The Student Room Group

Differentation help

So for the ENGAA 2016 Section 2 Q2d, I don't really understand why it says "You may find it helpful to use the fact that any value of R_3 that maximises P also minimises 1/P"
Differentiating P (from part 2c) or 1/P and making them = 0 both give you the same answer that P is a minimum when R_3 = (R_1)/2, so why would that statement be helpful? Thanks in advance


Links to solutions and questions
https://www.physicsandmathstutor.com/admissions/engaa/solutions-2016/

https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FAdmissions%2FENGAA%2FPapers%2FENGAA%25202016%2520Section%25202.pdf
Reply 1
Original post by HorizonOasis
So for the ENGAA 2016 Section 2 Q2d, I don't really understand why it says "You may find it helpful to use the fact that any value of R_3 that maximises P also minimises 1/P"
Differentiating P (from part 2c) or 1/P and making them = 0 both give you the same answer that P is a minimum when R_3 = (R_1)/2, so why would that statement be helpful? Thanks in advance


Links to solutions and questions
https://www.physicsandmathstutor.com/admissions/engaa/solutions-2016/

https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FAdmissions%2FENGAA%2FPapers%2FENGAA%25202016%2520Section%25202.pdf


Looks like for P youd differentiate using the quotient rule then argue that numerator = 0. For 1/P you can do some algebraic simplification before the differentiation, as per the mark scheme, which is a bit simpler, but tbh there isnt much in it either way and as you say obviously you have to get the same answer.

A slightly different way to do the 1/P one is to note its simplified form is proportional to
(2sqrt(R3) + R1/sqrt(R3))^2
This is minimized when
2sqrt(R3) = R1 / sqrt(R3)
or R3 = R1/2. So it can be done calculus free but not sure its much easier.
(edited 8 months ago)
Reply 2
Original post by mqb2766
Looks like for P youd differentiate using the quotient rule then argue that numerator = 0. For 1/P you can do some algebraic simplification before the differentiation, as per the mark scheme, which is a bit simpler, but tbh there isnt much in it either way and as you say obviously you have to get the same answer.

A slightly different way to do the 1/P one is to note its simplified form is proportional to
(2sqrt(R3) + R1/sqrt(R3))^2
This is minimized when
2sqrt(R3) = R1 / sqrt(R3)
or R3 = R1/2. So it can be done calculus free but not sure its much easier.


Oh ok, thanks very much :smile:
Reply 3
Original post by HorizonOasis
Oh ok, thanks very much :smile:


Thinking about it a bit more, like the model soln I did the last one the "long" way (expand, flip then factorise) and the simplest way is divide num and denom by R3 at the start (keeping it in factorised form) to get
1/P = (2sqrt(R3)+R1/sqrt(R3))^2 / V^2
which is minimised when 2sqrt(R3)=R1/sqrt(R3). So you dont have to expand the quadratic then differentiating ... as the model solution does, though it does rely on the am-gm trick.
(edited 8 months ago)

Quick Reply

Latest