Water pours out of a conical tank of height 10 feet and radius 4 feet at a rate of 10 cubic feet per minute. How fast is the water level changing when it is 5 feet high?
To find how fast the water level is changing, we need to find the rate at which the height of the water is changing with respect to time (dh/dt).
Let's consider a cross-section of the water in the conical tank, which forms a smaller cone. The volume V of a cone is given by the formula V = (1/3) * π * r^2 * h, where r is the radius and h is the height.
We know that the rate at which the water is pouring out of the tank is 10 cubic feet per minute, so the rate at which the volume is changing is dV/dt = 10. We need to find dh/dt when the height is 5 feet.
First, let's express the volume V in terms of h using similar triangles.
Since the height of the cone is 10 feet and the radius is 4 feet, we can write:
h / 10 = r / 4
Rearranging this equation, we get r = (4/10) * h = (2/5) * h.
Substituting this expression for r into the volume formula, we have:
V = (1/3) * π * ((2/5) * h)^2 * h
V simplifies to V = (4/75) * π * h^3.
Next, let's differentiate V with respect to time t:
dV/dt = (4/75) * π * 3 * h^2 * dh/dt
Given dV/dt = 10, we can solve for dh/dt:
10 = (4/75) * π * 3 * h^2 * dh/dt
We can simplify this equation by canceling out the constants and rearranging it to solve for dh/dt:
dh/dt = 10 / (π * 3 * (h^2 / (4/75)))
Simplifying further, we get:
dh/dt = (750 / (πh^2)) ft/min
Now, we can plug in the value h = 5 to find the rate at which the water level is changing when the height is 5 feet:
dh/dt = (750 / (π * (5^2))) ft/min
dh/dt = (750 / (25π)) ft/min
Calculating the numerical value:
dh/dt ≈ 9.55 ft/min
Therefore, the water level is changing at a rate of approximately 9.55 feet per minute when the height is 5 feet.