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# Chain rule of differentiation problem watch

1. OK for once I'm stuck. Here's the problem.

V = 200t and V = 4πr³/3

Firstly find dV/dt and dV/dr, which I did and got:

dV/dt = 200
dV/dr = 4πr²

Then find an expression for dr/dt if dV/dt = dV/dr*dr/dt

This I did and got dr/dt = 50/πr²

Now my problem is that I don't have an expression in terms of t but in r. Either I've gone wrong or it's OK for this to happen, but the next question says find dr/dt when t=1...

Help?
2. (Original post by XTinaA)
OK for once I'm stuck. Here's the problem.

V = 200t and V = 4πr³/3

Firstly find dV/dt and dV/dr, which I did and got:

dV/dt = 200
dV/dr = 4πr²

Then find an expression for dr/dt if dV/dt = dV/dr*dr/dt

This I did and got dr/dt = 50/πr²

Now my problem is that I don't have an expression in terms of t but in r. Either I've gone wrong or it's OK for this to happen, but the next question says find dr/dt when t=1...

Help?
Purely random (As I haven't done any of this stuff)

V = 200t and V = 4πr³/3

200t = 4πr³/3
t = (4πr³/3)/200

Now sub this back into something?
3. (Original post by bono)
Purely random (As I haven't done any of this stuff)

V = 200t and V = 4πr³/3

200t = 4πr³/3
t = (4πr³/3)/200

Now sub this back into something?
Decided it was too unwieldy and thought there has to be another way...
4. (Original post by XTinaA)
Decided it was too unwieldy and thought there has to be another way...
You have a solution for t though, surely you could then use this to find what you need?

*pretends to have any vague idea despite never having done this before*
5. (Original post by bono)
You have a solution for t though, surely you could then use this to find what you need?

*pretends to have any vague idea despite never having done this before*
Have you not done the chain rule? It's P2...
6. (Original post by XTinaA)
Have you not done the chain rule? It's P2...
Its P3 on our exam board..
7. (Original post by XTinaA)
Have you not done the chain rule? It's P2...
Not on AQA B - We were briefly taught the most absolute basic concept and use of it additionally by our teacher, but not for this differentiation stuff that you are doing.

Not on my syllabus.
8. Good old OCR. P2 in that syllabus. From what I see, theres nothing wrong. But then again, you didn't state the question. And thats an important bit.
9. (Original post by bono)
Not on AQA B - We were briefly taught the most absolute basic concept and use of it additionally by our teacher, but not for this differentiation stuff that you are doing.

Not on my syllabus.
Well allow me to demonstrate for you.

The chain rule (or function of a function rule) enables you to differentiate composite functions. The rule goes:

dy/dx = dy/du.du/dx

Example:
y = (2x+1)³
Let u = 2x+1, so y = u³
then du/dx = 2 and dy/du = 3u²
So dy/dx = 6(2x+1)²
10. (Original post by 2776)
Good old OCR. P2 in that syllabus. From what I see, theres nothing wrong. But then again, you didn't state the question. And thats an important bit.
It's just that I need an expression in t so I can work out the rate of change of r when t=1.
11. t= (pi r^3)/150
12. (Original post by 2776)
t= (pi r^3)/150
How?
13. Eliminating V, r = (150/pi)^(1/3) t^(1/3).

So dr/dt = (1/3) (150/pi)^(1/3) t^(-2/3).

So dr/dt = (1/3) (150/pi)^(1/3) when t = 1.

---
OR
---

As you said, dr/dt = 50/(pi r^2).

But when t = 1 we have V = 200, so 200 = (4/3) pi r^3,
so r = (150/pi)^(1/3), so

dr/dt
= 50/(pi (150/pi)^(2/3))
= 50/(pi^(1/3) 150^(2/3))
= (50/150^(2/3)) pi^(-1/3)
= (50*150^(1/3) / 150) pi^(-1/3)
= (150^(1/3) / 3) pi^(-1/3).

Same!
14. (Original post by XTinaA)
How?
Sim eqn with the 2 eqns
15. (Original post by 2776)
Sim eqn with the 2 eqns
Sorry, I was confused as to the 150 bit but I got it.

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