# 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.

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.