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Two boys start walking from the same point. One walks south at a rate 4 $\mathrm{m} / \mathrm{sec}$ . And the other walks
west at rate 3 $\mathrm{m} / \mathrm{sec}$ . At what rate is distance between the two boys increasing 2 seconds later.

2 seconds $\rightarrow$

$x=-6 \quad, y=-8$

$\frac{d x}{d t}=-3 \quad \frac{d y}{d t}=-4, S=5$

$s^{2}=x^{2}+y^{2}=(9)+(16) \rightarrow s=\sqrt{25}=5$

$S^{2}=x^{2}+y^{2} \longrightarrow 2 s \frac{d s}{d t}=2 x \frac{d x}{d t}+2 y \frac{d y}{d t}$

$S=5(2)=10$

$2(10) \frac{d s}{d t}=2(-6) \frac{d x}{d t}+2(-8) \frac{d y}{d t}$

$\rightarrow 20 \frac{d s}{d t}=2(-6)(-3)+2(-8)(-4)$

$d s / d t=5 m / s e c$

A snow ball is melting at a rate of 0.03 $\mathrm{f} t^{3} / \mathrm{h} \mathrm{r}$ . At what rate is the surface
area changing when the volume of the ball is 36$\pi ?$

$\frac{d v}{d t}=-0.03$

$V=\frac{4}{3} \pi r^{3}$

$36 \pi=\frac{4}{3} \pi r^{3} \rightarrow r^{3}=\frac{(36)(3)}{4}=27$

$\sqrt[3]{r^{3}}=\frac{(36) \cdot(3)}{4}=\sqrt[3]{27} \rightarrow r=3$

$\frac{d v}{d t}=4 \pi r^{2} \frac{d r}{d t} \longrightarrow-0.03=4 \pi(9) \frac{d r}{d t}$

$\rightarrow \frac{d r}{d t}=\frac{-0.03}{4 \pi(9)}=\frac{-1}{1200 \pi}$

$S=4 \pi r^{2} \longrightarrow \frac{d s}{d t}=8 \pi r \frac{d r}{d t}$

$\longrightarrow \frac{d s}{d t}=8 \pi(3) \cdot \frac{-1}{1200 \pi}=\frac{-1}{50} f t^{2} / \mathrm{hr}$