SOLUTION 18: Since $\displaystyle{ dx \over dt } = 1 \ mile/min.$ and the cars start at $x=0$, it follows that $x=t$ determines the $x$-value at time $t$ minutes for each car. Let $L$ be the distance between the cars after $t$ minutes and assume that $x$ and $L$ are both functions of time $t$ minutes.

GIVEN: $\ \ \ \displaystyle{ dx \over dt } = 1 \ mile/min.$

FIND: $\ \ \ \displaystyle{ dL \over dt }$ when $\ \$ a.) $\ t=1 \ min.$ $\ \$ b.) $\ t=3 \ min.$

The distance between the two cars after $t$ minutes is

$$L = e^x-(3x-2) \ \ \ \ \longrightarrow$$ $$L = e^x-3x+2$$

Differentiating this distance equation, we get

$$D \{ L \} = D \{ \displaystyle{ e^x-3x+2 } \} \ \ \ \ \longrightarrow$$

$\Big($Recall that $D \{ e^{f(t)} \} = e^{f(t)} \cdot f'(t).\Big)$

$$(**) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \displaystyle{ dL \over dt } = e^x \cdot \displaystyle{ dx \over dt } - 3 { dx \over dt } \ \ \ \ \longrightarrow$$

$\Big($a.) Now let $\displaystyle{ dx \over dt } = 1$ and $\ t=1$, so that $\ x=1$ in equation (**).$\Big)$

$$\displaystyle{ dL \over dt } = e^{1}(1) - 3 (1) \ \ \ \ \longrightarrow$$ $$\displaystyle{ dL \over dt } = e-3 \approx -0.282 \ miles/min.$$

$\Big($ b.) Now let $\displaystyle{ dx \over dt } = 1$ and $\ t=3$, so that $\ x=3$ in equation (**).$\Big)$

$$\displaystyle{ dL \over dt } = e^{3}(1) - 3 (1) \ \ \ \ \longrightarrow$$ $$\displaystyle{ dL \over dt } = e^3-3 \approx 17.086 \ miles/min.$$