SOLUTION: Sand is being poured at a rate of 0.50 m^3/min onto a conical pile whose radius is always equal to 2/3 of its height. At one point in time, an observer notices that the height of t

Question 1192719: Sand is being poured at a rate of 0.50 m^3/min onto a conical pile whose radius is always equal to 2/3 of its height. At one point in time, an observer notices that the height of the sand pile is increasing at a rate of 8.0 cm/min. What was the height of the pile at this instant?
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

r = radius in meters
h = height in meters
t = time in minutes

dr/dt = change in radius over change in time, ie speed at which the radius changes
dh/dt = change in height over change in time, ie speed at which the height changes
Each speed is measured in meters per minute.

100 cm = 1 m
8.0 cm/min = (8.0 cm/min)*(1 m/100 cm) = 0.08 meters per minute

The sentence
"height of the sand pile is increasing at a rate of 8.0 cm/min"
is equivalent to
"height of the sand pile is increasing at a rate of 0.08 m/min".
This is the dh/dt value for this snapshot in time.

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r = (2/3)h since the radius is 2/3 the height
That rearranges into
3r = 2h
h = 3r/2
h = 1.5r

Apply the derivative with respect to t to both sides
h = (3/2)r
dh/dt = (3/2)*dr/dt
dh/dt = (3/2)*dr/dt

Plug in dh/dt = 0.08 mentioned earlier and solve for dr/dt
dh/dt = (3/2)*dr/dt
dr/dt = (2/3)*dh/dt
dr/dt = (2/3)*0.08
dr/dt = (2/3)*(8/100)
dr/dt = 16/300
dr/dt = 4/75

------------------------------------------------------

We have this so far:
h = 3r/2 = 1.5r
dr/dt = 4/75

Recall the volume of a cone with radius r and height h is
V = (1/3)*pi*r^2*h

Subsitution allows us to say
V = (1/3)*pi*r^2*h
V = (1/3)*pi*r^2*(3r/2)
V = (1/3)*(3/2)*pi*r^3
V = (1/2)*pi*r^3

Apply the derivative with respect to t
V = (1/2)*pi*r^3
dV/dt = d/dt[ (1/2)*pi*r^3 ]
dV/dt = (1/2)*pi*d/dt[ r^3 ]
dV/dt = (1/2)*pi*(3r^2 * dr/dt)
Don't forget about the chain rule.
The radius r is a function of t.

We'll plug in the rate of change of the volume dV/dt = 0.50 m^3/min along with the rate of change in the radius dr/dt = 4/75
Then we isolate r.
dV/dt = (1/2)*pi*(3r^2 * dr/dt)
0.50 = (1/2)*pi*(3r^2 * 4/75)
0.50 = (2/25)*pi*r^2
(2/25)*pi*r^2 = 0.50
pi*r^2 = 0.50*(25/2)
pi*r^2 = 6.25
r^2 = (6.25)/pi
r = sqrt(6.25/pi)

This leads to
h = (3/2)r
h = (3/2)sqrt(6.25/pi)
which is the exact height in meters.

This approximates to (3/2)sqrt(6.25/pi) = 2.1157109 meters

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