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.
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.
To find the points on the curve where the normal is parallel to the y-axis, we need to find the values of t that make the derivative of y with respect to x equal to zero.
First, let's find the derivative of y with respect to x using the chain rule.
dy/dx = (dy/dt)/(dx/dt)
We have the parametric equations:
x = t^2 - t,
y = t^3 - 3t.
Differentiating x with respect to t, we get:
dx/dt = 2t - 1.
Differentiating y with respect to t, we get:
dy/dt = 3t^2 - 3.
Now we can find dy/dx:
dy/dx = (dy/dt)/(dx/dt) = (3t^2 - 3)/(2t - 1).
The normal to the curve is perpendicular to the tangent line. So, the condition for the normal to be parallel to the y-axis is that dy/dx = 0.
Setting dy/dx = 0, we have:
(3t^2 - 3)/(2t - 1) = 0.
To satisfy this equation, either the numerator (3t^2 - 3) must equal zero, or the denominator (2t - 1) must equal zero.
For the numerator, we have:
3t^2 - 3 = 0,
t^2 - 1 = 0,
(t - 1)(t + 1) = 0.
This gives us two possible values for t: t = 1 and t = -1.
For the denominator, we have:
2t - 1 = 0,
2t = 1,
t = 1/2.
So, we have three possible values of t: t = 1, t = -1, and t = 1/2. These are the values at which the normal to the curve is parallel to the y-axis.
To find the corresponding coordinates (x, y), we substitute these t-values back into the parametric equations.
For t = 1:
x = 1^2 - 1 = 0,
y = 1^3 - 3(1) = -2.
For t = -1:
x = (-1)^2 - (-1) = 2,
y = (-1)^3 - 3(-1) = -2.
For t = 1/2:
x = (1/2)^2 - (1/2) = 1/4 - 1/2 = -1/4,
y = (1/2)^3 - 3(1/2) = 1/8 - 3/2 = -23/8.
Thus, the coordinates of the points on the curve where the normal is parallel to the y-axis are (0, -2), (2, -2), and (-1/4, -23/8).
To find the coordinates of the point(s) on the curve for which the normal to the curve is parallel to the y-axis, we need to find the points where the derivative of y with respect to x is undefined.
Given that the parametric equations for the curve are:
x = t^2 - t
y = t^3 - 3t
We need to find dy/dx, the derivative of y with respect to x.
To find dy/dx, we can use the chain rule:
dy/dx = (dy/dt) / (dx/dt)
Let's find dx/dt and dy/dt:
dx/dt = d/dt(t^2 - t) = 2t - 1
dy/dt = d/dt(t^3 - 3t) = 3t^2 - 3
Now, we can find dy/dx:
dy/dx = (dy/dt) / (dx/dt) = (3t^2 - 3) / (2t - 1)
For our purpose, we need to find the points where dy/dx is undefined. This happens when the denominator is zero.
So, 2t - 1 = 0
t = 1/2
Now, substitute t = 1/2 back into the parametric equations to find the corresponding x and ycoordinates:
x = (1/2)^2 - 1/2 = 1/4 - 1/2 = -1/4
y = (1/2)^3 - 3(1/2) = 1/8 - 3/2 = -13/8
Therefore, the point on the curve for which the normal is parallel to the y-axis is (-1/4, -13/8).