Aspire Faculty ID #16421 · Topic: NIMCET 2009 · Just now
NIMCET 2009

Water runs into a conical tank of radius $5$ ft and height $10$ ft at a constant rate of $2\text{ ft}^3/\text{min}$. How fast is the water level rising when the water is $6$ ft deep?

Solution

Cone volume: $V = \dfrac{1}{3}\pi r^2 h$ 
 Similarity: $r = \dfrac{h}{2}$ 
 So, $V = \dfrac{\pi}{12}h^3$ 
 Differentiate: $\dfrac{dV}{dt} = \dfrac{\pi}{4}h^2 \dfrac{dh}{dt}$ 
 Put values: $2 = \dfrac{\pi}{4}(36)\dfrac{dh}{dt}$ $2 = 9\pi\dfrac{dh}{dt}$ $\displaystyle \dfrac{dh}{dt} = \dfrac{2}{9\pi}$

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