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Suppose water was being drained at from a conical tank. The radius is decreasing at a rate of 2 inch per minute and the height is decreasing at 4 inches per minute. Find the change in the volume when the height is 5 inches and the radius is 6 inches.

Sagot :

The change in volume is -88π

Volume?

The volume is nothing but the amount of space occupied by a three-dimensional figure as measured in cubic units.

What is volume change?

Volume Change is defined as the variation in the existing volume of the service type which is provided by the Supplier under the Agreement

Given:

The radius is decreasing at 2 inches per minute

The height is given as decreasing at 4 inches per minute

So let [tex]\frac{dr}{dt}=-2[/tex] and

[tex]\frac{dh}{dt}=-4[/tex]

and we have radius r=6,

                     height h=5

The volume of the cone is represented by V as V=[tex]\frac{1}{3}\pi r^{2}h[/tex]

Then

[tex]\frac{dv}{dt}=\frac{\pi }{3}[r^{2}\frac{dh}{dt} +h*2r\frac{dr}{dt}][/tex]

[tex]=\frac{\pi }{3}[6^{2}(-4)+5*2(6)(-2)][/tex]

[tex]=\frac{\pi }{3}[36(-4)+5(12)(-2)]\\ =\frac{\pi }{3}[-144+(-120)]\\ =\frac{-264}{3}\pi \\ =-88\pi[/tex]

So the volume change is [tex]-88\pi[/tex]

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