C1 Help!

Keep getting stuck on these types of questions:

10b:

https://a9497d7f220e174d38d014b2b1d86a68bedc42a8.googledrive.com/host/0B1ZiqBksUHNYNExoNEtSS1ViN2c/for-Edexcel/Solomon%20L%20QP%20-%20C1%20Edexcel.pdf

Could someone please explain?

Thanks

what's the gradient of the equation you got for part A?


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But what confuses me is if the tangent of the curve has a gradient 3, and this line is parallel so also has a gradient 3, and when normal to the curve it will be-1/3 why do u put it equal to the derivative of the curve?

But what confuses me is if the tangent of the curve has a gradient 3, and this line is parallel so also has a gradient 3, and when normal to the curve it will be-1/3 why do u put it equal to the derivative of the curve?

Thanks for the reply everyone, makes a lot more sense but the part I don't understand what does that point at B represent, does it represent the gradient of the line M and the gradient of the curve? Just getting a bit confused why would you put -1/3 = dy/dx

Thanks for the reply everyone, makes a lot more sense but the part I don't understand what does that point at B represent, does it represent the gradient of the line M and the gradient of the curve? Just getting a bit confused why would you put -1/3 = dy/dx

1. The tangent and normal are perpendicular to one another.

2. The point B is where the normal to the curve intersects the curve.

3. dy/dx gives you the gradient of the tangent to the curve at the point x.

4. You have the gradient of the normal .

5. You have the dy/dx

6. You know dy/dx = gradient of tangent.

7. You know gradient of tangent = -1/gradient of normal.

8. dy/dx = -1/gradient of normal.

Thanks for the reply everyone, makes a lot more sense but the part I don't understand what does that point at B represent, does it represent the gradient of the line M and the gradient of the curve? Just getting a bit confused why would you put -1/3 = dy/dx

The normal of a curve is a line that crosses through a point on the curve (in this case point B) perpendicular to the curve.

If the gradient of the normal was 2 for example then the gradient of the curve is -1/2

the normal is travelling upwards with a positive gradient at this point then the curve HAS to be travelling downwards in order for them to be perpendicular.

The gradient of the curve is also always 1/ the normal, just look at the graph you're given and think about it!

The gradient of the curve at any point is equal to dy/dx and because dy/dx will always have unknown x values in it, if you equate it to a gradient then it gives you the x co-ordinates of where the curve is that gradient