Question

An investment of $8,000 earns interest at an annual rate of 7% compounded continuously. Complete parts (A) and (B) below. Click the icon to view the derivatives of exponential and logarithmic functions.
(A) Find the instantaneous rate of change of the amount in the account after 2 year(s). $ (Round to two decimal places as needed).
(B) Find the instantaneous rate of change of the amount in the account at the time the amount is equal to $12,000. $ (Round to two decimal places as needed).

Answer 1

Answer:

A.    $\mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}$

B.    $\mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }$

Step-by-step explanation:

Given that:

An investment of  Amount = $8000

earns  at an annual rate of interest = 7% = 0.07 compounded continuously

The objective is to :

A)  Find the instantaneous rate of change of the amount in the account after 2 year(s).

we all know that:

$A = Pe^{rt}$

where;

$A = (8000) \ e ^{0.7t}$

The instantaneous rate of change = $\dfrac{dA}{dt}$

$\dfrac{dA}{dt} = \dfrac{d}{dt}(8000 \ e ^{0.07t} )$

$= 8000 \dfrac{d}{dt}e^{0.07 \ t}$

$\dfrac{dA}{dt}= 8000 (0.07)e^{0.07 \ t}$

$\dfrac{dA}{dt}= 560 e^{0.07 \ t}$

At t = 2 years; the instantaneous rate of change is:

$\dfrac{dA}{dt}|_{t=2}= 560 e^{0.07 \times 2}$

$\mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}$

(B) Find the instantaneous rate of change of the amount in the account at the time the amount is equal to $12,000.

Here the amount = 12000

$12000 = (8000)e^{0.07 \ t}$

$\dfrac{12000 }{8000}= e^{0.07 \ t}$

$1.5= e^{0.07 \ t}$

㏑(1.5) = 0.07 t

0.405465 = 0.07 t

t = 0.405465 /0.07

t = 5.79

$\dfrac{dA}{dt}= 560 e^{0.07 \ t}$

At t = 5.79

$\dfrac{dA}{dt}|_{t = 5.79}= 560 e^{0.07 \times 5.79}$

$\mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }$