The parameterisation of of a curve

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Question from Eunice, a student:

Let C be the path along the curve given by y−80=−5x^2 that moves from the point (5,−45) to the point (0,80).
Find r(t) the parameterisation of C in that direction as t∈[0,5]. How am I suppose to find the parametric of both x and y?
can I let x=t, then y=-5t^2+80? thanks

Hi Eunice,

The parameterisation $x = t$ and $y = -5 t^2 + 80$ almost works. With this parameterisation when $t = 0$ you are at $(0, 80)$ and at $t = 5$ you are at $(5, -45)$ so you have moved along the curve in the wrong direction. Do you see how to modify the parameterisation so that as $t$ moves from 0 to 5 you move along the curve in the opposite direction?

Penny

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.