x Turn on thread page Beta
 You are Here: Home >< Maths

# Edexcel FP2 Official 2016 Exam Thread - 8th June 2016 watch

1. (Original post by kkboyk)
I would probably do this by taking everything over to the left and looking for critical values
2. Hey. I was wondering if someone could clear something up for me regarding Taylor series. When using (x - a), do you sub in positive a or negative (-a) into your differentials? With the two examples in the book one does positive an the other does negative, but from most exam questions I've seen you sub in positive a. Thanks.
3. (Original post by economicss)
Please could anyone do me a massive favour and post a worked solution to question 16b as I always seem to just get stuck at the same stage in these types, thanks Attachment 543353

I tried but I don't think my answer is right. I got 8u^2 + 8v^2 =0 then gave up xD

What is the correct answer I ll have another go
4. (Original post by Cpj16)

I tried but I don't think my answer is right. I got 8u^2 + 8v^2 =0 then gave up xD

What is the correct answer I ll have another go
Thank you! The answer is (u-1/2)^2 + (v-1/2)^2= 1/2 thanks
5. (Original post by economicss)
Thank you! The answer is (u-1/2)^2 + (v-1/2)^2= 1/2 thanks
I m sorry I cant get it I have tried 4 times I m so close

will try again later
6. (Original post by economicss)
Thank you! The answer is (u-1/2)^2 + (v-1/2)^2= 1/2 thanks
I'll try now.

Posted from TSR Mobile
7. (Original post by LikeASomebody)
Hey. I was wondering if someone could clear something up for me regarding Taylor series. When using (x - a), do you sub in positive a or negative (-a) into your differentials? With the two examples in the book one does positive an the other does negative, but from most exam questions I've seen you sub in positive a. Thanks.
If its (x-a), you substitute "a", that is the positive value. If its (x+a), then it becomes "-a", since that would made it (x+a) ; this is when they ask you to expand it in ascending powers of (x-a).
8. (Original post by techfan42)
If its (x-a), you substitute "a", that is the positive value. If its (x+a), then it becomes "-a", since that would made it (x+a) ; this is when they ask you to expand it in ascending powers of (x-a).
Thank you
9. Attachment 543385543387
(Original post by economicss)
Please could anyone do me a massive favour and post a worked solution to question 16b as I always seem to just get stuck at the same stage in these types, thanks Attachment 543353
Edit: can you read it alright?
Attached Images

10. (Original post by economicss)
Thank you! The answer is (u-1/2)^2 + (v-1/2)^2= 1/2 thanks

Posted from TSR Mobile
11. (Original post by Rkai01)
Attachment 543385543387

Edit: can you read it alright?
Thank you, that's great! How do we know that the real and imaginary parts are the same please?
12. Thanks so much
13. (Original post by economicss)
Thank you, that's great! How do we know that the real and imaginary parts are the same please?
From the first part. When you remove the modulus sign and expand, you find out that x=y
14. (Original post by economicss)
Thank you, that's great! How do we know that the real and imaginary parts are the same please?
Because locus of p is y=x and transformation maps this meaning all real parts equal imaginary parts and you use what z equals in terms of u and v then equate
15. (Original post by Rkai01)
Because locus of p is y=x and transformation maps this meaning all real parts equal imaginary parts and you use what z equals in terms of u and v then equate
I see, thanks so much!
16. (Original post by techfan42)
From the first part. When you remove the modulus sign and expand, you find out that x=y
Got it, thank you
17. (Original post by Cpj16)
I m sorry I cant get it I have tried 4 times I m so close

will try again later
Thanks so much for trying, appreciate it
18. Anyone know how to solve the polar equation where its r squared and you have to find the coordinates of perpendicular or parallel line?
19. (Original post by fpmaniac)
Anyone know how to solve the polar equation where its r squared and you have to find the coordinates of perpendicular or parallel line?
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
20. (Original post by fpmaniac)
Anyone know how to solve the polar equation where its r squared and you have to find the coordinates of perpendicular or parallel line?
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

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 24, 2017
Today on TSR

### Complete university guide 2019 rankings

Find out the top ten here

### Can I go to freshers even if I'm not at uni?

Poll
Useful resources

Can you help? Study help unanswered threadsStudy Help rules and posting guidelinesLaTex guide for writing equations on TSR

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE