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    Keep getting stuck on these types of questions:

    10b:

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

    Could someone please explain?

    Thanks
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    (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).
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    what's the gradient of the equation you got for part A?


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


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    I got 3
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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?
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    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
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    (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
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    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
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    (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.
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    (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.
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    (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.
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    (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!!!
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    (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
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    (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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