Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

You could use implicit differentiation but normally I would just square root to get plus minus r and go from there like usual, hope that helps

I think you do this

find x^2 or y^2, where x^2= r^2sin^2(theta) and y^2=r^2sin^2(theta)

then you can differentiate implicitly, so for example you'd have 2y dy/d(theta) = f'(theta) => dy/d(theta) = f'(theta)/2y, and since it equals 0 the numerator must equal zero so you just put f'(theta)=0 and forget about the 2y

I think you do this

find x^2 or y^2, where x^2= r^2sin^2(theta) and y^2=r^2sin^2(theta)

then you can differentiate implicitly, so for example you'd have 2y dy/d(theta) = f'(theta) => dy/d(theta) = f'(theta)/2y, and since it equals 0 the numerator must equal zero so you just put f'(theta)=0 and forget about the 2y

I think you do this

find x^2 or y^2, where x^2= r^2sin^2(theta) and y^2=r^2sin^2(theta)

then you can differentiate implicitly, so for example you'd have 2y dy/d(theta) = f'(theta) => dy/d(theta) = f'(theta)/2y, and since it equals 0 the numerator must equal zero so you just put f'(theta)=0 and forget about the 2y

How would you solve this question:
Find the polar coordinates of the points on r²=a²sin2θ where the tangent is perp. to the initial line

Hope this helps, sorry it's so messy! Didn't know what the range of theta values was so haven't written a final answer, let me know if you need any more help

Theyve got the answers for θ to be pi/6 and pi/2. Thats what you got but how do you know which values to choose. Also what did you do with r =,-a root (sin2θ). Thanks

I can give a decent stab at them but would you say drawing polar graphs is fairly unlikely?

Would you happen to know how to go about 11b. I feel dumb because it's a hence question but I can't integrate with the sqrt and by substitution becomes awkward.

Not easy to explain other than looking at what they've done.

the 'middle term' for r = k is cancelled out by combining the 'last term' in r = k-1 and the 'first term' in r = k +1, so you work through up to r = 4 and you see that the only term at the start that isn't cancelling out is 1/2.

Then you work towards the end (starting from n-2).

So for r=n-2 everything cancels out (as you have the 'first term' below it and the 'last term' above it so the middle term is cancelled out, the 'first term' in r=n-2 is used up in cancelling out the previous 'middle term' and the last one is used in cancelling out the 'middle term' in r = n-1.



Similar logic in r=n-1 for the first and second term, but the third term is only half of the middle term in r = n so it only 'half cancels' it.

which is why you get 1/(n+1) - 2/(n+1) = -1/(n+1) as one of the terms in the summation.



and then for r=n the 'middle term' half cancels (as before) and the first one is used up, and the last one does not do anything as there needs to be something below it.

Which IAL papers are available??

Is it just
june 14
june 15
jan 16

https://8dedc505ac3fba908c50836f59059ccce5cd0f1e.googledrive.com/host/0B1ZiqBksUHNYdHIxUkJmdndfMlE/June%202009%20QP%20-%20FP2%20Edexcel.pdf

Question 6b, so |z| = 3 maps to the circle and now we have to sketch |z| < 3 - if anything id thought itd be inside the circle but its every but inside. Could someone explain this please?

Hi guys, I don't understand question 6a of June 2014 R. I've looked at the mark scheme but I still don't get it. Could someone show me the working please? I would be extremely grateful

Hi guys, I don't understand question 6a of June 2014 R. I've looked at the mark scheme but I still don't get it. Could someone show me the working please? I would be extremely grateful

This is what I think you're meant to do since it asks to estimate.