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    why do we substitute t values back into P(at^2,2at)


    3E question 4b
    https://644625398389466aee0063322305...0NRNU0/CH3.pdf
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    I don't understand your question. However, this may help.

    If we take a curve which is double valued in y such as a circle the the formula is y=\pm \sqrt{r^2-x^2} which is not a function as it is double valued, i.e each value of x does not lead to only one value of y

    Introducing a parameter to give us  a=r \cos t, \; y=r \sin t gives us two single valued functions and by using values of t 0<t<=360 we can plot the function.

    The same is the case for a parabola which we swap a sqrt form (not a function) for the parameterised version of (say) x=at^2, y=at

    This enables us to find proper functions in t for the gradient etc using the chain rule.

    Does that answer in any way what you are asking?
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    Soory, forgot to address the specific question. By eliminating t we end up with a cartesian form with y^2 in it. which we root out to find the form y=f(x) (the cartesian form) which is not as useful in a mathematical sense but at least has the benefit of getting us a mark or two at FP1
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    (Original post by Acrux)
    FP1
    why do we substitute t values back into P(at^2,2at)


    3E question 4b
    https://644625398389466aee0063322305...0NRNU0/CH3.pdf
    Because it wants the co-ordinates, not the parameter. The coordinates depend on t, so to find the coordinates you plug t.
 
 
 
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