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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$$\pi ?$$

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

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

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

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

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

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

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

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

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