Air is being pumped into a spherical balloon so that its volume increases at a rate of 90cm3/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=43πr3 and surface area S=4πr2.
Given that the volume of the balloon is increasing at a rate of 90 cm^3/s, we have dV/dt = 90 cm^3/s.
We need to find the rate at which the surface area (S) of the balloon is increasing when its radius (r) is 6 cm. We are asked to find dS/dt when r = 6 cm.
First, we need to find the relation between V and r using the given formula for the volume of a sphere: V = (4/3)πr^3.
Differentiating V with respect to t, we get:
dV/dt = (d/dt)((4/3)πr^3)
We can rewrite this in terms of dr/dt using the chain rule:
dV/dt = (dV/dr)(dr/dt)
Now substitute the given rate of volume increase:
90 = (dV/dr)(dr/dt)
We need to find dS/dt, so we will differentiate the given formula for the surface area of a sphere: S = 4πr^2.
Differentiating S with respect to t, we get:
dS/dt = (d/dt)(4πr^2)
We can rewrite this in terms of dr/dt using the chain rule:
dS/dt = (dS/dr)(dr/dt)
To determine the value of dS/dr, differentiate the formula for S with respect to r:
dS/dr = (d/dt)(4πr^2) = 8πr
Now, substitute the given radius of 6 cm:
dS/dr = 8π(6) = 48π cm^2/cm
Returning to the equation dV/dt = (dV/dr)(dr/dt), we can now solve for dr/dt:
90 = (dV/dr)(48π)
To determine dV/dr, differentiate the formula for V with respect to r:
dV/dr = (d/dr)((4/3)πr^3) = 4πr^2
Now, substitute the given radius of 6 cm:
dV/dr = 4π(6)^2 = 144π cm^3/cm
Returning to the equation 90 = (dV/dr)(48π), we can now solve for dr/dt:
90 = (144π)(48π)
dr/dt = 90 / (144π)(48π) ≈ 0.00002024 cm/s
Therefore, the surface area of the balloon is increasing at a rate of approximately 0.00002024 cm/s when its radius is 6 cm.