Graphing Astroid y(t) & Calculating Tangent Vector

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SUMMARY

The discussion centers on graphing the Astroid defined by the parametric equations y(t) = (cos³(t), sin³(t)) and calculating its tangent vector. The tangent vector, derived as y'(t) = (-3cos²(t)sin(t), 3sin²(t)cos(t)), equals zero at points where sin(t) = 0 or cos(t) = 0, indicating the presence of cusps at (0, 1) and (0, -1). The confusion arises from the distinction between the tangent vector and the conditions for vertical and horizontal tangents, which are clarified through the derivative dy/dx = (dx/dt)/(dz/dt). This understanding confirms the existence of four cusps in the astroid's graph.

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Homework Statement



Sketch the Astroid y(t) = (cos³(t), sin³(t)). Calculate its tangent vector at each point. At which points is the tangent vector 0?

2. The attempt at a solution

I'm not sure how to go about graphing parametric equations. I calculated the tangent vector:

y'(t) = (-3cos²(t)sin(t), 3sin²(t)cos(t)) and this equals 0 whenever sin(t) = 0 or cos(t) = 0. I have the solution to this question and it says that the points at which y'(t) = 0 correspond to the 4 cusps of the astroid. But I don't even see why this is true. Why is the y'(t) = 0 at (0,1) and (0,-1)? Shouldn't they be undefined?

So could someone please help me out? Maybe tell me what is the general way of graphing parametric equations?

Thanks!
 
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I've been reading another source and they seem to calculate dy/dx = (dx/dt)/(dz/dt) and state there is a vertical tangent when dz/dt = 0 and dx/dt is nonzero and a horizontal tangent when dx/dt = 0 and dz/dt is nonzero. Using this I do see why the astroid has four cusps. But using the tangent vector which is defined as simply the first derivative doesn't seem to help me. Using the other definition however, I can go about graphing this. Can anyone help me clear up this confusion?
 
Anyone?
 

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