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matrenka [14]
3 years ago
5

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

Mathematics
1 answer:
Sphinxa [80]3 years ago
4 0

Here is the solution for this specific problem:

<span>Based from the graph, the curve will intersect itself at the y-axis, i.e. x = 0. </span><span>

</span><span>t^3 - 6t = 0 </span><span>
<span>t(t^2 - 6) = 0 </span>
<span>t = 0 or t = ± √6 </span>

<span>dx/dt = 3t^2 - 6 </span>
<span>dy/dt = 2t </span>

<span>dy/dx = 2t/(3t^2 - 6) </span>

<span>@ t = 0, dy/dx = 0. </span>
<span>x = 0, y = 0 </span>

<span>y = 0 </span>

<span>@ t = √6, dy/dx = 2√6/12 = √6/6 </span>
<span>x = 0, y = 6 </span>

<span>y - 6 = (√6/6) x </span>
<span>y = (√6/6)x + 6 </span>

<span>@ t = -√6, dy/dx = -2√6/12 = -√6/6 </span>
<span>x = 0, y = 6 </span>

<span>y - 6 = (-√6/6) x </span>
y = (-√6/6)x + 6</span>

So the equations of the tangent line at the point where the curve crosses itself are: <span>y = (√6/6)x + 6 and </span>y = (-√6/6)x + 6. I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

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