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Question

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Water is pouring into an inverted cone at the rate of 6 cubic meters per minute. If the height of the cone is 10 meters and the radius of its base is 6 meters, how fast is the water level rising when the water is 3-meter deep.

Answer 1

Water is pouring into an inverted cone at the rate of 6 m^3/min =the derivative of Volume = V' = dV/dt.

height of the cone = h = 10 m, radius of base = r = 6 m,

how fast is the water level rising when water is 3 meters deep? h=3

Volume of water in the cone = V = (1/3)Ah = pi(r)^2(h)/3

where A=Area of the base = pi(r)^2

and h= height of the water level

when r=6, h=10

r =6h/10 = .6h

V = (pi/3)(.6h)^2(h) = pi/3(.6)^2(h)^3

V = .12pi(h^3)

take the derivative with respect to time

V' = .36pi(h^2)h'

plug in 3 for h

V'(3) = .36pi(9)h' = 6

h' = 6/3.24pi = 1/.54pi = 100/54pi = 50/27pi = meters per minute

h'= about 0.59 meters per minute = rate of rising water level when height of water = 3 meters

h' = about 0.6 m/min

Answer 2

The RATE of the incoming water is dV/dt (so dV/dt = 6 m³/min)

Volume of a cone is calculated from V = 1/3πr²h

We are given information about the cone however that helps us relate radius (r) and height (h). Looking at this cross section:

We can say r/h = 6/10 and cross multiplying yields 10r = 6h or r = 3/5h

So we can substitute the expression "3/5h" into our volume equation in place of "r":

V = 1/3πr²h

V = 1/3π(3/5h)²(h)

V = 3/25πh³

Taking the derivative with respect to time (using the chain rule):

Plugging in dV/dt = 6, h = 3 we get:

6 = 9/25π(3²)(dh/dt)

6 = 81/25π(dh/dt)

6 = 10.178(dh/dt)

dh/dt = 6/10.178

dh/dt = 0.59 meters/minute

Answer 3

Start with V= πr²h/3 use similar triangles to find a relationship between r and h such that h/r= 10/6 so r= 3h/5. Use this in the volume formula so that V= π(9h²/25)h/3 so V=π3h³/25 now dV/dt = 9πh²dh/25dt. using the information in the problem 6=9π9dh/25dt solving for dh/dt = 150/81π m/sec.