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Miscellaneous Exercise 2(II) · Q113

Q.A water tank in the form of an inverted cone is being emptied at the rate of 2 cubic feet per second. The height of the cone is 8 feet and the radius is 4 feet. Find the rate of change of the water level when the depth is 6 feet.

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Since the full cone has height 88 ft and radius 44 ft, at any water depth hh the water surface radius rr satisfies (by similar triangles) rh=48=12⇒r=h2\dfrac{r}{h}=\dfrac{4}{8}=\dfrac12 \Rightarrow r=\dfrac{h}{2}.

V=13πr2h=13π(h24)h=πh312V=\dfrac13\pi r^2h=\dfrac13\pi\left(\dfrac{h^2}{4}\right)h=\dfrac{\pi h^3}{12}.

dVdt=π12(3h2)dhdt=πh24dhdt\dfrac{dV}{dt}=\dfrac{\pi}{12}(3h^2)\dfrac{dh}{dt}=\dfrac{\pi h^2}{4}\dfrac{dh}{dt}. …

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