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    Well, in the 2 mocks I've done (one was modified), I've got a U in each. I've got a couple of little questions...

    1) What exactly is d²y/dx²? I've forgotton
    2)A curve has the equation y=x+(3/x), x≠0
    The point P on the curve has x-co-ordinate 1. (a)Show that the gradient of the curve at P is -2.(b)Find the equation for the normal to the curve at P, giving your answer in the form y=mx+c. (c)Find the co-ordinates of the point where the normal to the curve at P intersects the curve again.

    question 2 has me at a loss
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    (Original post by Norm)
    Well, in the 2 mocks I've done (one was modified), I've got a U in each. I've got a couple of little questions...

    1) What exactly is d²y/dx²? I've forgotton
    2)A curve has the equation y=x+(3/x), x≠0
    The point P on the curve has x-co-ordinate 1. (a)Show that the gradient of the curve at P is -2.(b)Find the equation for the normal to the curve at P, giving your answer in the form y=mx+c. (c)Find the co-ordinates of the point where the normal to the curve at P intersects the curve again.

    question 2 has me at a loss
    1. d²y/dx² is when you differentiate the equation twice with respect to x, such as if i had the equation y = x³ + 4x², then differentiating once (with respect to x) would give me dy/dx = 3x² + 8x, and differentiating again would give d²y/dx² = 6x + 8.

    2. a) remember you can write y=x+(3/x) as y = x + 3x-1 , hence differntiating once would give you the gradient of the curve, ie. dy/dx = 1 - 3x-2, which is the same as dy/dx = 1 - 3/x². remember that the gradient at x=1 is simply substituting x=1 into the dy/dx equation that you got, which gives you dy/dx = 1 - 3 = -2, hence you've shown that the gradient of the curve at point P = -2.

    b) the gradient of the normal to the curve at point P is just 1/(dy/dx). i'm sure that you can figure the rest out from here.

    hope that this helps.
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    ^ The gradient of the normal is m = -\frac{1}{(\frac{dy}{dx})}
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    (Original post by AlphaNumeric)
    ^ The gradient of the normal is m = -\frac{1}{(\frac{dy}{dx})}
    whoops, i made a typing error! sorry to the OP.
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    Thankies

    I spent ages last night revising all the formulas. Learnt everything imaginable about APs and then couldn't answer the question on them D'oh. At least I'm good at the other 3 subjects I'm doing.
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    All you need to do to get a good mark in C1 is do all the past papers you can lay your hands on. Do the real ones and the solomon ones and you'll definitely get a good mark.
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    (Original post by Godsize)
    All you need to do to get a good mark in C1 is do all the past papers you can lay your hands on. Do the real ones and the solomon ones and you'll definitely get a good mark.
    Well I have the Solomon papers, and can download a fair few off the internet. Let's hope I get a good (better than before) mark.
 
 
 
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