Asked by Daksh
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.
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.
Answers
Answered by
Steve
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.
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.
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