Hard Calculus Question?

[匿名]

2017-03-29 2:52 pm

Hello, please help me with this difficult calculus problem:

Water is being removed from the vertex at the bottom of a conical tank at the rate of 32^pi cubic feet per minute. The diameter at the base of the tank is 30 feet and the height of the tank is 50 feet.
a) Determine an expression for the volume of the water in the tank in terms of the radius at the surface of the water.
b) At what rate is the radius of the water in the tank changing when the radius is 12 feet?
c) At what rate is the height of the water in the tank changing at the instant that the radius is 12 feet?

回答 (1)

wanszeto

2017-03-29 3:17 pm

✔ 最佳答案

a)
V ft³ : volume of the water
r ft : radius of the water
h ft : height of the water

The ratio of r : h is constant i.e.
r : h = 30 : 50
30h = 50r
h = (5/3)r

V = (1/3)πr²h
V = (1/3)πr² * (5/3)r
V = (5/9)πr³

b)
V = (5/9)πr³
dV/dt = d[(5/9)πr³]/dt
dV/dt = (5/9)π * (3r²)(dr/dt)
dV/dt = (5/3)πr²(dr/dt)

When dV/dt = 32 ft³/min and r = 12 ft :
32 = (5/3)π(12)(dr/dt)
dr/dt = 32*3/(5π*12)
dr/dt = 1.6/π ≈ 0.51 (to 2 sig. fig.)

The rate of change of the radius of the water when the radius of water is 12 ft = 1.6/π ft/mim ≈ 0.51 ft/min

c)
From a), h = (5/3)r
dh/dt = (5/3)(dr/dt)

When dr/dt = 1.6/π :
dh/dt = (5/3)*(1.6/π)
dh/dt = 8/(3π) ≈ 0.85 (to 2 sig. fig.)

The rate of change of the height of the water when the radius of water is 12 ft = 8/(3π) ft/mim ≈ 0.85 ft/min

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