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Math Help - differentiation past exam question

  1. #1
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    differentiation past exam question

    hello everybody!

    1. if y= x(x^2 + 4) find dy/dx
    2. then find the equation of the tangent to the curve where x=1 giving answer in the form y=mx+c
    3. the normal to the curve y=x^3 + ax at coordinates (2,b) has gradient 1/2. find the values of the constants a and b.

    1. dy/dx = 3x^2 + 4

    2. I assume we would use y-y1 = m(m-m1) where the 1 is at the bottom of the y and m, but we only have one coordinate, which is x=1. So I am not sure about this

    3. I haven't got a clue what the 'normal to the curve is'
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  2. #2
    Senior Member JaneBennet's Avatar
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    2. Substitute x=1 into the equation y=x(x^2+4). So your other co-ordinate is y=5.

    3. The normal to a curve at a given point P is the straight line perpendicular to the tangent at P and passing through P. If the tangent has gradient m\ne0, the gradient of the normal is -\,\frac{1}{m}.

    In your example, the normal to the curve y=x^3+ax at x=2 is \frac{1}{2}; hence \frac{\mathrm{d}y}{\mathrm{d}x}=-2 at x=2. You should be able to find a from here. Once you’ve found a, substitute x=2 into y=x^3+ax to find the y-coordinate, which is b.
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  3. #3
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    Quote Originally Posted by JaneBennet View Post

    In your example, the normal to the curve y=x^3+ax at x=2 is \frac{1}{2}; hence \frac{\mathrm{d}y}{\mathrm{d}x}=-2 at x=2.

    I don't understand the above, if the normal is -1/m and m=1/2 then the normal is -2, right?

    And what have you differentiated to give dy/dx = -2 ?
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  4. #4
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    Here's what I got anyway,

    for y=x^3 + ax

    the normal = -1/m = -1/(1/2) = -2 = a

    so, y= b = x^3 + 2x

    b= 2^3 + (2*-2) = 4
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