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# C4 Integration Question watch

1. I already have answers for these, I'm just doubtful about the way I've worded them and it seems kinda wrong to me :/

Could someone work through these and show me how they'd do them? Thx in advance for any help

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2. ..
3. (Original post by Louisb19)
Rate of concentration is given by , the concentration is given by .

Since goes up as goes down (negatively proportional) the relationship can be modeled by the differential equation

, where k is a negative constant.

So the relationship can therefore me modelled by where k is positive constant.

ii)

Yeah, thanks for that. That's what I did woo

iii) Not sure exactly how you are supposed to handle the A and k constants together (if you are supposed to merge them or what) so I'll leave this for someone else.
Damn, this was the bit I needed the help with... I'm having the same problem in that you can't really define k due to that extra constant :/
4. Can you post your workings because for the general solution I got C=Ae^(-kt)
5. But anyway if at t=0 C=C_0
This means that the constant you have in front of the exponential term will be C_0.
At t=4, C= 0.1C_0
Then just solve for k using logarithms.
6. When I I integrate (cos (x))^4.
Why cant i use inverse chain rule and why cant i use inverse chain rule if i want to integrate,

(Cos (4x))^4
7. (Original post by justthatboy)
When I I integrate (cos (x))^4.
Why cant i use inverse chain rule and why cant i use inverse chain rule if i want to integrate,

(Cos (4x))^4
By using reverse chain rule here you will get a function that includes the product of cosines and sine. But if you differentiate a function that is a product of dimes and cosines you is the product rule - not the chain rule. So you would have to use the reverse product rule (integration by parts or use trig identities or substitution).
8. (Original post by Louisb19)
ii)
(Original post by Alexion)
Yeah, thanks for that. That's what I did woo
That's not correct, unfortunately, you get , so .

You know that , so .

Now, plug in the info for t=4 and you should be able to solve for k.
9. (Original post by Zacken)
That's not correct, unfortunately, you get , so .

You know that , so .

Now, plug in the info for t=4 and you should be able to solve for k.
Ah. Oops. Just a coincidence that Louis and I made the same mistake then haha

Thanks for the help
10. (Original post by Alexion)
Ah. Oops. Just a coincidence that Louis and I made the same mistake then haha

Thanks for the help
Louis Van Gaal?
11. (Original post by TeeEm)
Louis Van Gaal?
@Louisb19 ya silly poop

But who's to say that Van Gaal didn't make the same mistake in a maths A-level?
12. (Original post by Alexion)
@Louisb19 ya silly poop

But who's to say that Van Gaal didn't make the same mistake in a maths A-level?

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