Air is being pumped into a spherical balloon so that its volume increases at a rate of 90cm^3/s. How fast is the surface area of the balloon increasing when its radius is 6cm? Recall that a ball of radius r has volume V=43πr^3 and surface area S=4πr^2.
To find the rate at which the surface area of the balloon is increasing, we need to find the derivative of the surface area with respect to time.
Given that the volume of the balloon is increasing at a rate of 90 cm^3/s, we can use the formula for the volume of a sphere to relate the rate of change of the volume to the rate of change of the radius.
V = (4/3)πr^3
Taking the derivative of both sides with respect to time, we get:
dV/dt = 4πr^2(dr/dt)
We are given that dV/dt = 90 cm^3/s and r = 6 cm. Plugging these values into the equation, we can solve for dr/dt:
90 cm^3/s = 4π(6^2)(dr/dt)
90 cm^3/s = 144π(dr/dt)
dr/dt = 90 cm^3/s / (144π) cm^2
dr/dt = 90 / (144π) cm/s
Now that we have the rate of change of the radius, we can find the rate of change of the surface area using the formula for the surface area of a sphere.
S = 4πr^2
Taking the derivative with respect to time, we get:
dS/dt = 8πr(dr/dt)
Plugging in r = 6 cm and dr/dt = 90 / (144π) cm/s, we can solve for dS/dt:
dS/dt = 8π(6)(90 / (144π)) cm^2/s
dS/dt = (6)(90) / 18 cm^2/s
dS/dt = 540 / 18 cm^2/s
dS/dt = 30 cm^2/s
Therefore, the surface area of the balloon is increasing at a rate of 30 cm^2/s when its radius is 6 cm.