Maths

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a curve ahs parametric equations x=t^2
and y= 1-1/2t for t>0.

i)find the co-ordinates of the point P where the curve cuts the x-axis which i found to be P(1/4, 0)

the next part i cant do
ii) find the gradient of the curve at this point.

So far, I have the gradient to be:
-2/4t^3

is that gradient right and how do I get the value at the point P.

  • Maths -

    you could change it to cartesian form ...

    t^2 = x
    t = √x

    y = 1 - (1/2)t
    2y = 2 - t
    t = 2 - 2y

    so √x = 2 - 2y
    √x + 2y = 2
    to cut the x-axis, y = 0
    √x=2
    x=4

    so the x-intercept is (4,0)

    >b>Unless ... you meant to type y = 1 - 1/(2t)
    in that case the point is correct and the cartesian equation would be
    √x = 1/(2-2y)

    (but you didn't type it that way)

    In either case, differentiate the correct equation implicitly, and sub in the corresponding x-intercept.

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