To find the rate at which the surface area of the cylinder is increasing, we need to differentiate the surface area formula with respect to time, and then substitute the given values to find the rate.
The surface area of a right circular cylinder is given by the formula:
A = 2Ï€r^2 + 2Ï€rh
Where:
A = Surface area
r = Radius of the base
h = Height
Given:
Radius (r) = 5 cm
Height (h) = 2r = 2 * 5 cm = 10 cm
Now, let's differentiate the surface area equation with respect to time (t) to find the rate of change of surface area (dA/dt) when the volume is increasing at 5Ï€ cm/sec.
Differentiating both sides with respect to time, we get:
dA/dt = 2Ï€ * (2r * dr/dt) + 2Ï€ * (r * dh/dt)
Where:
dr/dt = Rate of change of radius
dh/dt = Rate of change of height
Given:
dv/dt = 5Ï€ cm/sec (Rate of change of volume)
The volume of a cylinder is given by the formula:
V = πr^2h
Differentiating with respect to time, we have:
dv/dt = π * (2r * dr/dt * h + r^2 * dh/dt)
Now we can substitute the given values into the equations and solve for the rate at which the surface area is increasing.
Given:
Radius (r) = 5 cm
Height (h) = 10 cm
dv/dt = 5Ï€ cm/sec
Substituting these values into the volume equation, we have:
5π = π * (2 * 5 * dr/dt * 10 + 5^2 * dh/dt)
Simplifying this equation, we get:
5 = 10 * dr/dt + 25 * dh/dt
Now, substituting this expression for dh/dt into the surface area equation, we get:
dA/dt = 2Ï€ * (2 * 5 * dr/dt) + 2Ï€ * (5 * (5 - 10 * dr/dt))
Simplifying further, we have:
dA/dt = 20Ï€ * dr/dt + 10Ï€ * (5 - 10 * dr/dt)
Finally, substituting dv/dt = 5Ï€ into the equation:
dA/dt = 20 * 5Ï€ + 10Ï€ * (5 - 10 * dr/dt)
Simplifying the equation:
dA/dt = 100Ï€ + 50Ï€ - 100Ï€ * dr/dt
dA/dt = 150Ï€ - 100Ï€ * dr/dt
Therefore, the rate at which the surface area of the cylinder is increasing when the radius is 5 cm and the volume is increasing at 5Ï€ cm/sec is 150Ï€ - 100Ï€ * dr/dt.