SOLUTION 6: Draw a right triangle with dimensions $x, y,$ and $13$, and assume that $x$ and $y$ are functions of time $t$.

GIVEN: $\ \ \ \displaystyle{ dx \over dt } = -2 \ in/min.$ (It is negative since $x$ is decreasing.)

FIND: $\ \ \ \displaystyle{ dy \over dt }$ when $\ \ \ a.) \ x = 5 \ ft.$ $\ \ \ b.) \ x = 1 \ ft.$

Using the Pythagorean Theorem we get that $$x^2+y^2=13^2=169$$

Now differentiate this length equation with respect to time $t$, getting

$$D \{ x^2+y^2 \} = D \{ 169 \} \ \ \ \longrightarrow$$ $$2 x \displaystyle{ dx \over dt } + 2y \displaystyle{ dy \over dt } = 0 \ \ \ \longrightarrow$$ $$(**) \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \displaystyle{ dx \over dt } + y \displaystyle{ dy \over dt } = 0 \ \ \ \longrightarrow$$

$\Big($ a.) Now let $\displaystyle{ dx \over dt } = -2$ and $x =5$ in equation (**). The Pythagorean Theorem equation becomes $\ 5^2+y^2=169 \ \ \rightarrow \ y^2=144 \ \ \rightarrow \ y=12$ in equation (**). $\Big)$

$$(5)(-2)+ (12) \displaystyle{ dy \over dt } = 0 \ \ \ \longrightarrow$$ $$12 \displaystyle{ dy \over dt } = 10 \ \ \ \longrightarrow$$ $$\displaystyle{ dy \over dt } = {10 \over 12} \ \ \ \longrightarrow$$ $$\displaystyle{ dy \over dt } = { 5 \over 6} \ ft/sec. \approx 0.833 \ ft/sec.$$

$\Big($ b.) Now let $\displaystyle{ dx \over dt } = -2$ and $x =1$ in equation (**). The Pythagorean Theorem equation becomes $\ 1^2+y^2=169 \ \ \rightarrow \ y^2=168 \ \ \rightarrow \ y=\sqrt{168}$ in equation (**). $\Big)$

$$(1)(-2)+ (\sqrt{168}) \displaystyle{ dy \over dt } = 0 \ \ \ \longrightarrow$$ $$\sqrt{168} \displaystyle{ dy \over dt } = 2 \ \ \ \longrightarrow$$ $$\displaystyle{ dy \over dt } = { 2 \over \sqrt{168}} \ ft/sec. \approx 0.154 \ ft/sec.$$