Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    Offline

    11
    ReputationRep:
    (Original post by kkboyk)
     \frac{x}{2-x} > \frac{x+3}{x}
    I would probably do this by taking everything over to the left and looking for critical values
    Offline

    2
    ReputationRep:
    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.
    Offline

    12
    ReputationRep:
    (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
    hi what is the answer

    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
    Offline

    3
    ReputationRep:
    (Original post by Cpj16)
    hi what is the answer

    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
    Offline

    12
    ReputationRep:
    (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
    Offline

    17
    ReputationRep:
    (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
    Offline

    2
    ReputationRep:
    (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).
    Offline

    2
    ReputationRep:
    (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
    Offline

    6
    ReputationRep:
    Attachment 543385543387Name:  image.jpg
Views: 96
Size:  525.2 KB
    (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
     
    Offline

    17
    ReputationRep:
    (Original post by economicss)
    Thank you! The answer is (u-1/2)^2 + (v-1/2)^2= 1/2 thanks
    Name:  1465070002791.jpg
Views: 86
Size:  21.9 KB

    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by Rkai01)
    Attachment 543385543387Name:  image.jpg
Views: 96
Size:  525.2 KB

    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?
    Offline

    3
    ReputationRep:
    Thanks so much
    Offline

    2
    ReputationRep:
    (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
    Offline

    6
    ReputationRep:
    (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
    Offline

    3
    ReputationRep:
    (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!
    Offline

    3
    ReputationRep:
    (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
    Offline

    3
    ReputationRep:
    (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
    Offline

    2
    ReputationRep:
    Anyone know how to solve the polar equation where its r squared and you have to find the coordinates of perpendicular or parallel line?
    Offline

    3
    ReputationRep:
    (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
    Offline

    11
    ReputationRep:
    (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
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: February 24, 2017
Poll
Do you like carrot cake?

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

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.