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1. Keep getting stuck on these types of questions:

10b:

https://a9497d7f220e174d38d014b2b1d8...%20Edexcel.pdf

Could someone please explain?

Thanks
2. (Original post by ryandaniels2015)
Keep getting stuck on these types of questions:

10b:

https://a9497d7f220e174d38d014b2b1d8...%20Edexcel.pdf

Could someone please explain?

Thanks
If the gradient of normal is some number a. Then the gradient of the tangent will be -1/a. You know the gradient of m is something, so do -1/something to get the gradient of the tangent.

So the derivative of the curve evaluated at the coordinate x of B should give you -1/a.

What is the derivative of the curve? Set that (the derivative) equal to -1/a. (a will be a number, I'm just using it as a letter here).
3. what's the gradient of the equation you got for part A?

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4. (Original post by bubblegumcat)
what's the gradient of the equation you got for part A?

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I got 3
5. 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?
6. 10 a) ok so what you need to do is find the gradient of L by subbing x=0 (the x co-ordinate of point A) into the dy/dx of the curve. then find the equation of the line using y-y1=x-x1 (y1 can be found by subbing the x value of A which is 0 into the equation of the line)

b) You know the gradient of L, this will be the same gradient of M because the lines are parallel

Because M is the normal of the curve at B the gradient of the curve at B is the negative reciprocal of the gradient of M (change the sign and 1/)

Now you know the gradient of the curve at B you can make this value = to dy/dx of the curve. solve this (I'm sure you'll figure this bit out) and you should get and x value.

sub this X value back into the equation of the curve to get the corresponding Y value. and there you have the co-ordinates of B.

If you need any help just ask, hope this is clear enough for you
7. (Original post by ryandaniels2015)
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?
Because this is the gradient of the curve at B and the derivative is equal to the gradient of the curve at any point when its given different x values
8. 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
9. (Original post by ryandaniels2015)
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?
it's at a different point, at this point the normal is not -1/3 as it doesn't cut the y-axis.
- The tangent of the curve at point B would be -1/3.
10. (Original post by ryandaniels2015)
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.
11. (Original post by ryandaniels2015)
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
point B Is just a point on the curve where the tangent grad is -1/3 so therefore the normal to the tangent grad is 3.
12. (Original post by Zacken)
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 clarification, makes more sense, I think I'll have a look again tomorrow hopefully it will make perfect sense then!

Thanks everyone who helped!!!
13. (Original post by ryandaniels2015)
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
14. (Original post by splashywill)
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
Thanks!! I think it makes sense now...

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