Gravel is being dumped from a conveyor belt at a rate of 15 ft^3/hr and its coarseness is such that is forms a pile in the shape of an inverted right cone whose height is three times its base radius. How fast is the height of the pile increasing when the pile has a height of 10 feet?
height ------- h
base radius --- h/3
V = (1/3)π r^2 h
= (1/3)π (h^2/9)(h) = (π/27) h^3
dV/dt = (π/9) h^2 dh/dt
for the given data:
15 = π/9 (100)dh/dt
dh/dt = 135/(100π) ft/hr
To solve this problem, we need to use the concept of related rates, which involves differentiating both sides of an equation with respect to time. In this case, we'll differentiate the equation that relates the height, radius, and volume of the cone with respect to time.
Let's denote the height of the cone as h (in feet), the radius as r (in feet), and the volume as V (in cubic feet).
We are given that the rate at which gravel is being dumped is 15 ft^3/hr, which is equal to the rate of change of volume with respect to time: dV/dt = 15 ft^3/hr.
We also know that the height is three times the base radius: h = 3r.
The volume of a cone can be calculated using the formula V = (1/3)πr^2h.
Differentiating both sides of this equation with respect to time t gives:
dV/dt = (1/3)[2πrh(dr/dt) +πr^2(dh/dt)].
Since we're interested in finding dh/dt (the rate at which the height is changing), we rearrange the equation and substitute the given values to solve for dh/dt.
Given:
dV/dt = 15 ft^3/hr
h = 10 ft (the height of the pile)
We need to find dh/dt when h = 10 ft.
Now, we can substitute the given values into the equation:
15 ft^3/hr = (1/3)[2πr(3dr/dt) + πr^2(dh/dt)].
Since we're interested in dh/dt, we can rearrange the equation:
15 ft^3/hr = (2/3)πr(3dr/dt) + (1/3)πr^2(dh/dt).
We are given that the height is three times the radius, so we can substitute 3r for h in the equation:
15 ft^3/hr = (2/3)πr(3dr/dt) + (1/3)πr^2(dh/dt).
Substituting the given values and solving for dh/dt, we have:
15 ft^3/hr = (2/3)πr(3dr/dt) + (1/3)π(10 ft)(dh/dt).
Simplifying further:
15 ft^3/hr = 2πr(dr/dt) + (10/3)π(dh/dt).
Now, we can solve for dh/dt by isolating the term:
(10/3)π(dh/dt) = 15 ft^3/hr - 2πr(dr/dt).
Finally, we can divide both sides by (10/3)π and substitute the given values to find dh/dt when h = 10 ft:
dh/dt = (15 ft^3/hr - 2πr(dr/dt)) / ((10/3)π).
Now you can substitute the appropriate values into the equation and calculate dh/dt accordingly.