gcn_zelda

10-17-2005, 05:16 PM

We're learning related rates, and I'm still trying to completely understand them, so I'm not sure how correct my answer to this question is, and I just wanted some help on it.

A container has the shape of an open right circular cone. The height of the cone is 10 cm and the diameter of the opening is 10 cm. Water in the container is evaporating so that its depth h is changing at the constant rate of -3/10 cm/hr.

Find the rate of change of the volume of water in the container with respect to time, when h = 5 cm.

How I think it should be done:

let h = depth of water in container

let r = radius of surface of water

V(cone) = 1/3(pi)(r^2)(h)

h/r = 10/5 = 2, so r = h/2

Since I'm trying to relate dh/dt and dV/dt, I solved for r and will plug that into the formula for volume of a cone.

V = 1/3(pi)[(h^2)/4](h)

V = [(pi)(h^3)]/12

I'll differentiate that:

dV/dt = [2h^2(pi)/3](dh/dt)

Substitute 5 cm for h and -3/10 cm/hr for dh/dt.

dV/dt = [10(pi)/3](-3/10)

Thus, dV/dt = -(pi)

Did I do this about right?

A container has the shape of an open right circular cone. The height of the cone is 10 cm and the diameter of the opening is 10 cm. Water in the container is evaporating so that its depth h is changing at the constant rate of -3/10 cm/hr.

Find the rate of change of the volume of water in the container with respect to time, when h = 5 cm.

How I think it should be done:

let h = depth of water in container

let r = radius of surface of water

V(cone) = 1/3(pi)(r^2)(h)

h/r = 10/5 = 2, so r = h/2

Since I'm trying to relate dh/dt and dV/dt, I solved for r and will plug that into the formula for volume of a cone.

V = 1/3(pi)[(h^2)/4](h)

V = [(pi)(h^3)]/12

I'll differentiate that:

dV/dt = [2h^2(pi)/3](dh/dt)

Substitute 5 cm for h and -3/10 cm/hr for dh/dt.

dV/dt = [10(pi)/3](-3/10)

Thus, dV/dt = -(pi)

Did I do this about right?