SEARCH HOME
Math CentralQuandaries & Queries

search

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

About Math Central
 

 


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
Quandaries & Queries page Home page University of Regina PIMS