Parametric Curves: Finding Tangents at (0,0) Using Lissajous Figure Equations

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SUMMARY

The discussion centers on finding the equations of the tangents to the Lissajous figure defined by the parametric equations x=sin(2t) and y=cos(t) at the point (0,0). The slopes of the tangents at this point are determined to be 1/2 and -1/2, occurring at t = π/2 and t = -π/2, respectively. The user seeks assistance in deriving the tangent equations from these slopes, indicating a need for clarity on the application of calculus in parametric equations.

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  • Understanding of parametric equations
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Students and educators in mathematics, particularly those focusing on calculus and parametric equations, as well as anyone interested in the geometric properties of Lissajous figures.

ziddy83
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URGENT help needed please...

I have been having problems with this problem...it says...
A graph of the Lissajous figure is given by the paraetric equations:

x=sin2t and y=cost

Show that the curve has two tangents at the point (0,0) and find their equations

Can someone please help me? I've been trying to figure this out for the past two days. I got the slope to be

-sint/cos(2t)*2 I am not sure where to go from there...so please..if anyone can help that would be fantastic. :confused:
 
Last edited:
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anyone...??
 
Origin means x=y=0. They are zero for t = pi/2 and -pi/2. For these you get two values of the slope, 1/2 and -1/2.


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