The volume of a sphere is changing at a rate of 8pi cm/sec. What is the rate of change of its surface area when the radius is 1? (the volume of a sphere is given by V=4/3pir^3 and its surface area, by A=4pi r ^2).
V = (4/3)π r^3
dV/dt = 4π r^2 dr/dt
given: dV/dt = 8π , r = 1
8π = 4π(1) dr/dt
dr/dt = 8π/4π = 2
A = 4πr^2 ----> (did you notice that surface area equals the derivative of volume ??)
dA/dt = 8πr dr/dt
when r = 1
dA/dt = 8π(1)(1) = 8π
the surface area is changing at 8π cm/sec
To find the rate of change of the surface area of a sphere, we need to differentiate the surface area formula with respect to time and then substitute the given values into the equation.
Let's start by differentiating the surface area formula, A = 4πr^2, with respect to time (t):
dA/dt = d/dt(4πr^2)
To find the derivative, we'll need to use the chain rule. Since r is a function of time, we differentiate r^2 with respect to t and multiply by the derivative of r with respect to t:
dA/dt = 8πr(d