if the surface area of a sphere is given by the formula SA=4pir^2, find the rate of increase of the surface area if the radius is 4 cm and is increasing at the rate of 0.5 cm/min.
To find the rate of increase of the surface area of a sphere, we need to take the derivative of the surface area formula with respect to time and substitute the given values.
Given:
Surface area formula: SA = 4πr^2
Radius of the sphere: r = 4 cm
Rate of increase of radius: dr/dt = 0.5 cm/min
Step 1: Find the derivative of the surface area formula with respect to time (t).
dSA/dt = d(4πr^2)/dt
Step 2: Apply the power rule for differentiation to the formula.
dSA/dt = 2 × 4πr × dr/dt
Step 3: Substitute the given values into the equation.
dSA/dt = 2 × 4π × 4 × 0.5
Step 4: Simplify the equation.
dSA/dt = 8π cm^2/min
Therefore, the rate of increase of the surface area of the sphere is 8π square centimeters per minute.
given: dr/dt = .5 cm/min
r = 4
SA = 4πr^2
d(SA)/dt = 8πr dr/dt
= 8π(4)(.5)
= 16π cm^2/min