Question

help please. a spherical balloon filled with 2.5 liters of gas per second. how fast will the balloon's radius increasing when r = 5 cm. the lesson is about integrals.

Answer 1

volume of ballon is given by:\[V=\frac{4}{3}\pi r^3\]so rate of increase of volume with respect to time would be:\[\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=2.5\]giving:\[\frac{dr}{dt}=\frac{2.5}{4\pi r^2}\]I believe you should be able to solve from here...

Answer 2

can u please continue so i check my answer ?

Answer 3

what answer did you get?

Answer 4

209.3333 sec

Answer 5

do i have to change the unit of the radius from cm to m ?

Answer 6

BTW: The volume increaase was given as 2.5 L/s = 2.5 * 1000 cm^3/s

Answer 7

so equation to solve should be:\[\frac{dr}{dt}=\frac{2.5*1000}{4\pi r^2}=\frac{625}{\pi r^2}\]

Answer 8

you are being asked to find the "rate of increase of the radius" when the value of the radius is 5cm. so need to set r=5 in the equation I gave you and the answer will be in cm/s

Answer 9

im i not supposed to integrate the equation to get t

Answer 10

no

Answer 11

aaa ok now i get it :D

Answer 12

good.

Answer 13

ty for the help :)

Answer 14

yw

Answer 15

the answer will be 7.9 cm per sec right ?

Answer 16

but the volume of the sphere is 4/3*pay*r^2
where did the 1/3 disappear ?

Answer 17

\[Dr[\frac{4}{3}r^3]=\frac{4}{\cancel{3}}\cancel{3}r^2\]

Answer 18

aha ok ty thought the rule was r^2

Answer 19

\[dr=\frac{2500}{4 \pi 5^2}=\frac{25}{\pi }=\text{Round}\left[\frac{25}{\pi },0.01\right]=7.96 \text{cm}/\sec \]