x = t^2-t
dx/dt = 2t-1
y = t^3-3t
dy/dt = 3t^2-3
dy/dx = (dy/dt)/(dx/dt) = (3t^2-3)/(2t-1)
If the normal to the curve is vertical, the tangent is horizontal. So, dy/dx = 0.
That happens when t=±1
So, the normal is vertical at (2,2) and (0,-2)
See the graph at
http://www.wolframalpha.com/input/?i=parametric+plot+x+%3D+t^2-t%2C+y+%3D+t^3-3t
But note also that at t=2, (2,2) is again on the curve, but the normal has slope -1/3.
A curve is defined by the parametric equations: x = t2 – t and y = t3 – 3t
Find the coordinates of the point(s) on the curve for which the normal to the curve is parallel to the y-axis.
You must use calculus and clearly show your working, including any derivatives you need to find when solving this problem.
1 answer
1 answer