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(C) Let $h$ be the depth of water,$r$ be the radius of the water surface,and $	heta$ be the semi-vertical angle. Given $	an 	heta = \frac{r}{h} = \

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Let $h$ be the depth of water,$r$ be the radius of the water surface,and $	heta$ be the semi-vertical angle. Given $	an 	heta = \frac{r}{h} = \frac{3}{4}$,so $r = \frac{3}{4}h$.The volume of water is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{3}{4}hight)^2 h = \frac{3 \pi}{16} h^3$.Differentiating with respect to time $t$,we get $\frac{dV}{dt} = \frac{3 \pi}{16} \cdot 3h^2 \frac{dh}{dt} = \frac{9 \pi}{16} h^2 \frac{dh}{dt}$.Given $\frac{dV}{dt} = 6 	ext{ m}^3/	ext{hr}$. At $h = 4 	ext{ m}$,$6 = \frac{9 \pi}{16} (4)^2 \frac{dh}{dt} = 9 \pi \frac{dh}{dt}$,so $\frac{dh}{dt} = \frac{6}{9 \pi} = \frac{2}{3 \pi} 	ext{ m/hr}$.The slant height is $\ell = \sqrt{r^2 + h^2} = \sqrt{(\frac{3}{4}h)^2 + h^2} = \sqrt{\frac{9}{16}h^2 + h^2} = \sqrt{\frac{25}{16}h^2} = \frac{5}{4}h$.The curved surface area is $S = \pi r \ell = \pi (\frac{3}{4}h) (\frac{5}{4}h) = \frac{15 \pi}{16} h^2$.Differentiating with respect to $t$,$\frac{dS}{dt} = \frac{15 \pi}{16} \cdot 2h \frac{dh}{dt} = \frac{15 \pi}{8} h \frac{dh}{dt}$.Substituting $h = 4$ and $\frac{dh}{dt} = \frac{2}{3 \pi}$,we get $\frac{dS}{dt} = \frac{15 \pi}{8} (4) (\frac{2}{3 \pi}) = \frac{15 \pi}{8} \cdot \frac{8}{3 \pi} = 5 	ext{ m}^2/	ext{hr}$.

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