A hemispherical container with a radius of 2m.
When the depth of the liquid is h metres, the volume of the liquid, V m^3 is given:
V= [ pih^2(6-h) ]/3

If the liquid is being pumped out at a constant rate of 0.25m^3/min, find the rate at which h is falling at the instant that h=1. (answer to nearest mm/minute)

Also, prove that the given formula for V to be correct.

Question

A hemispherical container with a radius of 2m.
When the depth of the liquid is h metres, the volume of the liquid, V m^3 is given:
V= [ pih^2(6-h) ]/3

If the liquid is being pumped out at a constant rate of 0.25m^3/min, find the rate at which h is falling at the instant that h=1. (answer to nearest mm/minute)

Also, prove that the given formula for V to be correct.

anonymous · Answer

Integral of rotation is a bit tedious, but doable

anonymous · Answer

Look I'll give you an outline @Kystal :
You have a function$$V(h) = K* (6 h^2 -h^3)]$$

anonymous · Answer

this gives you the derivative

anonymous · Answer

$$\frac{dV}{dh}$$

anonymous · Answer

Now using the "inverse function derivative" theorem compute$$\frac{ dh }{ dV }$$

anonymous · Answer

Now the rate of flow is in fact the rate of change of VOLUME$$\frac{ dV }{ dt }$$

anonymous · Answer

erm... why is it K∗(6h2−h3)]? where the K comes from?

anonymous · Answer

Multiply and use the chain rule$$\frac{dh}{dV}*\frac{dV}{dt} = \frac{dh}{dt}$$

anonymous · Answer

$$K = \frac{\pi}{6}$$

anonymous · Answer

sorry pi/3

anonymous · Answer

This is a 35 minutes (at min) work

anonymous · Answer

That counts the proof of volume formula

anonymous · Answer

Is there a picture or diagram with this?

anonymous · Answer

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