Rate of change of base area & height, find inst. rate of

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bobers

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if you have the rate of change of the area of a base of a cone and the rate of change of its height, how would you find the instantaneous rate of change of the area of its base with respect to its height?

V=1/3(pi)r^2h
A=(pi)r^2
would it be V=1/3Ah?
 
bobers said:
if you have the rate of change of the area of a base of a cone and the rate of change of its height, how would you find the instantaneous rate of change of the area of its base with respect to its height?

V=1/3(pi)r^2h
A=(pi)r^2
would it be V=1/3Ah?
Click to expand...

Please post an actual problem - where this situation arises.
 
Volume V of a cone is increasing at a rate of 28pi cubic units per second. At the instant when radius r of the cone is 3 units, its volume is 12pi cubic units and the radius is increasing at 0.5 units per second.

a.) At the instant when the radius of the cone is 3 units, what is the rate of change of the area of its base?
b.) At the instant when the radius of the cone is 3 units, what is the rate of change of its height?
c.) At the instant when the radius of the cone is 3 units, what is the rate of change of the area of its base with respect to its height.

I got
a.) 6pi units^2/sec
b.) 8 units/sec
c.)don't know how to get it
 
bobers said:
Volume V of a cone is increasing at a rate of 28pi cubic units per second. At the instant when radius r of the cone is 3 units, its volume is 12pi cubic units and the radius is increasing at 0.5 units per second.

a.) At the instant when the radius of the cone is 3 units, what is the rate of change of the area of its base?
b.) At the instant when the radius of the cone is 3 units, what is the rate of change of its height?
c.) At the instant when the radius of the cone is 3 units, what is the rate of change of the area of its base with respect to its height.

I got
a.) 6pi units^2/sec<<< Incorrect

given at r = 3 we have dr/dt = 0.5

A = pi * r^2

dA/dt = 2 * pi * r * dr/dt = 2 * pi * 3 * 0.5 = 3 * pi unit^2/sec


b.) 8 units/sec<<< Incorrect - please show work

c.)don't know how to get it
Click to expand...
 
b) V=1/3(pi)r^2h

dv/dt=1/3pi(2rh(dr/dt)+r^2(dh/dt))
28pi=1/3pi(2(3)(4)(o.5)+(9)(dh/dt)
84=(12+9dh/dt)
72=9dh/dt
dh/dt=8
 
bobers said:
b) V=1/3(pi)r^2h

dv/dt=1/3pi(2rh(dr/dt)+r^2(dh/dt))
28pi=1/3pi(2(3)(4)(o.5)+(9)(dh/dt)
84=(12+9dh/dt)
72=9dh/dt
dh/dt=8 <<<< You are correct - I had an arithmetic problem
Click to expand...
 
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