The Student Room Group

C1 maths help:

The only thing I know I should do is integrate the dy/dx equation, but then I don't really know what to do afterwards :/

(edited 9 years ago)
I think you've uploaded the wrong image. The stuff you've typed does not match the image(text is integration, image is surds).
Original post by morgan8002
I think you've uploaded the wrong image. The stuff you've typed does not match the image(text is integration, image is surds).


Oops sorry, it's fixed now :smile:
To find the equation of a tangent, you need to find the gradient of the tangent and have a point on the line. Does this help?


The first step is to find the equation of the tangent. Integrating comes later in the question.
Original post by morgan8002
To find the equation of a tangent, you need to find the gradient of the tangent and have a point on the line. Does this help?


The first step is to find the equation of the tangent. Integrating comes later in the question.


I managed to get the gradient but I didn't know what the y coordinate would be so I could form and equation?
Original post by creativebuzz
I managed to get the gradient but I didn't know what the y coordinate would be so I could form and equation?


The question says that the tangent passes through the origin, so you can use (0, 0) as a point on the tangent.

You need to use all of the information that the question gives you, because they will never give you more information than you need.
Reply 6
F'(x) = 3(x)^2 - 4x -1

F(x) (which is equal to y) is just the integral of f'(x)

Then set up an equation involving y and sub in x=2 to get the y value innit :yes:

Posted from TSR Mobile
Original post by Fermions
F'(x) = 3(x)^2 - 4x -1

F(x) (which is equal to y) is just the integral of f'(x)

Then set up an equation involving y and sub in x=2 to get the y value innit :yes:

Posted from TSR Mobile

You are left with an arbitrary constant of integration, so you have to use the other information given in the question to find the value of the constant and thus the equation of the line.
Reply 8
Original post by morgan8002
You are left with an arbitrary constant of integration, so you have to use the other information given in the question to find the value of the constant and thus the equation of the line.


Dayuum :frown:

Posted from TSR Mobile


That is exactly ​where I got stuck, I then thought of using the (0,0) coordinates with that method but that didn't give me the right answer and I don't understand why it didn't.. :/
(edited 9 years ago)
Original post by creativebuzz
That is exactly ​where I got stuck

So have you managed to find the equation of the tangent?
Original post by morgan8002
So have you managed to find the equation of the tangent?


Well my initial method was to
1) find gradient = 3

2) then the eq of the tangent: y - 0 = 3(x -0)
to give me y=3x

but I wasn't confident in my answer :/
To do this you must first find the equation of the tangent. You would try to find the gradient l. You will find this to be 3 i think. You would use 3 and also the point (0,0) to form an equation of the tanget. You will find this to be y=3x. Next you would sub in 2 to get a y value of 6. Then you would integrate the given equation and subsitute in 2 as the x value and 6 as the y value to find the constant.

Hope this helps :smile:

Posted from TSR Mobile
Original post by creativebuzz
Well my initial method was to
1) find gradient = 3

2) then the eq of the tangent: y - 0 = 3(x -0)
to give me y=3x

but I wasn't confident in my answer :/

Yes that is correct. Now use this to find the y coordinate at x = 2, then use this point to find an equation of the curve.
Original post by ImNervous
To do this you must first find the equation of the tangent. You would try to find the gradient l. You will find this to be 3 i think. You would use 3 and also the point (0,0) to form an equation of the tanget. You will find this to be y=3x. Next you would sub in 2 to get a y value of 6. Then you would integrate the given equation and subsitute in 2 as the x value and 6 as the y value to find the constant.

Hope this helps :smile:

Posted from TSR Mobile


Yes, then substitute this constant into the equation of the curve to find the final answer.
Original post by ImNervous
To do this you must first find the equation of the tangent. You would try to find the gradient l. You will find this to be 3 i think. You would use 3 and also the point (0,0) to form an equation of the tanget. You will find this to be y=3x. Next you would sub in 2 to get a y value of 6. Then you would integrate the given equation and subsitute in 2 as the x value and 6 as the y value to find the constant.

Hope this helps :smile:

Posted from TSR Mobile


Brilliant! Positive rating given for thanks :smile::smile:

Quick Reply

Latest