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Water is poured into the top half of a spherical tank at a constant rate. If W(t) is the rate of increase of the depth of the water, then W is:

User Jcmitch
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Answer with explanation:

Let V be the volume of the tank.

Water is being poured into the spherical tank at a constant rate.

Let r, be the radius of spherical tank.


(dr)/(dt)=\text{Constant}=r

As, radius is constant. --------------------------------(1)

Volume of tank will increase with time.

So,⇒ Height of sphere=Radius of sphere

Volume of Spherical tank


V=(4\pi r^3)/(3)\\\\(dv)/(dt)=4\pi r^2(dr)/(dt)\\\\=4\pi r^2* r

-----------------[using (1)]

So,


W(t)=4\pi r^3

User Chmanie
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