Question

A balloon is being filled with helium at a rate of 4 ft^3/min. The rate, in square feet per minute, at which the surface are is increasing when the volume is 32(pi)/ 3 is...

Answer 1

you accidently the problem

Answer 2

what?

Answer 3

is it a spherical balloon?

Answer 4

yeah and i know that the Volume of a sphere is
\[4/3 \pi r^2 \]

Answer 5

A = 4*Pi*r^2

Answer 6

so thats the surface area...

Answer 7

volume is (4/3)pi r ^3 but that's not what the question is asking

Answer 8

oh hold on a sec

Answer 9

dA/dt = dV/dt * dA/dV

Answer 10

Volume = Area * r/3

Answer 11

Area = 3*Volume/r

Answer 12

r being rate or radius?

Answer 13

radius

Answer 14

okay... well i don't know the radius..

Answer 15

need to replace that as well

Answer 16

oh okay wait, i know how i can find it

Answer 17

volume = (4/3)pi r ^3

r = (3/(4Pi) * V)^(1/3)

Answer 18

Area = 3*Volume/(3/(4Pi) * V)^(1/3)

Answer 19

area = 3*Volume^(2/3)/(3/(4Pi)^(1/3)

Answer 20

cancelled volume from the denominator

Answer 21

now find the derivative to get dA/dV

Answer 22

to clean it up I'll combine all the numbers

Answer 23

3/(3/(4Pi))^(1/3)=6^(2/3)*Pi^(1/3)

Answer 24

area = 6^(2/3)*Pi^(1/3)*V^(2/3)

Answer 25

dA/dV = (2/3)*6^(2/3)*Pi^(1/3)*V^(-1/3)

Answer 26

dA/dV = 1 when V= 32*Pi/3

Answer 27

dA/dt = 4*1 = 4 f^2/min

Answer 28

well that cleaned up well

Answer 29

haha, thanks!

Answer 30

checking to see if i made a mistake anywhere

Answer 31

seems correct

Answer 32

you didn't, it was multiple choice and that was an answer.. and the answer key says its right, i just needed to know how to get the answer haha

Answer 33

that was a difficult question

Answer 34

yeah, its from my practice AP calc book