# Rates of change exam question

#### Sophie02

##### New member
Hi,

I was wondering how to do this question? Everything I seem to try, the result never seems to be the same as the equation that you need to prove? And then for part b, I don’t know what bits of data I would be using to go where?

thankyou

#### Attachments

• 199.5 KB Views: 17

#### lex

##### Full Member
Do you want to post an example of what you are doing for (a) and we can see what is happening? It may simply be a matter of getting your work to 'look like' their answer.

#### HallsofIvy

##### Elite Member
In order to get that equation, the balloon is a sphere and the volume of a sphere of radius r is $$\displaystyle V= (4/3)\pi r^3$$. Then $$\displaystyle dV/dr= 4\pi r^2$$ and $$\displaystyle dV/dt= 4\pi r^2 dr/dt$$. If that is a constant, C, the $$\displaystyle 4\pi r^2 dr/dt= C$$ so that $$\displaystyle dr/dt=C/4\pi r^2= k/r^2$$ with $$\displaystyle k= C/4\pi$$ a constant.

#### lex

##### Full Member
...so that $$\displaystyle dr/dt=C/4\pi r^2= k/r^2$$ with $$\displaystyle k= C/4\pi$$ a constant.
$$\displaystyle C$$ is negative, but $$\displaystyle k$$ is positive. So $$\displaystyle k= \boldsymbol{-C}/4\pi \hspace3ex$$( and $$\displaystyle \tfrac{dr}{dt}=-\tfrac{k}{r^2}$$)

Last edited:

#### Sophie02

##### New member
In order to get that equation, the balloon is a sphere and the volume of a sphere of radius r is $$\displaystyle V= (4/3)\pi r^3$$. Then $$\displaystyle dV/dr= 4\pi r^2$$ and $$\displaystyle dV/dt= 4\pi r^2 dr/dt$$. If that is a constant, C, the $$\displaystyle 4\pi r^2 dr/dt= C$$ so that $$\displaystyle dr/dt=C/4\pi r^2= k/r^2$$ with $$\displaystyle k= C/4\pi$$ a constant.
Ohh that makes more sense now because i was getting 4pi and I didn’t know how to get rid of it. But it’s just because it’s a constant included within k

#### Sophie02

##### New member
How would you go about doing part b?

#### Subhotosh Khan

##### Super Moderator
Staff member
How would you go about doing part b?
What have YOU tried for this EXAM problem?

Are you allowed to ask for external assistance?

#### lex

##### Full Member
Ohh that makes more sense now because i was getting 4pi and I didn’t know how to get rid of it. But it’s just because it’s a constant included within k
Do you want to post an example of what you are doing for (a) and we can see what is happening? It may simply be a matter of getting your work to 'look like' their answer.
That's what I suspected and why I offered to look at your work if you posted it.

#### Sophie02

##### New member
What have YOU tried for this EXAM problem?

Are you allowed to ask for external assistance?
Yes I’m allowed external help. I’m just doing practise papers for revision!

#### Subhotosh Khan

##### Super Moderator
Staff member
Yes I’m allowed external help. I’m just doing practise papers for revision!
OK

So where is your work/thoughts regarding this problem?

#### Sophie02

##### New member
but as you can see I don’t have an equation so I’ve definitely don it wrong

#### Subhotosh Khan

##### Super Moderator
Staff member
but as you can see I don’t have an equation so I’ve definitely don it wrong
V= (4/3)*π*r^3 → dV/dr = 4 * π * r^2

so the DE is:

(-80*π) = 4 * π * r^2 * dr/dt

There is your equation (DE) ...... continue......

#### Sophie02

##### New member
Hi, can anyone tell me whether this is correct for part b? It doesn’t look quite right!

#### Attachments

• 199.5 KB Views: 2
• 186.7 KB Views: 2

#### Sophie02

##### New member
V= (4/3)*π*r^3 → dV/dr = 4 * π * r^2

so the DE is:

(-80*π) = 4 * π * r^2 * dr/dt

There is your equation (DE) ...... continue......
is this correct?

#### Attachments

• 186.7 KB Views: 3

#### Subhotosh Khan

##### Super Moderator
Staff member
is this correct?
Should the 'r' (radius) increase or decrease as 't' (time) increases?

Do you get r = 20 at t = 5 from the equation you have derived?

and

Do you get r = 40 at t = 0 from the equation you have derived?