A conical tank has a circular base with radius 5 ft and height 12 ft. If water is flowing out of the tank at a rate of 3 ft^3/min, how fast is the height of the water changing when the height is 7 ft?

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Answer 1 · 2018-03-08T01:15:09

#0.112253 #ft /sec. [to #6# decimal places]

We are given that #dV#/dt#=3#...........#[1]# [ the rate of change of volume with respect to time, t.

Assuming the cone is uniform with straight sides, then although #r and h# are variables their ratio will remain constant,i.e.

#r/5=h/12#, so #r=[5h]/12#.............#[2]#

Volume of such a cone is #V=1/3pir^2h#.........#[3]#, Substituting #r #from ......#[2],# in......... #[3]# we obtain,

Volume of cone #=pi/3[[5h]/12]^2h,# = #[25pih^3]/[3[144]]#.......#[4]#

differentiating #[4]# with respect to time #[t]# implicitly,

#dV/dt=[25pih^3]/[3[144]##[dh/dt][h^3]#=#[[75pih^2]/[3[144]]]dh/dt#, since #d/dt[h^3]=3h^2#.

We know #dV/dt=3#, So #3=[75pih^2]/[3[144]##dh/dt#, and when #h=7#

#3=[[75]49pi/[3[144]]dh/dt# and so #dh/dt=9[144]/[[75][49pi]#

#dh/dt=0.112253# ft /sec#.