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# Stuck on dy/dx question?

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1. 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??
2. (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?)
3. (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?
4. (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?
5. (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
6. (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?
7. (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.
8. (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
9. (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".
10. (Original post by Zacken)
I already said this...

I already found the paper and answered the OP... hence the "in the future".
oops nvm
11. (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 ^
12. (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

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