Q. 16. - Calculus

Question

Suppose your bank account grows at $3$ percent interest yearly, so that your bank balance after $t$ years is $B\left(t\right)=B_0{(1.03)}^t$.

 

(a) Show that your bank balance grows at a rate proportional to the amount of the balance. 

(b) What is the proportionality constant for the growth rate, and what is the corresponding differential equation for the exponential growth model of $B\left(t\right)$?

Step-by-Step Solution

Answer

Ans:  

(a)    The growth rate of bank balance is proportional to the balance at any time $t$.

(b)    The differential equation defining the exponential growth model is $\frac{dB}{dt}=0.02956B\left(t\right)$

Step 1. Given information.

given,

    $B\left(t\right)=B_0{(1.03)}^t$

Step 2. (a) Differentiating both sides of the equation with respect to t ,

    $\begin{array}{r}\frac{dB}{dt}=B_0(1.03{)}^t\ln (1.03)\\ =B_0\ln \left(1.03\right)\cdot (1.03{)}^t\\ =\ln \left(1.03\right)\cdot \left[B_0(1.03{)}^t\right]\\ =\ln \left(1.03\right)B\left(t\right)\end{array}$

The result replies that the growth rate of the bank balance is a constant multiple of the balance at any time $t$. That is

       $\begin{array}{r}\frac{dB}{dt}=K\cdot B\left(t\right)\\ \alpha B\left(t\right)\end{array}$

Therefore,  The growth rate of bank balance is proportional to the balance at any time  $t$.

Step 3. (b) The proportionality constant k in the result obtained in part (a) is equal to ln ⁡ ( 1.03 ) = 0.02956 .

And the differential equation defining the exponential growth model is $\frac{dB}{dt}=0.02956B\left(t\right)$.