Related Rates - I know! It's horrible! I should probably be able to do this problem with google, but I haven't had any luck.
It's a three part problem, and I think I have parts A and B down, but am stuck on C. Thanks for any help.
The volume of a cone is increasing at 28pi cubic units per second. When radius r is 3 units, its volume is 12pi cubic units and the radius is increasing at 1/2 unit per second.
A) When r of cone = 3, what is the rate of change in the area of its base?
Area of base = (pi)r^2
dA/dt = 2 * (pi) * r * (dr/dt)
When r = 3 and dr/dt = 1/2, dA/dt = 6(pi) * (1/2) = 3(pi)
B) When r of cone = 3, what is the rate of change of its height h?
dV/dt = 28(pi)
V = (1/3) * (pi) * r^2 * h
dV/dt = (2/3) * (pi) * r * (dr/dt) * (dh/dt)
When r = 3, dr/dt = 1/2
dV/dt = (2/6) * (pi) * r * (dh/dt)
28(pi) = (6/6) * (pi) * (dh/dt)
dh/dt = 28
Does this make any sense? It seems too simple.
C) When r of cone = 3, what is the instantaneous rate of change of area of its base with respect to its height h?
I don't see a way to find dA/dh. Where can I go from here?
It's a three part problem, and I think I have parts A and B down, but am stuck on C. Thanks for any help.
The volume of a cone is increasing at 28pi cubic units per second. When radius r is 3 units, its volume is 12pi cubic units and the radius is increasing at 1/2 unit per second.
A) When r of cone = 3, what is the rate of change in the area of its base?
Area of base = (pi)r^2
dA/dt = 2 * (pi) * r * (dr/dt)
When r = 3 and dr/dt = 1/2, dA/dt = 6(pi) * (1/2) = 3(pi)
B) When r of cone = 3, what is the rate of change of its height h?
dV/dt = 28(pi)
V = (1/3) * (pi) * r^2 * h
dV/dt = (2/3) * (pi) * r * (dr/dt) * (dh/dt)
When r = 3, dr/dt = 1/2
dV/dt = (2/6) * (pi) * r * (dh/dt)
28(pi) = (6/6) * (pi) * (dh/dt)
dh/dt = 28
Does this make any sense? It seems too simple.
C) When r of cone = 3, what is the instantaneous rate of change of area of its base with respect to its height h?
I don't see a way to find dA/dh. Where can I go from here?