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C1 Question Help

Hi, could someone please help with this differentiation C1 question?

"Find the equation of the tangent to the curve at the given point:
c) y=1/(3x^2) at (1/3, 3)

I'm not sure where I'm going wrong!

when I make y=(3x^2)^-1
then so y = (1/3)x^-2
so dy/dx =(-2/3)x^-3
= -2/3x^3

subbing x=1/3 into the derivative I get
dy/dx = -2/1
= -2 ????????????

y - 3 = -2(x-1/3)
y - 3 = -2x +2/3
y = (11/3)- 2x

or 3y + 2x = 11 ?

But the answer in the book is y + 18x = 9 ?!?

It's from OCR (non MEI) exc 8A 3c (c1)

Attached the question as well

Thanks in advanced
(edited 8 years ago)
12166043_10206761993781526_1951470620_n.jpg55.2KB
Original post by Jitesh
Hi, could someone please help with this differentiation C1 question?

"Find the equation of the tangent to the curve at the given point:
c) y=1/(3x^2) at (1/3, 3)

I'm not sure where I'm going wrong!

when I make y=(3x^2)^-1
then so y = (1/3)x^-2
so dy/dx =(-2/3)x^-3
= -2/3x^-3

subbing x=1/3 into the derivative I get
dy/dx = -2/1
= -2 ????????????

y - 3 = -2(x-1/3)
y - 3 = -2x +2/3
y = (11/3)- 2x

or 3y + 2x = 11 ?

But the answer in the book is y + 18x = 9 ?!?

It's from OCR (non MEI) exc 8A 3c (c1)

Attached the question as well

Thanks in advanced


It's where you've subbed x into dy/dx - the expression for dy/dx is correct but what you got out isn't.
Reply 2
Avatar for Jitesh
Jitesh
OP
14
Original post by SeanFM
It's where you've subbed x into dy/dx - the expression for dy/dx is correct but what you got out isn't.


Was it the -2/3x^-2 bit? I realised I think I typed that wrong as it should have been -2x^-2/3 or -2/3x^2

I think I did that in my working out, I'll double check now
Reply 3
Avatar for Jitesh
Jitesh
OP
14
Wait I see where I went wrong, redid the calculation and got the gradient as -6, but wouldn't the gradient be -18 so is my dy/dx expression wrong?
Original post by Jitesh
Was it the -2/3x^-2 bit? I realised I think I typed that wrong as it should have been -2x^-2/3 or -2/3x^2

I think I did that in my working out, I'll double check now


Not at all. dy/dx is exactly what you've written. It's just that you've put in x = 1/3, which is also correct, but got the wrong number out.

It may help to break it down and show your steps as to how you put in x = 1/3.
Original post by Jitesh
Wait I see where I went wrong, redid the calculation and got the gradient as -6, but wouldn't the gradient be -18 so is my dy/dx expression wrong?


Your expression isn't wrong - please show how you put in x = 1/3 and we'll see what's going on, as you should indeed be getting -18 out and the only thing that's going wrong is the actual evaluation.
Reply 6
Avatar for Jitesh
Jitesh
OP
14
Original post by SeanFM
Not at all. dy/dx is exactly what you've written. It's just that you've put in x = 1/3, which is also correct, but got the wrong number out.

It may help to break it down and show your steps as to how you put in x = 1/3.


Oh thanks Sean12166423_10206762118584646_424537295_n.jpg
I tried skipping steps and did x^2 which was giving me 1/9 in my head

Thanks again
Original post by Jitesh
Oh thanks Sean12166423_10206762118584646_424537295_n.jpg
I tried skipping steps and did x^2 which was giving me 1/9 in my head

Thanks again


Perfect! Well done Jitesh :borat:
Reply 8
Avatar for Jitesh
Jitesh
OP
14
Original post by SeanFM
Perfect! Well done Jitesh :borat:


Thank you! Sorry about that rookie error!

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