C1 Homework helping

How would I find the equation for l. I have part a but I need b and c.

(edited 7 years ago)

So you're finding the gradient of the curve at the origin, (0, 0)
You have a point, you have the gradient of the line, that's all you need to find the equation of the line

\ell

Reply 2

Youngey4

Original post by B_9710

So you're finding the gradient of the curve at the origin, (0, 0)
You have a point, you have the gradient of the line, that's all you need to find the equation of the line

\ell

How do I find the gradient with only one point?

Reply 3

username2339433

If I remember correctly you differentiate twice then sub in the x co-ordiante to get the gradient of the curve at that point.

Reply 4

Youngey4

Original post by TheStudent19

If I remember correctly you differentiate twice then sub in the x co-ordiante to get the gradient of the curve at that point.

I know how to differentiate but this is only c1 chapters 1-5 so we havent done differentiation in core, just in FP1. Thats what confused me.

Reply 5

username2339433

Original post by Youngey4

I know how to differentiate but this is only c1 chapters 1-5 so we havent done differentiation in core, just in FP1. Thats what confused me.

Hmm good point mate. What year paper is this? Is it edexcel?

Reply 6

Youngey4

Original post by TheStudent19

Hmm good point mate. What year paper is this? Is it edexcel?

Yeah its edexcel but i'm not sure what year as its just a booklet our teacher put together to do over half term

Reply 7

username2339433

The only way I would go about doing b is to do c first. To do c you make the two equations equal and then find the co-ordinates I think.

EDIT: Just realised that you can't do that. Give me another minute :biggrin:

Do you have the answers?

Reply 8

B_9710

Original post by Youngey4

I know how to differentiate but this is only c1 chapters 1-5 so we havent done differentiation in core, just in FP1. Thats what confused me.

There is another way to do it without differentiation, although I do think that this question is designed so that you use differentiation.
You have

y=x^3+3x^2-4x

and suppose we have a line through the origin

y=ax

. Then solving these simultaneously you get

x^3+3x^2-(4+a)x=0

right.
Now we know that the line is tangent to the curve at the origin.So you can write

x^3+3x^2-(4+a)x

in the form

x^2(x-c)

since you know that there is a repeated root at the origin - which is where the

x^2

comes from.
So this is an identity,

x^3+3x^2-(4+a)x\equiv x^2(x-c)

.
Should be straight forward to solve for a and c and so both parts of the question are done.
This way is similar to when you use the discriminant to find points where a line is tangent to a parabola.
In fact for this question you could use the discriminant of a cubic to get the answer but it's quite long and most people don't know it off the top of their heads I'd imagine.