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?

1 Answer
Mar 8, 2018

0.112253 ft /sec. [to 6 decimal places]

Explanation:

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#.