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Rate of growth C4 questions

I'm doing Q7bi and 7bii on the following paper:

QP- http://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2013/january/AQA-MPC4-QP-JAN13.PDF

Ms- http://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2013/january/AQA-MPC4-W-MS-JAN13.PDF

I understand the first line of the working out on 7bi but can't follow the algebra.. I get that you're working towards the form given in the question but I can't even get to an equivalent form :frown:

I guess this is just my algebra rather than my methods but I'd appreciate if anyone could try and explain what the mark scheme is doing!

I also don't understand the method on 7bii- if it is the maximum rate of growth, what does this mean in terms of the equation of dN/dt?

Thanks to anyone who can help at all :smile:

Reply 1

RDKGames

Study Forum Helper

Original post by jazz_xox_

I guess this is just my algebra rather than my methods but I'd appreciate if anyone could try and explain what the mark scheme is doing!

Second line: from the eq. for

N

, you get that

(1+9e^{-t/8})N = 500

hence

1+9e^{-t/8} = \frac{500}{N}

. So from the first line you use this and sub it in for the bracket with exponent of -2. Also, taking the last from the first sentence here, you can say that

9e^{-t/8} = \frac{500}{N}-1

hence

\frac{9}{8}e^{-t/8} = \frac{1}{8}(\frac{500}{N}-1)

so you can replace this into the other bracket.

After that it should be clear (?)

I also don't understand the method on 7bii- if it is the maximum rate of growth, what does this mean in terms of the equation of dN/dt?

Thanks to anyone who can help at all :smile:

You're maximising a rate of growth, so you need to differentiate the rate of growth and set the derivative = 0 for the extremals. The solution yields a max point.

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