Parametric Equations: Find the equation of the tangent line at the point where the curve crosses itself. x = t^3-6t, y = t^2

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Parametric Equations: Find the equation of the tangent line at the point where the curve crosses itself. x = t^3-6t, y = t^2

Mathematics
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I'm particularly stuck on how to get the points. In a previous problem I just made a chart where I plugged in values for t in the x and y equation until I got two coordinates that were the same for 2 t-values. When I try to do the same method with this problem, I'm not able to get the same result.
you meant when x(t) = y(t) ?
there are 3 places where they intersect t = -2,0,3

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curve would cross itself at that point where the derivative would cease to exist .. as you see the derivative is 2t/(3t^2 - 6) .. hence for t = +- sqrt(2) derivative ceases to exist ... Hence the point where it crosses itself is (0,2)
no dumbcow he means that a curve in parametric form crosses itself .. what is the equation of the tangent ...
at the point where it crosses itself...
also you cannot find the slope at this point ... because the slope is not defined for THIS function at this point ...although you can find the slope of points with parameter slightly greater (or slightly smaller) than +-sqrt(2) and find the slope...
ok
If a curve crosses itself at any point, that point will have 2 tangents right?
correct.
@Mashy ... how will you define the derivative at that point ??? .. As I said you can find the slope by assumint t --> sqrt(2) - ... but at that point derivative won't exist (function cannot have two values for one variable --> function is not defined otherwise ) ...
\[t = \pm \sqrt{y}\] substituting into x(t) gives 2 equations \[x = y \sqrt{y} - 6\sqrt{y}\] and \[x = -y \sqrt{y} + 6\sqrt{y}\] setting them equal will give us the point where the curve crosses itself \[y \sqrt{y} -6\sqrt{y} = -y \sqrt{y} +6\sqrt{y}\] \[2y \sqrt{y} = 12\sqrt{y}\] \[y = 6\] \[x = 6\sqrt{6} - 6\sqrt{6} = 0\] point is (0,6) \[\frac{dx}{dy} = \pm (\frac{3\sqrt{y}}{2} -\frac{3}{\sqrt{y}})\] this is the derivative of the inverse, to find slope of curve at x=0 use: \[f'^{-1}(y) = \frac{1}{f'(x)}\] \[\rightarrow f'(x) = \frac{1}{f'^{-1}(y)}\] plug in y=6 into dx/dy \[dx/dy = \pm(\frac{3\sqrt{6}}{2} -\frac{3}{\sqrt{6}}) = \pm \sqrt{6} \] \[f'(0) = \pm \frac{1}{\sqrt{6}} = \pm \frac{\sqrt{6}}{6}\] tangent lines: \[y = \pm \frac{\sqrt{6}}{6}x + 6\] http://www.wolframalpha.com/input/?i=x+%3D+t%5E3+-6t+%2C+y+%3D+t%5E2
I'll have look this over a few times but thank you for your help.
Why would the curve intersect itself where x(t) = y(t)? The only thing I can imagine coming from this is a point graphed that is equal in distance from both axes in the first and third quadrant, e.g. a point on the line y = x, with some arbitrary constant added or subtracted from both sides. I completely understand why the derivative would not exist at the curve's intersection with itself (or another curve), but I fail to understand why x(t) and y(t) being equal means that the graph crosses itself.

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