instantaneous rate of change of this function at $t=4$ is evaluated by $$f'(4)=270e^{0.12}\approx304.42.$$
You can explore the difference between average and instantaneous rate of growth here (links to Desmos).
Here the black curve represents your function, the dashed line is the secant line representing the average rate of change for $t\in[0,4]$, and the red line is the tangent line, representing the instantaneous rate of change at exactly $t=4$.