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    Question 7 bii) on this paper here: http://filestore.aqa.org.uk/subjects...4-QP-JAN13.PDF

    I'm unsure as to why when you want to find the maximum rate of growth you don't but dN/dt = 0 and factorise it to find the stationary values of N, and substitute them back into the original equation. I know this is wrong because you end up with an e^-t/8 = 0, but I don't understand why.

    However according to the mark scheme you find the second differential and put that = 0 to find the value of N at which the maximum rate of growth occurs, could somebody please explain why this is the case?

    Thanks for any help anyone can give
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    (Original post by danbroown)
    Question 7 bii) on this paper here: http://filestore.aqa.org.uk/subjects...4-QP-JAN13.PDF

    I'm unsure as to why when you want to find the maximum rate of growth you don't but dN/dt = 0 and factorise it to find the stationary values of N, and substitute them back into the original equation. I know this is wrong because you end up with an e^-t/8 = 0, but I don't understand why.

    However according to the mark scheme you find the second differential and put that = 0 to find the value of N at which the maximum rate of growth occurs, could somebody please explain why this is the case?

    Thanks for any help anyone can give
    The keyword is 'maximum' in 'maximum rate of change'. The rate of change can take a range of values, given by dN/dt, but the maximum will be where the rate of change of the rate of change (the second differential) is equal to 0.
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    (Original post by SeanFM)
    The keyword is 'maximum' in 'maximum rate of change'. The rate of change can take a range of values, given by dN/dt, but the maximum will be where the rate of change of the rate of change (the second differential) is equal to 0.
    Thank you, that helps
 
 
 
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