Water is leaking out of an inverted conical tank at a rate of 9300 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height13 meters and the diameter at the top is 7meters. If the water level is rising at a rate of 21 centimeters per minute when the height of the water is 4meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
Recall that the volume of a right circular cone with height h and radius of the base r is given by V=1/3πr^2h
using similar triangles, note that if the water depth is h, then the radius of the surface of the water is 7/26 h
So, now we have
v = π/3 (7h/26)^2 * h = 49π/2028 h^3
so that means that
dv/dt = 49π/676 h^2 dh/dt - 9300 = 49π/676 * 400^2 * 21 - 9300 = 755835 cm^3/min
check my math
This looks wrong
To solve this problem, we need to find the rate at which water is being pumped into the tank. We are given the rate at which water is leaking out of the tank and the rate at which the water level is rising.
Let's first find the rate at which the volume of water in the tank is changing with respect to time. This can be done using the formula for the volume of a cone:
V = (1/3)πr^2h
Differentiating both sides of the equation with respect to time t, we get:
dV/dt = (1/3)π(2rhdr/dt + r^2dh/dt)
Here, r is the radius of the cone, h is the height of the water level, dr/dt is the rate of change of radius, and dh/dt is the rate of change of height.
Given:
dh/dt = 21 cm/min (rate at which the water level is rising)
h = 4 m (height of the water)
Now, we need to find the rate of change of radius dr/dt. To do this, we can use similar triangles.
The large triangle formed by the height of the tank, the radius of the tank, and the radius of the water is similar to the small triangle formed by the change in height, the change in radius, and the radius of the water.
Using the similar triangles, we can write:
dr/dh = (R - r)/h
Where R is the radius of the tank and r is the radius of the water.
Substituting r = (h/13)R (since the water level is 4m and the tank height is 13m), we get:
dr/dh = ((13 - h)/13)R/h
Now, we have expressions for dh/dt and dr/dt. Substituting these into the equation for dV/dt, we get:
dV/dt = (1/3)π[(2rh((13 - h)/13)R/h) + (h/13)^2(21)]
Simplifying this equation, we get:
dV/dt = (1/3)π[(2R(13 - h)) + (h/13)^2(21)]
Finally, substituting the given values of R = 7/2 m and h = 4 m into the equation, we can find the rate at which water is being pumped into the tank.
dV/dt = (1/3)π[(2(7/2)(13 - 4)) + (4/13)^2(21)]
dV/dt = (1/3)π[(7(13 - 4)) + (16/169)(21)]
dV/dt = (1/3)π[(7(9)) + (336/169)]
dV/dt = (1/3)π[(63) + (336/169)]
dV/dt = (1/3)π[(423/169)+(336/169)]
dV/dt = (1/3)π(759/169)
dV/dt approx. 14.24π cubic meters per minute
Therefore, the rate at which water is being pumped into the tank is approximately 14.24π cubic meters per minute.