A spherical weather balloon is being inflated. The radius of the balloon is increasing at the rate of 7 cm/s. Express the surface area of the balloon as a function of time t (in seconds). (Let S(0) = 0.)
S(t) = cm?

Question

A spherical weather balloon is being inflated. The radius of the balloon is increasing at the rate of 7 cm/s. Express the surface area of the balloon as a function of time t (in seconds). (Let S(0) = 0.)
S(t) = cm?

campbell_st · Answer

Use $$dSA/dt = dSA/dr \times dr/dt$$

dr/dt =7

$$SA = 4 \pi r^2$$

$$d(SA0=) = 8 \pi r$$

$$d(SA)?dt = 8 \pi r \times 7$$

campbell_st · Answer

then $$d(SA)/dt = 56 \pi r$$

anonymous · Answer

nope

campbell_st · Answer

if dr/dt = 7 then by integrating r = 7t + c

$$SA = 4 \pi r^2$$

SA = $$SA = 4 \pi ( 7t + c)^2$$