Question

The value V of an item t years after it is purchased is given below.V = 9000exp(-0.6301t), 0 ≤ t ≤ 10Part b only-456.21 wrong-129.40 wrong

Answer 1

Answer:

• V'(4)=-456.10

,

• V'(6)=-129.35

Explanation:

Given the function V which is defined as follows:

$V=9000e^{-0.6301t},0\leq t\leq10$

First, find the derivative of V with respect to time.

$\begin{gathered} \frac{dV}{dt}=9000\frac{d}{dt}(e^{-0.6301t}) \\ \text{ The derivative of }e^{kt}=ke^{kt} \\ V^{\prime}(t)=9000(-0.6301)e^{-0.6301t} \end{gathered}$

I. When t=4

$\begin{gathered} V^{\prime}(t)=9000(-0.6301)e^{-0.6301t} \\ V^{\prime}(4)=9000(-0.6301)e^{-0.6301\times4} \\ =-456.10 \end{gathered}$

The rate of change when t=4 is -456.10 (correct to 2 decimal places).

II. When t=6

$\begin{gathered} V^{\prime}(t)=9000(-0.6301)e^{-0.6301t} \\ V^{\prime}(4)=9000(-0.6301)e^{-0.6301\times6} \\ =-129.35 \end{gathered}$

The rate of change when t=6 is -129.35 (correct to 2 decimal places).