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A cone has a radius of 5 ft and a height of 15 ft. It is empty and is being filled with water at a constant rate of 15 15 ft 3 ft3 / sec sec . Find the rate of change of the radius of the surface of the water when the radius of the surface of the water is 2 2 ft. (You must also include the units)

User Firebush
by
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1 Answer

5 votes

Answer:


(dr)/(dt)=0.11062ft/sec

Explanation:

From the question we are told that:

Radius
r=5ft

Height
H=15ft

Rate
R=15ft/3sec =5ft/s

Surface Radius
R_(surf)=2.2ft

Generally the equation for Volume is mathematically given by


V=(1)/(3)\pi*r^2h

Since radius to height ratio gives


(r)/(h)=(5)/(15)


(r)/(h)=(1)/(3)


h=3r

Therefore


V=(1)/(3)\pi*r^2(3r)


V=\pi r^3

Generally the equation for Change of Volume is mathematically given by


(dv)/(dt)=\pi (d)/(dt)(r^3)


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


(dv)/(dt)=\pi 3*(2.2)^2 (dr)/(dt)


(dr)/(dt)=(5)/(45.62)


(dr)/(dt)=0.11062ft/sec

User Mixja
by
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