Air is being pumped into a spherical balloon so that its volume increases at a rate of 80 cm^3 per second. How fast is the surface area of the balloon increasing when its radius is 9cm?
* ball of radius r has volume V=(4/3)pi r^3 and surface area S=4pir^2
** I try to do it twice and my answer is 72pi(80/324pi) which is wrong
dV/dt = 4πr^2 dr/dt
80 = 4π(81) dr/dt
dr/dt = 20/(81π)
SA = 4πr^2
dSA/dt = 8πr dr/dt
= 8π(9)(20/(81π))
= 16/9
To find the rate of change of the surface area of the balloon, you can use the given formulas for the volume and surface area of a sphere:
Volume of sphere (V) = (4/3)πr^3
Surface area of sphere (S) = 4πr^2
First, let's find the relationship between the volume and the radius by differentiating the volume equation with respect to time:
dV/dt = d((4/3)πr^3)/dt
Using the power rule of differentiation, the equation becomes:
dV/dt = (4/3)π * 3r^2 * (dr/dt)
Given that dV/dt = 80 cm^3/s, we can substitute this value into the equation and solve for dr/dt:
80 = (4/3)π * 3(9^2) * (dr/dt)
80 = 36π * (dr/dt)
dr/dt = 80 / (36π)
dr/dt ≈ 0.707 cm/s
Now, let's find the rate of change of the surface area of the balloon by differentiating the surface area equation with respect to the radius:
dS/dr = d(4πr^2)/dr
Using the power rule again, the equation becomes:
dS/dr = 4π * 2r
Substituting the radius value of 9 cm, we get:
dS/dr = 4π * 2(9)
dS/dr = 72π
Therefore, the rate of change of the surface area of the balloon when its radius is 9 cm is 72π cm^2.