anonymous12443535
16.01.2020 •
Mathematics
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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Ответ:
Here is the solution for this specific problem:
Based from the graph, the curve will intersect itself at the y-axis, i.e. x = 0.
t^3 - 6t = 0
t(t^2 - 6) = 0
t = 0 or t = ± √6
dx/dt = 3t^2 - 6
dy/dt = 2t
dy/dx = 2t/(3t^2 - 6)
@ t = 0, dy/dx = 0.
x = 0, y = 0
y = 0
@ t = √6, dy/dx = 2√6/12 = √6/6
x = 0, y = 6
y - 6 = (√6/6) x
y = (√6/6)x + 6
@ t = -√6, dy/dx = -2√6/12 = -√6/6
x = 0, y = 6
y - 6 = (-√6/6) x
y = (-√6/6)x + 6
So the equations of the tangent line at the point where the curve crosses itself are: y = (√6/6)x + 6 and 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.
Ответ: