At the instant when the radius of a cone is 3 inches, the volume of the cone is increasing at the rate of 9 pi or 28.27433388 cubic inches per minute. If the height is always 3 times the radius, find the rate of change of the radius at that instant?
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To find the rate of change of the radius at a given instant, we can use related rates and the formula for the volume of a cone.
Given:
- The volume of the cone is increasing at a rate of 9π cubic inches per minute.
- The radius is 3 inches at the instant of interest.
- The height is always 3 times the radius.
Let's denote:
- V as the volume of the cone (in cubic inches)
- r as the radius of the cone (in inches)
- h as the height of the cone (in inches)
We know the formula for the volume of a cone is:
V = (1/3)πr²h
Since we are given that the height is always 3 times the radius, we have:
h = 3r
Now, we can differentiate the formula for the volume with respect to time (t) to relate the rate of change of volume to the rates of change of the radius and height:
dV/dt = (1/3)π(2rh(dr/dt) + r²(dh/dt))
Substituting the given values and differentiating h = 3r with respect to time:
9π = (1/3)π(2(3r)(dr/dt) + r²(3(dr/dt)))
Simplifying:
9 = 2(3r)(dr/dt) + 3r²(dr/dt)
Now we can solve for the rate of change of the radius, (dr/dt), at the given instant when the radius is 3 inches:
9 = 6r(dr/dt) + 3r²(dr/dt)
9 = 9r(dr/dt)(2 + r)
Dividing both sides by 9:
1 = r(dr/dt)(2 + r)
At the given instant, when the radius is 3 inches, we have:
1 = 3(dr/dt)(2 + 3)
1 = 3(dr/dt)(5)
1 = 15(dr/dt)
Now we can find the rate of change of the radius, (dr/dt):
(dr/dt) = 1/15
Therefore, the rate of change of the radius at the instant when the radius is 3 inches is 1/15 inches per minute.
To find the rate of change of the radius at the instant when the volume of the cone is increasing at the given rate, we can use related rates. First, let's establish the given information:
Given:
- The radius of the cone is 3 inches.
- The volume of the cone is increasing at a rate of 9π or 28.27433388 cubic inches per minute.
- The height of the cone is always 3 times the radius.
To solve this problem, we need to find a relationship between the radius, height, and volume of the cone. The formula for the volume of a cone is:
V = (1/3)πr²h
Given that the height is always 3 times the radius, we can substitute h = 3r into the formula:
V = (1/3)πr²(3r)
V = πr³
Now we have the volume formula in terms of the radius. Differentiate both sides of the equation with respect to time (t) to find the rates of change:
dV/dt = d(πr³)/dt
dV/dt = 3πr²(dr/dt)
Here, dV/dt represents the rate of change of volume with respect to time (given as 9π or 28.27433388 cubic inches per minute), and dr/dt represents the rate of change of the radius with respect to time (what we're trying to find).
Substituting the given values, we can solve for dr/dt:
9π = 3π(3²)(dr/dt)
9π = 27π(dr/dt)
Now, we can solve for dr/dt:
dr/dt = 9π / 27π
dr/dt = 1/3
Therefore, the rate of change of the radius at that instant is 1/3 inches per minute.