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?

Answer 1

Use \[dSA/dt = dSA/dr \times dr/dt\]

dr/dt =7

\[SA = 4 \pi r^2\]

\[d(SA0=) = 8 \pi r\]

\[d(SA)?dt = 8 \pi r \times 7\]

Answer 2

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

Answer 3

nope

Answer 4

if dr/dt = 7 then by integrating r = 7t + c

\[SA = 4 \pi r^2\]

SA = \[SA = 4 \pi ( 7t + c)^2\]