Question

A spherical balloon is inflated so that its volume is increasing at the rate of 3.8 ft3/min. How rapidly is the diameter of the balloon increasing when the diameter is 1.8 feet?

Answer 1

seems like we want a formula for the colume of a sphere to play with

Answer 2

*volume

Answer 3

4/3*pi*r^2

Answer 4

^3

Answer 5

ya

Answer 6

^3 but, now lets work this for diameter. d = 2r, (2r)^2 = 4r^2

Answer 7

we dont need to convert, lets just dig in with the derivative ...

Answer 8

ok

Answer 9

so what is our derivative of the volume equation ... ill prolly work it with the conversion afterwards just for fun tho

Answer 10

Surface area no?
that's what we need to differentiate

Answer 11

dV/dt=4*pi*r^2dr/dt

Answer 12

we have a rate of change of volume in the information, so lets play with that :)
V' = 4 pi r^2 r' yes

Answer 13

ok so we just sub in 3.8 for r?

Answer 14

the rate of change of the diameter is just the rate of change of the radius

Answer 15

4r^2 = (2r)^2 = d^2

V' = pi d^2 r'


d = 2r

d' = 2r' ; r' = d'/2


V' = pi d^2 d'/2

Answer 16

input the values and solve for d'

Answer 17

what do i put for v'

Answer 18

v' is defined as the rate of change of volume ...

Answer 19

so 3.8

Answer 20

yep

Answer 21

they gave it to you the volume is increasing with the rate of 3.8

Answer 22

d'=.747

Answer 23

V = 4r^3 pi/3

r = d/2

V = 4(d/2)^3 pi/3

V = 4d^3/8) pi/3

V = d^3 pi/6 is our 'adjusted' volume


V' = 3d^2 d' pi/6

V' = d^2 d' pi/2 just like before lol

Answer 24

.747?

Answer 25

3.8 = (1.8)^2 pi d'/2

2(3.8)/((1.8)^2 pi) = d' = .7466

http://www.wolframalpha.com/input/?i=2%283.8%29%2F%28%281.8%29%5E2+pi%29

Answer 26

is that the final answer?

Answer 27

what was it they were looking for?

Answer 28

the diameter is increasing at ___ ft/min

Answer 29

then yeah, we determine the rate of change of the diameter

Answer 30

sweet thanks!!

Answer 31

yw