Sand falling from a hopper at 10 pi ft3/s forms a conical sand pile whose radius is always equal to its height. How fast is the radius increasing when the radius is 5ft?

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The sand falling from a hopper at `10*pi` ft^3/s forms a conical pile where the radius is always equal to the height.

The volume of a cone if the radius r is also equal to the height is given by `V = (1/3)*(pi*r^2)*r = (1/3)*pi*r^3`

`(dV)/(dr) = pi*r^2`

`(dV)/(dt) = 10*pi`

`((dV)/(dt))/((dr)/(dt)) = pi*r^2`

=> `(10*pi)/(pi*r^2) = (dr)/(dt)`

=> `(dr)/(dt) = 10/(r^2)`

At r = 5

`(dr)/(dt) = 10/25 = 2/5` ft/s

**The radius is increasing at `2/5` ft/s**

Volume of the cone can be obtained by 1/3r^3pi where r = radius and the height of the cone

dV/dt = 1/3 X 3r^2pi X dr/dt = pi r^2 X dr/dt = 10 pi

When r=5,

pi 5^2 X dr/dt = 10 pi

Therefore, dr/dt = 10/25 = 0.4 (ft/s)

Radius is increasing at 0.4 ft/s

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