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    I was looking at past papers for my c2 exam next week, and I was stuck on question 7 of the june 2012 ocr mei paper.

    I've got the dy/dx as 4x^3/2 - 5x + c
    and the equation of the line as y - 20 = 7 (x-4)
    which equals y=7x-8

    and my final answer as 4x^3/2 - 5x - 8

    but the mark scheme says it's 4x^3/2 -5x + 8
    I don't understand how it's +8??
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    (Original post by alevelnerd123)
    I was looking at past papers for my c2 exam next week, and I was stuck on question 7 of the june 2012 ocr mei paper.

    I've got the dy/dx as 4x^3/2 - 5x + c
    and the equation of the line as y - 20 = 7 (x-4)
    which equals y=7x-8

    and my final answer as 4x^3/2 - 5x - 8

    but the mark scheme says it's 4x^3/2 -5x + 8
    I don't understand how it's +8??
    are you sure you didn't make a small mistake before you did this?(such as swapping signs?)
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    (Original post by Steelmeat)
    are you sure you didn't make a small mistake before you did this?(such as swapping signs?)
    it says the point is (4,20) when dy/dx is 6x^1/2 - 5 and when i put 4 into the equation the gradient was 7?
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    (Original post by alevelnerd123)
    I was looking at past papers for my c2 exam next week, and I was stuck on question 7 of the june 2012 ocr mei paper.

    I've got the dy/dx as 4x^3/2 - 5x + c
    and the equation of the line as y - 20 = 7 (x-4)
    which equals y=7x-8

    and my final answer as 4x^3/2 - 5x - 8

    but the mark scheme says it's 4x^3/2 -5x + 8
    I don't understand how it's +8??
    Could you post a link to the paper?
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    (Original post by alevelnerd123)
    I was looking at past papers for my c2 exam next week, and I was stuck on question 7 of the june 2012 ocr mei paper.

    I've got the dy/dx as 4x^3/2 - 5x + c
    and the equation of the line as y - 20 = 7 (x-4)
    which equals y=7x-8

    and my final answer as 4x^3/2 - 5x - 8

    but the mark scheme says it's 4x^3/2 -5x + 8
    I don't understand how it's +8??
    First of all, please link papers in the future or post a picture of the question.

    Second of all, there's no line involved here.

    You have y = 4x^(3/2) - 5x + c

    You know that (4, 20) lies on this curve. Plug it in.

    20 = 4 * 4^(3/2) - 5(4) + c

    Now solve for c. You'll find that you get 20 = 12 + c
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    (Original post by alevelnerd123)
    I was looking at past papers for my c2 exam next week, and I was stuck on question 7 of the june 2012 ocr mei paper.

    I've got the dy/dx as 4x^3/2 - 5x + c
    and the equation of the line as y - 20 = 7 (x-4)
    which equals y=7x-8

    and my final answer as 4x^3/2 - 5x - 8

    but the mark scheme says it's 4x^3/2 -5x + 8
    I don't understand how it's +8??
    Y=(4x^(3/2) -5x +c) substitute X and y
    20=4(8)-20 +c , c=8
    Y=(4x^(3-2) -5x +8?
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    Very Important Poster
    (Original post by alevelnerd123)
    I was looking at past papers for my c2 exam next week, and I was stuck on question 7 of the june 2012 ocr mei paper.

    I've got the dy/dx as 4x^3/2 - 5x + c
    and the equation of the line as y - 20 = 7 (x-4)
    which equals y=7x-8

    and my final answer as 4x^3/2 - 5x - 8

    but the mark scheme says it's 4x^3/2 -5x + 8
    I don't understand how it's +8??
    Please post a link or photo of the question - it's more convenient for people and it's not entirely clear what the question is asking.
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    (Original post by alevelnerd123)
    it says the point is (4,20) when dy/dx is 6x^1/2 - 5 and when i put 4 into the equation the gradient was 7?
    lol there's no line??? just sub the co-ordinates back into your expression for y= whatever +c
    (Original post by Someboady)
    Could you post a link to the paper?
    http://www.mei.org.uk/files/papers/c2_june_2012.pdf
    (Original post by Zacken)
    First of all, please link papers in the future or post a picture of the question.
    http://www.mei.org.uk/files/papers/c2_june_2012.pdf took me some time to find xD
    (Original post by SeanFM)
    Please post a link or photo of the question - it's more convenient for people and it's not entirely clear what the question is asking.
    http://www.mei.org.uk/files/papers/c2_june_2012.pdf
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    (Original post by Steelmeat)
    lol there's no line??? just sub the co-ordinates back into your expression for y= whatever +c
    I already said this...

    I already found the paper and answered the OP... hence the "in the future".
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    (Original post by Zacken)
    I already said this...



    I already found the paper and answered the OP... hence the "in the future".
    oops nvm
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    (Original post by Steelmeat)
    lol there's no line??? just sub the co-ordinates back into your expression for y= whatever +c

    http://www.mei.org.uk/files/papers/c2_june_2012.pdf

    http://www.mei.org.uk/files/papers/c2_june_2012.pdf took me some time to find xD

    http://www.mei.org.uk/files/papers/c2_june_2012.pdf
    Integrate and you should get:

    y = 4x ^ 3/2 -5x + c
    Substitute your value of x in and your value of y.
    Then you should get c as 20 - 12 = 8.
    You probably made a silly error in subbing in.
    Never mind, Zacken explained it above ^
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    (Original post by Someboady)
    Integrate and you should get:

    y = 4x ^ 3/2 -5x + c
    Substitute your value of x in and your value of y.
    Then you should get c as 20 - 12 = 8.
    You probably made a silly error in subbing in.
    Never mind, Zacken explained it above ^
    i didn't get the mistake xD
    i'm ok at C2 integration :angelwings:
 
 
 
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