Cardiod parametric equation problem

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Homework Statement .
Let ##C_a## a cardioid given in polar coordinates by ##r_a= a+cos(\theta)## with a being a parameter, and ##\theta## ##\in [0,2\Pi]##

a)Prove that, for a>1, ##C_a## is a smooth curve.
b)Calculate the tangent line to the curve ##C_a## in the cartesian coordinates point ##(x,y)=(0,a)##

My problem is I don't understand how this curve is, I mean, I googled and saw how it looks like, but I don't know the parameterization.

By the way, I accidently pressed enter, so I opened an empty thread and I couldn't check the spell in the title, it says "cardiod"", when it should be "cardioid"", does anyone know how can I correct the title?
 
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  • #2
LCKurtz
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Homework Statement .
Let ##C_a## a cardioid given in polar coordinates by ##r_a= a+cos(\theta)## with a being a parameter, and ##\theta## ##\in [0,2\Pi]##

a)Prove that, for a>1, ##C_a## is a smooth curve.
b)Calculate the tangent line to the curve ##C_a## in the cartesian coordinates point ##(x,y)=(0,a)##

My problem is I don't understand how this curve is, I mean, I googled and saw how it looks like, but I don't know the parameterization.

By the way, I accidently pressed enter, so I opened an empty thread and I couldn't check the spell in the title, it says "cardiod"", when it should be "cardioid"", does anyone know how can I correct the title?

I wouldn't worry to much about the misspelling in the title but you can ask a moderator to do it for you if it bothers you. Not sure what is bothering you about the parameterization. It is just a polar equation in the usual form ##r = f(\theta)##. Surely you have studied polar coordinates and their graphs??
 
  • #3
STEMucator
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Homework Statement .
Let ##C_a## a cardioid given in polar coordinates by ##r_a= a+cos(\theta)## with a being a parameter, and ##\theta## ##\in [0,2\Pi]##

a)Prove that, for a>1, ##C_a## is a smooth curve.
b)Calculate the tangent line to the curve ##C_a## in the cartesian coordinates point ##(x,y)=(0,a)##

My problem is I don't understand how this curve is, I mean, I googled and saw how it looks like, but I don't know the parameterization.

By the way, I accidently pressed enter, so I opened an empty thread and I couldn't check the spell in the title, it says "cardiod"", when it should be "cardioid"", does anyone know how can I correct the title?

For part (a), does the tangent vector vanish at all or is it continuous?
 
  • #4
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I wouldn't worry to much about the misspelling in the title but you can ask a moderator to do it for you if it bothers you. Not sure what is bothering you about the parameterization. It is just a polar equation in the usual form ##r = f(\theta)##. Surely you have studied polar coordinates and their graphs??

I got confused, the parameterization in polar coordinates is ##ψ(t)=(f(\theta)cos(\theta),f(\theta)sin(\theta))##
 
  • #5
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For part (a), does the tangent vector vanish at all or is it continuous?

Sorry, I don't understant the question. The tangent vector doesn't vanish if the curve is smooth, I'm not sure if that what you've asked.
 
  • #6
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Sorry, I don't understant the question. The tangent vector doesn't vanish if the curve is smooth, I'm not sure if that what you've asked.

Exactly, if the tangent vector doesn't vanish anywhere, then it is continuous. That is the exact condition you need for the curve to be smooth.
 
  • #7
LCKurtz
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I got confused, the parameterization in polar coordinates is ##ψ(t)=(f(\theta)cos(\theta),f(\theta)sin(\theta))##

Use either ##\theta## or ##t## for the parameter, not both. But yes, for a curve given as ##r = f(\theta)##, that is the x-y parameterization$$
\Psi(\theta) = \langle f(\theta)\cos(\theta), f(\theta)\sin(\theta)\rangle$$You can get the slope by$$
\frac{dy}{dx}= \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$
 

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