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Find out the x-y coordinates of the points in which the following parametric equations will have horizontal or vertical tangents.
x = t3 - 3t
y = 3t2 - 9
Solution
We'll first require the derivatives of the parametric equations.
dx/dt = 3t2 - 3 = 3 (t2 -1) dy/dt = 6t
Horizontal Tangents
We'll have horizontal tangents in which,
6t = 0 ⇒ t = 0
Now here, this is the value of t that gives the horizontal tangents and we were asked to find out the x-y coordinates of the point. To get these we just only need to plug t into the parametric equations. Hence, the just only horizontal tangent will take place at the point (0,-9).
Vertical Tangents
In this case we require to solve,
3(t2 -1) = 0 ⇒ t = ≠1
The two vertical tangents will take place at the points (2,-6) and (-2,-6). On behalf of completeness and at least partial verification here is the drawing of the parametric curve.
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