A cone of fixed height 12 inches is changing in shape through the change in
the radius of the base. What rate of increase of the radius will make the lateral
surface area of the cone increase at the rate of 10 !
square inches per minute
when the radius of the base is 5 inches?
To solve this problem, we need to use the formula for the lateral surface area of a cone. The formula is given by:
Lateral Surface Area = π * r * L
where r is the radius of the base and L is the slant height of the cone.
We are given that the height of the cone is fixed at 12 inches. In this case, using the Pythagorean theorem, we can find the slant height L in terms of the radius r and the height h:
L^2 = r^2 + h^2
L^2 = r^2 + 12^2
L = sqrt(r^2 + 144)
Now, let's differentiate both sides of the equation with respect to time t:
d/dt(Lateral Surface Area) = d/dt(π * r * L)
d/dt(Lateral Surface Area) = π * (dr/dt * L + r * dL/dt)
We are given that d/dt(Lateral Surface Area) = 10 square inches per minute. We also know that L is a function of r, so we can express dL/dt in terms of dr/dt using the chain rule:
dL/dt = dL/dr * dr/dt
Substituting this back into the equation, we get:
10 = π * (dr/dt * sqrt(r^2 + 144) + r * dL/dr * dr/dt)
Now we need to find dr/dt when r = 5 inches. To do this, we can rearrange the equation and solve for dr/dt:
10 - π * r * dL/dr * dr/dt = π * dr/dt * sqrt(r^2 + 144)
dr/dt * (10 - π * r * dL/dr) = π * dr/dt * sqrt(r^2 + 144)
dr/dt * (10 - π * r * dL/dr - π * sqrt(r^2 + 144)) = 0
Now we have an equation that relates dr/dt to other variables. We can solve this equation to find the rate of increase of the radius that will make the lateral surface area of the cone increase at the rate of 10 square inches per minute when the radius of the base is 5 inches.
To solve this problem, we can use the formula for the lateral surface area of a cone:
Lateral surface area = πrℓ
Where r is the radius of the base, and ℓ is the slant height of the cone.
We are given that the height of the cone is fixed at 12 inches. Let's denote the change in radius as Δr and the rate of increase of the radius as dr/dt.
We are also given that the lateral surface area is increasing at a rate of 10 square inches per minute, so dA/dt = 10.
To find the rate of increase of the radius that will result in a lateral surface area increase of 10 square inches per minute, we need to find the derivative of the lateral surface area with respect to the radius and set it equal to dA/dt:
dA/dt = d(πrℓ)/dt
Since the height of the cone is fixed, we can express ℓ in terms of r using the Pythagorean theorem:
ℓ^2 = r^2 + h^2
ℓ^2 = r^2 + 12^2
ℓ = √(r^2 + 144)
Now, let's differentiate the lateral surface area equation with respect to time:
d(πrℓ)/dt = π(d(rℓ)/dt)
d(πrℓ)/dt = π(dr/dt * ℓ + r * dℓ/dt)
Since the height is fixed, dℓ/dt is 0:
d(πrℓ)/dt = π(dr/dt * ℓ + r * 0)
d(πrℓ)/dt = π(dr/dt * ℓ)
Now, substitute the values provided into the equation:
10 = π(dr/dt * √(5^2 + 144))
10 = π(dr/dt * √(25 + 144))
10 = π(dr/dt * √169)
10 = π(dr/dt * 13)
Now, isolate dr/dt:
dr/dt = 10 / (π * 13)
Simplifying further:
dr/dt = 10 / (13π)
Therefore, the rate of increase of the radius that will result in a lateral surface area increase of 10 square inches per minute when the radius of the base is 5 inches is 10 / (13π) inches per minute.