Question
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
* 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
Answers
Reiny
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
80 = 4π(81) dr/dt
dr/dt = 20/(81π)
SA = 4πr^2
dSA/dt = 8πr dr/dt
= 8π(9)(20/(81π))
= 16/9