Question

Show that if the volume of a balloon is decreasing at a rate proportional to its surface area, the radius of the balloon is shrinking at a constant rate.

Answer 1

- hope you know that a balloon has a geometric form of a sphere
- so in this case you need to know formula for the volume of a sphere and formula for surface area of a sphere

- using these two formulas you need prove this rate proportional

Answer 2

My calculus is a bit rusty so bear with me

Since we’re dealing with rates of change, we’ll be using derivatives

dV/dt gives the volume change wrt t

Since the rate of decrease is proportional to surface area, dV/dt = k(SA)

Surface area of a sphere is 4(pi)(r)^2

Therefore dV/dt = k * 4pi(r^2)

Find dr/dt through implicit differentiation wrt t and you’ll notice that dr/dt ends up being a constant if you do it correctly

Answer 3

Then what variable would you be solving for?

Answer 4

It’s not asking for a variable, it’s asking to prove the radius rate of change is a constant

Answer 5

@Vocaloid yes i agree it but i start - begin - it always that who asked dont learned again about derivare - how calcule the derivative

Answer 6

so in this situation how you solve it ?

Answer 7

How would you approach this without derivatives ? (Asking our of curiosity)

Answer 8

You would have to solve using a equation and @jhonyy9 knows how to do that equation.

Answer 9

- this is true that in my birth country i ve learned about derivatives in my 3rd class of my high school but here in Hungary dont learn about derivare just in university or faculty - college

Answer 10

Ok so the first thing is you need to have a example equation to reference off of because that will help you a lot.

Answer 11

Also what @Vocaloid said above about having a constant if you dont have a constant then every time you will get the answer wrong.

Answer 12

@jhonyy9 would you mind sharing the non calculus solution sometime? (You can pm me if you don’t want to give away the answer)

Answer 13

V = (4pi*r^3)/3

A = 4pi*r^2

\[\frac{ V }{ A } = \frac{ \frac{ 4\pi*r^3 }{ 3 } }{ 4\pi*r^2 }\]

\[\frac{ V }{ A } = \frac{ r }{ 3}\]

@Vocaloid

Answer 14

That just shows that volume and surface area are proportional for constant r. I don’t see how that proves that r changes at a constant rate when the volume rate of change is proportional to SA.

Answer 15

maybe this with

\[\frac{ \delta V }{ \delta A } = \frac{ \delta r }{ 3 }\]

\[\frac{ V_2 - V_1 }{ A_2 - A_1 } = \frac{ (r_2 -r_1) }{ 3 }\]

Answer 16

so result that the constant is 1/3

Answer 17

@Vocaloid really i agree what you ve said above with derivative but how you solve this problem when the student who asked dont learn again about derivare ?

Answer 18

Ok thank you, I was genuinely just curious about your method

Answer 19

@Timmyspu

Answer 20

np

yw

like a first idea i ve thought it like you but i always i think in this second way what is in case then who asked dont learn again about calculus - in this case how maybe possibile solve the posted problem ???

thank you all your collaboration

Answer 21

No problem!

Answer 22

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
maybe this with

\[\frac{ \delta V }{ \delta A } = \frac{ \delta r }{ 3 }\]

\[\frac{ V_2 - V_1 }{ A_2 - A_1 } = \frac{ (r_2 -r_1) }{ 3 }\]
\(\color{#0cbb34}{\text{End of Quote}}\)
this is nonsense in so many ways

Answer 23

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
so result that the constant is 1/3
\(\color{#0cbb34}{\text{End of Quote}}\)
sheesh, so it's always 1/3, fixed in time forever?!

Answer 24

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
V = (4pi*r^3)/3

A = 4pi*r^2

\[\frac{ V }{ A } = \frac{ \frac{ 4\pi*r^3 }{ 3 } }{ 4\pi*r^2 }\]

\[\frac{ V }{ A } = \frac{ r }{ 3}\]

@Vocaloid
\(\color{#0cbb34}{\text{End of Quote}}\)
look here pls

Answer 25

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
maybe this with

\[\frac{ \delta V }{ \delta A } = \frac{ \delta r }{ 3 }\]

\[\frac{ V_2 - V_1 }{ A_2 - A_1 } = \frac{ (r_2 -r_1) }{ 3 }\]
\(\color{#0cbb34}{\text{End of Quote}}\)
and continue here

Answer 26

and from this result this 1/3 constant

Answer 27

you were supposed to show that the answer is k, the constant.

not 1/3. a very specific answer, don't you think?

Answer 28

in this case of this above posted problem the k constant will be always k= 1/3

check it please

Answer 29

\(V = {4 \over 3} \pi r^3 \qquad S = 4 \pi r^2\)

\(dV = 4 \pi r^2 ~ dr\qquad dS = 8 \pi r ~ dr\)

\(\frac{dV}{dt} = - k S\)

\(\implies \frac{4 \pi r^2 ~ dr}{dt} = - k (4 \pi r^2)\)

\(\implies \frac{ dr}{dt} = - k\)

That is what you are supposed to demonstrate.

What you are posting here is pure nonsense, sadly.

Answer 30

\( d\left( \dfrac{A}{B} \right) \ne \dfrac{dA}{dB} \)

think about it.