A conical water tank with vertex down has a radius of 12 feet at the top and is 23 feet high. If water flows into the tank at a rate of 20 {\rm ft}^3{\rm /min}, how fast is the depth of the water increasing when the water is 12 feet deep?
Let the radius of the water level be r
let the height of the water level be h
by ratios,
r/h = 12/23
12h = 23r
r = 12h/23
V = (1/3)πr^2 h
= (1/3)π(144h^2/529)h
= (48π/529) h^3
dV/dt = (144π/529) h^2 dh/dt
for the given data:
20 = (144π/529)(144) dh/dt
I will leave it up to you to do the buttonpushing to find dy/dt
To find the rate at which the depth of the water is increasing, we need to first understand the relationship between the variables involved: the radius, height, and volume of the water in the tank.
We know that the shape of the water tank is a cone and the vertex is pointing down. This means that the volume of the water in the tank can be determined by the formula for the volume of a cone: V = (1/3)πr^2h, where V is the volume, r is the radius, and h is the height.
However, we need to express the volume in terms of h to solve the problem. By using similar triangles, we can find the equation for the radius r in terms of the height h:
r / h = (12 / 23)
Simplifying this ratio, we get:
r = (12 / 23)h
Substituting this expression for r in the formula for volume, we have:
V = (1/3)π[(12 / 23)h]^2h
Now, differentiate both sides of this equation with respect to time (t):
dV / dt = (1/3)π[(-2)(12 / 23)h](dh / dt)h + (1/3)π[(12 / 23)^2](dh / dt)h^2
Simplifying this equation further, we get:
dV / dt = (1/3)π(-24 / 23)h^2(dh / dt) + (1/3)π(144 / 529)(dh / dt)h^2
Now, we can substitute the given values into this equation to find the rate at which the depth of the water is increasing when the water is 12 feet deep.
Given:
dh / dt = 20 ft^3/min (Rate of water flowing into the tank)
We need to find the value of d(h) / dt when h = 12 ft.
Substitute these values into the equation:
20 = (1/3)π(-24 / 23)(12)^2(dh / dt) + (1/3)π(144 / 529)(dh / dt)(12)^2
Now, we can solve this equation to find the value of dh / dt.
20 = (-8π)(dh / dt) + (48π / 529)(dh / dt)(144)
Simplifying further, we have:
20 = (-8π)(dh / dt) + (6912π / 529)(dh / dt)
Combining similar terms, we get:
20 = (6904π / 529)(dh / dt)
Finally, solving for dh / dt, the rate at which the depth of the water is increasing, we have:
dh / dt = (20 * 529) / (6904π)
Evaluate this expression, or use a calculator, to find the numerical value of dh / dt.