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Stuck on dy/dx question?

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??
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?)
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?
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?
Reply 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??


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
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?
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.
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
Reply 8
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".
Original post by Zacken
I already said this...



I already found the paper and answered the OP... hence the "in the future".


oops nvm :smile:
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 ^
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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