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 cubic inches per minute. If the height is always 3 times the radius, find the rate of change of the radius at that instant.
dV/dt = 9 = surface area * dh/dt
9 = (pi r^2)(dh/dt)
but
h = 3 r
so
dh/dt = 3 dr/dt
so
9 = (pi r^2)(2 dr/dt)
but r = 3
so
9 = 9 pi (2 dr/dt
dr/dt = 1/2pi
To find the rate of change of the radius at that instant, we can use the related rates formula. This formula relates the rates of change of the variables involved. Let's denote the radius of the cone as r (in inches) and the volume of the cone as V (in cubic inches).
Given:
- The radius of the cone (r) is 3 inches.
- The volume of the cone (V) is increasing at a rate of 9π cubic inches per minute.
- The height (h) is always 3 times the radius: h = 3r.
We can use the formula for the volume of a cone to relate the variables:
V = (1/3)πr^2h
Now, let's take the derivative of both sides of the equation with respect to time (t):
dV/dt = (1/3)(2πrh(dr/dt) + πr^2(dh/dt)
Here, dV/dt represents the rate of change of the volume with respect to time (9π in this case), dr/dt represents the rate of change of the radius with respect to time (what we are trying to find), and dh/dt represents the rate of change of the height with respect to time.
We are given that dh/dt is always 3 times dr/dt because the height is always 3 times the radius. So, we can substitute dh/dt = 3(dr/dt) in the equation:
9π = (1/3)(2πrh(dr/dt) + πr^2(3(dr/dt))
Simplifying and solving for dr/dt:
9π = (2πrh(dr/dt))/3 + (3πr^2(dr/dt))
Multiplying through by 3:
27π = 2πrh(dr/dt) + 9πr^2(dr/dt)
Dividing through by π:
27 = 2rh(dr/dt) + 9r^2(dr/dt)
Factoring out dr/dt:
27 = (2rh + 9r^2)(dr/dt)
Now, plugging in the given values at the instant when the radius is 3 inches, we have:
27 = (2(3)(3r) + 9(3r^2))(dr/dt)
27 = (18r + 27r^2)(dr/dt)
Dividing through by (18r + 27r^2):
27/(18r + 27r^2) = dr/dt
Simplifying further, we get:
dr/dt = 27/(9r + 9r^2)
Now, substituting the given radius value of r = 3 into the equation gives us:
dr/dt = 27/(9(3) + 9(3)^2)
dr/dt = 27/(27 + 81)
dr/dt = 27/108
dr/dt = 0.25 inches per minute
Therefore, at that instant, the rate of change of the radius is 0.25 inches per minute.
To find the rate of change of the radius at the instant when the volume of the cone is increasing at a rate of 9π cubic inches per minute, we can use the relationship between the volume of a cone and its radius given by the formula:
V = (1/3)πr^2h
where V is the volume, r is the radius, and h is the height.
Given that the height is always 3 times the radius, we can express h in terms of r:
h = 3r
Now, we differentiate both sides of the volume equation with respect to time (t):
dV/dt = (1/3)(2πr)(dh/dt) + (1/3)(πr^2)(dh/dt)
Since we are interested in finding the rate of change of the radius (dr/dt), we need to express dh/dt in terms of dr/dt. From the given information, we know that dh/dt = 3(dr/dt).
Substituting this value into the volume equation, we have:
9π = (1/3)(2πr)(3dr/dt) + (1/3)(πr^2)(3dr/dt)
Simplifying the equation:
9π = 2πr(dr/dt) + πr^2(dr/dt)
Dividing both sides by πr:
9 = 2r(dr/dt) + r^2(dr/dt)
Now, we can solve for dr/dt:
9 = (2r + r^2)(dr/dt)
Dividing both sides by (2r + r^2):
dr/dt = 9 / (2r + r^2)
We are given that at the instant when the radius is 3 inches, the volume of the cone is increasing at a rate of 9π cubic inches per minute. Therefore, we can substitute r = 3 into the equation to find the rate of change of the radius at that instant:
dr/dt = 9 / (2(3) + (3)^2)
dr/dt = 9 / (6 + 9)
dr/dt = 9 / 15
dr/dt = 0.6 inches per minute
Therefore, the rate of change of the radius at the instant when the volume of the cone is increasing at a rate of 9π cubic inches per minute is 0.6 inches per minute.