A volume of a cone is given by #V=(pir^2h)/3#. A particular cone has a height equal to five times the radius of its base: #h=5r#....? (continued below)

Question

(The volume of the cone may be expressed in terms of height as V=(pih^3)/75#). The cone is now inverted and is filled with water at the rate of 20 cubic centimetres per second. Determine the expression for the rate of change of the height of water.

I have to find #(dh)\/dt# but don't know where to start.

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Answer 1 · 2018-04-26T18:10:15

# (dh)/dt=500/(pih^2)" (cm)/sec"#.

#V=1/3pir^2h, and, h=5r. :. r=h/5 #.

#:. V=1/3pi(h/5)^2h=pi/75h^3#.

#:. d/dt(V)=d/dt(pi/75h^3)#.

#:. (dV)/dt=pi/75*d/dt(h^3)#,

#=pi/75*d/(dh)(h^3)*(dh)/dt.........[because," the Chain Rule]"#.

#:. (dV)/dt=(pih^2)/25*(dh)/dt#.

# rArr (dh)/dt=25/(pih^2)*(dV)/dt#,

#=25/(pih^2)*20.......[because, (dV)/dt=20"(cc)/sec]"#.

#:. (dh)/dt=500/(pih^2)" (cm)/sec"#.