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FP1
why do we substitute t values back into P(at^2,2at)


3E question 4b
https://644625398389466aee00633223056f55519d5fbc.googledrive.com/host/0B1ZiqBksUHNYQWY1UWtNa0NRNU0/CH3.pdf
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Avatar for nerak99
nerak99
13
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=±r2−x2y=\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=rcos⁡t,  y=rsin⁡t 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?
(edited 7 years ago)
Reply 2
Avatar for nerak99
nerak99
13
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
Reply 3
Avatar for Zacken
Zacken
22
Original post by Acrux


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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