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
is the answer 8/49pi= 0.052 meters per minute
I get dV/dt = pi/16 h^2 dh/dt
giving
2 = 49pi/16 dh/dt
or
dh/dt= 32/49pi
One of us has a factor of 2 out of place. Could be me ...
I did not get that
Let the height of the water be h m and the radius of the water level be r m
by ratios:
r/h = 5/20 = 1/4
r = h/4
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
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
You have found the rate at which the rate is changing.
Unfortunately, your question asked for how fast the height is changing.
should say:
You have found the rate at which the radius is changing.
Unfortunately, your question asked for how fast the height is changing.
To find the rate at which the water level is rising, we need to use the concept of related rates. Here's how we can solve it step by step:
1. First, let's visualize the problem. The tank is in the shape of an inverted circular cone, which means it is narrow at the top and wide at the base. The radius of the base is 5 meters, and the height is 20 meters.
2. Let's denote the radius of the water's surface as r (which changes with time) and the height of the water as h (which also changes with time).
3. We need to find the rate at which the water level is rising, i.e., dh/dt (the derivative of h with respect to time t), when the water is 7 meters deep.
4. The key idea is to relate the height of water to the radius of the water's surface using similar triangles.
Consider the triangle formed by the height of the cone (20 meters), the radius of the base (5 meters), and the height of the water (h meters). This triangle is similar to the triangle formed by the height of the cone (20 meters), the radius of the water's surface (r meters), and the height of the water (h meters).
Using the concept of similar triangles, we can write the following proportion:
(r - 5) / 20 = r / h
5. Rearranging the proportion, we get:
r = (h/20) * (r - 5)
6. We have the expression for the radius of the water's surface in terms of h.
7. Now, differentiate both sides of the equation with respect to time t:
dr/dt = [(1/20) * (r - 5)] * dh/dt
8. We want to find dh/dt when h = 7 meters.
9. Plug in the known values into the equation: r = 5 meters (given), h = 7 meters (given).
10. Using the given rate of water input, which is 2 cubic meters per minute, we can say that dh/dt = 2 m^3/min.
11. Substitute the known values into the equation:
dr/dt = [(1/20) * (5 - 5)] * (2)
dr/dt = 0
Therefore, the rate at which the water level is rising when the water is 7 meters deep is 0. This means that when the water is at a height of 7 meters, the water level is not changing or rising.