Jesse has constructed a huge cylindrical can with a diameter of 60 ft. The can is being filled with water at a rate of 450 ft3/min. How fast is the depth of the water increasing? (Hint: The volume of water in the cylinder is determined by πr2h where r is the radius and h is the depth of the water )
the rate of change of volume is the surface area times the rate of change of height
450 ft^3/min = surface area * dh/dt
surface area = pi r^2
450 = pi (30^2) dh/dt
dh/dt = .159 ft/min
You could do this by saying
V = pi r^2 h
dV/dh = pi r^2
dV/dh*dh/dt = pi r^2 dh/dt
chain rule
dV/dt = pi r^2 dh/dt
but most of us just look at the lake rising an inch and seeing that the added volume is the area times one inch
Thanks Damon, that really clears it up for me
To find the rate at which the depth of the water is increasing, we need to determine the derivative of the height with respect to time. Let's break down the problem step by step:
1. First, let's determine the radius of the cylindrical can. We are given the diameter, which is 60 ft. The radius (r) is half the diameter, so r = 60 ft / 2 = 30 ft.
2. Next, we need to find the volume of water in the cylindrical can. The volume of a cylinder is given by the formula V = πr^2h, where V is the volume, r is the radius, and h is the height (or depth) of the water.
3. We are told that the can is being filled with water at a rate of 450 ft^3/min. Therefore, the rate at which the volume of water is increasing is dV/dt = 450 ft^3/min.
4. Now, let's differentiate the volume formula with respect to time (t) to find an expression for dV/dt in terms of dh/dt, where h is the height of the water.
dV/dt = π * (2r) * h * dh/dt
Since we are looking for the rate at which the depth of the water is increasing (dh/dt), we need to solve for it.
5. Rearrange the equation to solve for dh/dt:
dh/dt = (dV/dt) / (π * (2r) * h)
6. Now, we can substitute the known values into the equation:
dh/dt = (450 ft^3/min) / (π * (2 * 30 ft) * h)
Simplifying further:
dh/dt = (450 ft^3/min) / (60π ft * h)
Therefore, the rate at which the depth of the water is increasing is (450 ft^3/min) / (60π ft * h).