물통 수면 높이 변화율

높이가 $h$일 때 수면의 반지름을 $r$, 부피를 $V$라 하자. 그림에서 $50 : 30 = ( 100 + x ) : x$ $\therefore x = 150$ $r : ( h + 150 ) = 50 : 250$ $\displaystyle \therefore r = \frac{1}{5} ( h + 150 )$ $\displaystyle V = \frac{1}{3} \pi \cdot \frac{1}{25} ( h + 150 ) ^{3} - \frac{1}{3} \pi \cdot 30 ^{2} \cdot 150$ $\displaystyle \frac{dV}{dt} = \frac{1}{25} \pi \left( h + 150 \right) ^{2} \cdot \frac{dh}{dt}$ $\displaystyle - 4 \sqrt{h} = \frac{1}{25} \pi \left( h + 150 \right) ^{2} \cdot \frac{dh}{dt}$ $\displaystyle \therefore \frac{dh}{dt} = \frac{- 100 \sqrt{h}}{\pi \left( h + 150 \right) ^{2}}$ $\displaystyle \therefore \left( \frac{dh}{dt} \right) _{h = 50} = \frac{- 100 \sqrt{50}}{\pi \cdot 200 ^{2}} = - \frac{5 \sqrt{2}}{4 \pi} \times 10 ^{- 2}$ [별해]

$30 : 150 = r : ( 150 + h )$ $\displaystyle \therefore r = 30 + \frac{h}{5}$ $\displaystyle V = \frac{\pi}{3} \left\{ \left( 30 + \frac{h}{5} \right) ^{2} ( h + 150 ) - 30 ^{2} \times 150 \right\}$ $\displaystyle \frac{dV}{dt} = \frac{\pi}{3} \left\{ 2 \left( 30 + \frac{h}{5} \right) \times \frac{1}{5} ( h + 150 ) \right.$$\displaystyle + \left( 30 + \frac{h}{5} \right) ^{2} \} \frac{dh}{dt}$ $h = 50$일 때, $\displaystyle - 4 \sqrt{50} = \frac{\pi}{3} \times 4800 \times \frac{dh}{dt}$ $\displaystyle \therefore \frac{dh}{dt} = - \frac{1}{\pi} \times \frac{4 \times 5 \times \sqrt{2}}{1600}$$\displaystyle = - \frac{5 \sqrt{2}}{4 \pi} \times 10 ^{-2}$