# 1) The value of a computer t years after purchase is v(t)=2000e^{-.35t} At what rate is the...

## Question:

1) The value of a computer {eq}t{/eq} years after purchase is {eq}v(t)=2000e^{-.35t}{/eq} At what rate is the computer's value falling after {eq}3{/eq} years?

2) The velocity of a parachutist during free fall is {eq}f(t)=70(1-e^{-.14t}){/eq} meters per second. Answer the following

questions using the graph. (Recall that acceleration is the derivative of velocity.)

3) Find the value of {eq}x{/eq} at which the function has a possible relative maximum or minimum point.

(Recall that {eq}e{/eq} Superscript {eq}xex{/eq} is positive for all {eq}x{/eq}.) Use the second derivative to determine the nature of the function at these points. {eq}f(x)=9-\frac{9x}{e^{4x}}{/eq}

## Differentiation:

In both the questions solved below, we need differentiation. We use the following property:

$$\begin{align} \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{ax} \right )=ae^{ax} \end{align} $$

Apart from this property, we also use the quotient rule of differentiation in the third part solved below.

## Answer and Explanation:

1) The rate of change in the computer's value is given by the derivative of {eq}v(t) {/eq}.

$$\begin{align} v'(t)&=\frac{\mathrm{d} }{\mathrm{d} t}\left ( 2000e^{-.35t} \right )\\[0.3cm] &=-2000*0.35e^{-0.35t}\\[0.3cm] &=-700e^{-0.35t}\\[0.3cm] \end{align} $$

When {eq}t=3 {/eq},

$$\begin{align} v'(3)&=-700e^{-0.35*3}\\[0.3cm] &\approx -244.96 \end{align} $$

Thus, the value is falling by {eq}244.96{/eq} after 3 years.

3) The critical point is solved by first equating the first derivative to zero.

$$\begin{align} f'(x)&=\frac{\mathrm{d} }{\mathrm{d} x}\left ( 9-\frac{9x}{e^{4x}} \right )\\[0.3cm] &=-\frac{9e^{4x}-36xe^{4x}}{e^{8x}}\\[0.3cm] &=\frac{-9+36x}{e^{4x}}\\[0.3cm] &\frac{-9+36x}{e^{4x}}=0\\[0.3cm] x&=0.25 \end{align} $$

We now know that there is a critical point at {eq}x=0.25{/eq}. Let's see its nature using the second derivative test.

$$\begin{align} f"(x)&=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{-9+36x}{e^{4x}} \right )\\[0.3cm] &=\frac{36e^{4x}-4(36x-9)e^{4x}}{e^{8x}}\\[0.3cm] &=\frac{72-144x}{e^{4x}}\\ \end{align} $$

At {eq}x=0.25{/eq},

$$\begin{align} f"(x)&=13.24>0 \end{align} $$

As the second derivative is positive, {eq}x =0.25 {/eq} is a minimum point.

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