Question

Hard calculus question, using related rates: A spherical balloon is inflated with helium at the rate of 80pi ft^3/min (pi=3.14, I just don't think there a key for it). At the moment the diameter is 8ft, how fast is the diameter of the balloon changing?

Answer 1

yeah, spherical ballons are almost none existent these days to compare to

Answer 2

we need a formula that relates volume to diameter

Answer 3

or at least to radius then we can interpolate for diameter

Answer 4

my professor wrote down: v=(4pi/3)(3r^2)(r)

Answer 5

there is a spurious 3 in there i think

Answer 6

it should be, if i aint mistaken
\[V_{sph}=\frac{4\pi}{3}r^3\]

Answer 7

is that the formula?

Answer 8

for the volume of a sphere; id say it is

Answer 9

can you find the derviative of that for me with respect to time?

Answer 10

typos not withstanding

Answer 11

ok, so based on my prof's work, he put 80pi=4pi/3(3)(4^2)(r1). Sorry if this is hard to read, I'm not sure how to get the pi symbol :(

Answer 12

lets forget about what your prof did; and lets see if we can discover this anew :)

Answer 13

lets start by deriving the volume of a sphere;
\[V=\frac{4pi}{3}r^3\]
we can use this becasue it relates volume to something akin to a diameter

Answer 14

can we replace r with d/2 ? since radius = half the diamter?

Answer 15

oh yea, that makes sense

Answer 16

\[V=\frac{4pi}{3}r^3\]
\[V=\frac{4pi}{3}(d/2)^3\]
\[V=\frac{4pi}{3}\frac{d^3}{8}\]
\[V=\frac{4pi}{24}d^3\]
\[V=\frac{pi}{6}d^3\]
this gets it all talking in V and d terms

Answer 17

let me know where this doesnt make sense at becasue its important that you know whats going on behind the scenes

Answer 18

\[V=\frac{pi}{6}d^3\]
\[\frac{d}{dt}[V=\frac{pi}{6}d^3]\]
\[\frac{d}{dt}[V]=\frac{d}{dt}[\frac{pi}{6}d^3]\]
\[\frac{dV}{dt}=\frac{pi}{6}\frac{d}{dt}[d^3]\]
\[\frac{dV}{dt}=\frac{pi}{6}\frac{dd}{dt}\ 3d^2\]
\[\frac{dV}{dt}=\frac{3pi\ d^2}{6} \frac{dd}{dt}\]
\[\frac{dV}{dt}*\frac{6}{3pi\ d^2}=\frac{dd}{dt}\]
\[\frac{dV}{dt}*\frac{2}{pi\ d^2}=\frac{dd}{dt}\]

Answer 19

No, i think i actually understand so far, but it doesn't seem related to what I have in my notes. But your version seems much clearer.

Answer 20

We now dV/dt = 80pi

and the diameter is 8 ft so lets plug them in:

\[\frac{dV}{dt}*\frac{2}{pi\ d^2}=\frac{dd}{dt}\]
\[80pi*\frac{2}{pi\ 8^2}=\frac{dd}{dt}\]
\[80*\frac{2}{8^2}=\frac{dd}{dt}\]
\[\frac{80*2}{8*8}=\frac{dd}{dt}\]
\[\frac{10*2}{8}=\frac{dd}{dt}\]
\[\frac{10}{4}=\frac{dd}{dt}=\frac{5}{2}\]

Answer 21

there are more than one way to resolve this problem; i perfer to do it the easiest way :)

Answer 22

Wow, thats amazing. Are you a genius or something, lol?

Answer 23

my kids would say: or something ;)

Answer 24

Do you teach math?

Answer 25

depends, did I teach you?

Answer 26

Just now, for sure :)

Answer 27

then maybe :) good luck with it all

Answer 28

thanks, I've been so stressed since i have an exam tmrw. but thanks for taking the time to help me out :)