A water dispenser in the shape of a cylinder with radius $10$ cm is filled with water. The water i

Question

A water dispenser in the shape of a cylinder with radius $10$ cm is filled with water. The water is dispensed at a steady rate into an empty cup in the form of an inverted cone of height $8$ cm and base radius $4$ cm. After $t$ seconds, the depth of water in the conical cup is $x$ cm, show that the volume of water in the cup is $\tfrac{\pi x^3}{12}$ cm$^3$.

If the depth of water in the cylinder decreases by $0.0025$ cms$^{-1}$, find the rate of change of volume of water in the conical cup, in terms of $\pi$. Hence, find, at the instant when the volume is $\frac{2}{3}\pi$ cm$^3$, the rate of increase of

the depth of the liquid in the conical cup,
the area of the horizontal surface of the liquid in the conical cup.

Accepted Answer

$V=\frac{\pi x^3}{12}$, $\frac{\mathrm dV}{\mathrm dt}=\frac{\pi}{4}\text{ cm}^3\text{ s}^{-1}$ (a) $\frac{\mathrm dx}{\mathrm dt}=\frac{1}{4}\text{ cm s}^{-1}$ (b) $\frac{\mathrm dA}{\mathrm dt}=\frac{\pi}{4}\text{ cm}^2\text{ s}^{-1}$

Connected Rate of Change involving Two Objects

Connected Rate of Change involving Two Objects

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