A tank has the shape of an inverted circular cone with a base radius of 5 meters
and a height of 20 meters. If water is being pumped into the tank at 2 cubic meters
per minute, find the rate at which the water level is rising when the water is 7
meters deep
WELL I HAVE THIS
v=1/3pi*r^2h
v=1/3pi*r2(4r)
v=4/3pi*r^3
dv/dt=4pi*r^2
2=4pi(7/4)^2
2=49pi/4 * dr/dt
dr/dt= 8/49pi = 0.052 meters per minute
but steve got
"I get dV/dt = pi/16 h^2 dh/dt
giving
2 = 49pi/16 dh/dt
or
dh/dt= 32/49pi
"
and reiny got
V = (1/3)πr^2 h = (1/3)π(h^2/16)(h)
= (1/48)π h^3
dV/dt = (1/16)π h^2 dh/dt
so when dV/dt = 2, h = 7
2 = (1/16)π(49) dh/dt
2(16)/(49π) = dh/dt
= 32/(49π)
= appr. .208 m/min
I answered in your other post on this
You have found the rate at which the radius is changing.
Unfortunately, your question asked for how fast the height is changing.
Did you notice that Steve and I both had the same answer?
It seems like there are multiple approaches and answers provided by different people. Let's break down the problem and go through each solution.
Given:
Base radius of the tank, r = 5 meters
Height of the tank, h = 20 meters
Rate of water being pumped into the tank, dV/dt = 2 cubic meters per minute
The goal is to find the rate at which the water level is rising when the water is 7 meters deep.
Solution 1:
Using the formula for the volume of a cone, V = (1/3) * π * r^2 * h
Differentiate both sides with respect to time (t):
dV/dt = (1/3) * π * 2r * dr/dt * h + (1/3) * π * r^2 * dh/dt
Substituting the known values:
2 = (1/3) * π * 2(5) * dr/dt * 7 + (1/3) * π * (5^2) * dh/dt
2 = 70π * dr/dt + 25π * dh/dt
To find dh/dt, we need to express dr/dt in terms of dh/dt. From the given information, we know that when the water is 7 meters deep, which is equivalent to (7/20) of the full height, the rate of change of radius is 8/(49π) meters per minute.
So, substituting dr/dt = 8/(49π) into the equation:
2 = 70π * (8/(49π)) + 25π * dh/dt
2 = (560/49) + 25π * dh/dt
25π * dh/dt = 2 - (560/49)
dh/dt = (2 - (560/49))/(25π)
dh/dt = 32/(49π)
The result is approximately 0.208 meters per minute.
Solution 2:
Using the same formula for the volume of the cone, V = (1/3) * π * r^2 * h
Rearranging the equation to isolate h, we get h = (3V)/(πr^2)
Differentiate both sides with respect to time (t):
dh/dt = (3/dt) * [(dV/dt)r^2 - V(2r(dr/dt))]
Substituting the known values:
dh/dt = (3/dt) * [(2)(5^2) * (dr/dt) - (V)(2r)(dr/dt)]
dh/dt = (3/dt) * [(50)(dr/dt) - (V)(10)(dr/dt)]
From the given information, we know that when the water is 7 meters deep, which is equivalent to (7/20) of the full height, V = (1/3) * π * (5^2) * 7 = (49/3)π.
Substituting the values into the equation:
dh/dt = (3/dt) * [(50)(dr/dt) - (49/3)π * (10)(dr/dt)]
dh/dt = (9/16)π * [(50 - (49/3)(10))] * dr/dt
dh/dt = (9/16)π * [(500 - (490/3))] * dr/dt
dh/dt = (9/16)π * [(500 - (1633/3))]
dh/dt = (9/16)π * (137/3)
dh/dt = 32/(49π)
The result is approximately 0.208 meters per minute.
Both solutions yield the same rate of the water level rising, approximately 0.208 meters per minute. It is important to note that the different approaches might seem different, but they ultimately provide the same answer.