SOLUTION 19: Consider the given diagram where $s(t)$ feet is the height of the boulder above the ground after falling for $t$ seconds and $L$ is the distance between the running you and the falling boulder. Assume that $x, L,$ and $s(t)$ are all functions of time $t$.

GIVEN: $\ \ \ \displaystyle{ dx \over dt } = 10 \ ft/sec.$

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

Recall from elementary physics that the height $s(t)$ in feet of a falling object under the influence of only gravity is

$$s(t) = -16t^2+ v_{o} t + s_{o}$$

where $s_{o}$ is the initial height and $v_{o}$ is the initial velocity of the falling object. For this problem $s_{o}=200$ is the initial height and $v_{o}=0$ is the initial velocity, so that

$$s(t) = -16t^2 + 200$$

Using the Pythagorean Theorem, we now get that

$$L^2 = x^2+ \{ s(t) \}^2 \ \ \ \ \longrightarrow$$ $$L^2 = x^2+ ( -16t^2+200 )^2$$

Differentiating this distance equation, we get

$$D \{ L^2 \} = D \{ x^2 + ( -16t^2+200 )^2 \} \ \ \ \ \longrightarrow$$ $$2L \displaystyle{ dL \over dt } = 2x \displaystyle{ dx \over dt } + 2( -16t^2+200) \cdot (-32t) \ \ \ \ \longrightarrow$$

$\Big($Multiply both sides of the equation by $\displaystyle{ 1 \over 2 }.\Big)$

$$L \displaystyle{ dL \over dt } = x \displaystyle{ dx \over dt } -32t ( -16t^2+200) \ \ \ \ \longrightarrow$$ $$(**) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ L \displaystyle{ dL \over dt } = x \displaystyle{ dx \over dt } + 512t^3-6400t \ \ \ \ \longrightarrow$$

$\Big($ a.) Now let $\displaystyle{ dx \over dt } = 10$ and $\ t=1$ in equation (**). Since you initially start at $x=12$, the distance $x$ feet after $t$ seconds is $x=10t+12$, so that $x=10(1)+12=22$ in equation (**). Since $s(1)=-16(1)^2+200=184$, it follows from the distance equation that $\ L^2=22^2+184^2 \ \rightarrow L^2=34,340 \ \rightarrow L=\sqrt{34,340} \approx 185.31$ in equation (**). $\Big)$

$$( \sqrt{34,340}) \displaystyle{ dL \over dt } = (22)(10) + 512 (1)^3- 6400(1) \ \ \ \ \longrightarrow$$ $$\sqrt{34,340} \displaystyle{ dL \over dt } = -5668 \ \ \ \ \longrightarrow$$ $$\displaystyle{ dL \over dt } = { -5668 \over \sqrt{34,340} } \approx -30.586 \ ft/sec.$$

$\Big($ b.) Now let $\displaystyle{ dx \over dt } = 10$ and $\ t=3$ in equation (**). Since you initially start at $x=12$, the distance $x$ feet after $t$ seconds is $x=10t+12$, so that $x=10(3)+12=42$ in equation (**). Since $s(3)=-16(3)^2+200=56$, it follows from the distance equation that $\ L^2=42^2+56^2 \ \rightarrow \ L^2=4900 \ \rightarrow \ L=\sqrt{4900} = 70$ in equation (**). $\Big)$

$$(70) \displaystyle{ dL \over dt } = (42)(10) + 512 (3)^3 - 6400 (3) \ \ \ \ \longrightarrow$$ $$70 \displaystyle{ dL \over dt } = -4956 \ \ \ \ \longrightarrow$$ $$\displaystyle{ dL \over dt } = { -4956 \over 70 } = -70.8 \ ft/sec.$$