A conical water tank with vertex down has a radius of 10 feet at the top and is 29 feet high. If water flows into the tank at a rate of 10 , how fast is the depth of the water increasing when the water is 17 feet deep?
I will be happy to critique your work.
relate volume to height
then
dv/dt= f'(h,t)
solve for dh/dt
To find how fast the depth of the water is increasing, we need to use related rates and the concept of similar triangles.
Let's denote the radius of the water at any given depth as r and the height of the water as h. We are given that the tank is conical with the vertex down, so we can imagine the tank as an upside-down cone.
First, let's find an expression for the radius r in terms of the height h. Since the tank is conical, we can use similar triangles to relate the radius at the top of the tank (10 feet) to the height of the tank (29 feet):
(10 feet) / (29 feet) = r / h
Now, we can solve this equation for r:
r = (10 feet * h) / (29 feet)
Next, we differentiate both sides of this equation with respect to time (t) to find how the radius is changing with respect to time:
dr/dt = (10 feet * dh/dt) / (29 feet)
The left side represents the rate at which the radius is changing, and the right side represents the rate at which the height is changing.
Given that the water is flowing into the tank at a rate of 10 ft^3/min, we can use this information to find how the height of the water is changing with respect to time. Since the tank is conical, the volume of water in the tank can be expressed as:
V = (1/3) * π * r^2 * h
Differentiating both sides of this equation with respect to time (t):
dV/dt = (1/3) * π * [2r * dr/dt * h + r^2 * dh/dt]
Substituting the relationship between r and h that we found earlier:
dV/dt = (1/3) * π * [2(10 feet * h) / (29 feet) * (10 feet * dh/dt) / (29 feet) * h + (10 feet * h)^2 / (29 feet)^2 * dh/dt]
Given that dV/dt = 10 ft^3/min and h = 17 feet, we can solve this equation for dh/dt, which represents the rate at which the height is changing:
10 ft^3/min = (1/3) * π * [2(10 feet * 17 feet) / (29 feet) * (10 feet * dh/dt) / (29 feet) * 17 feet + (10 feet * 17 feet)^2 / (29 feet)^2 * dh/dt]
Simplifying and solving for dh/dt:
dh/dt = (3 * 10 ft^3/min * (29 feet)^2) / [2 * π * (10 feet * 17 feet)^2 - 2 * (10 feet * 17 feet) * (10 feet)^2]
Evaluating this expression will give us the rate at which the depth of the water is increasing when the water is 17 feet deep.