Cardiod parametric equation problem

  • Thread starter mahler1
  • Start date
  • Tags
    Parametric
In summary, we have a cardioid ##C_a## given in polar coordinates by ##r_a= a+cos(\theta)## with a being a parameter, and ##\theta## ##\in [0,2\Pi]##. For part (a), we need to prove that for a>1, ##C_a## is a smooth curve. In part (b), we are asked to calculate the tangent line to the curve ##C_a## at the point (0,a) in cartesian coordinates. The parameterization for the curve in polar coordinates is given by ##ψ(t)=(f(\theta)cos(\theta),f(\theta)sin(\theta))##, where ##f(\theta) = a
  • #1
mahler1
222
0
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?
 
Last edited:
Physics news on Phys.org
  • #2
mahler1 said:
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??
 
  • Like
Likes 1 person
  • #3
mahler1 said:
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
LCKurtz said:
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
Zondrina said:
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
mahler1 said:
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
mahler1 said:
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$$
\frac{dy}{dx}= \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$
 

1. What is the "Cardiod parametric equation problem"?

The Cardiod parametric equation problem involves finding the parametric equations for a graph of a cardiod, which is a heart-shaped curve. These equations describe the relationship between the x and y coordinates of points on the graph, and are commonly used in mathematics and physics.

2. How do you solve the Cardiod parametric equation problem?

To solve the Cardiod parametric equation problem, you first need to understand the equation for a cardiod, which is x = a(1 - cos(t)) and y = a sin(t), where a is the radius of the circle and t is the angle. You can then use these equations to find the x and y coordinates for different values of t, and plot these points to create the cardiod graph.

3. What are the applications of the Cardiod parametric equation problem?

The Cardiod parametric equation problem has various applications in mathematics and physics. It can be used to describe the motion of a point on a rotating wheel, the orbit of a planet around a star, and the shape of a vibrating string. It is also used in computer graphics to create realistic images of heart-shaped objects.

4. What is the difference between a cardiod and a circle?

A cardiod is a heart-shaped curve, while a circle is a round curve. The equation for a circle is x² + y² = r², while the equation for a cardiod is x = a(1 - cos(t)) and y = a sin(t). A circle has a constant radius, while a cardiod has a varying radius that depends on the angle t.

5. Can the Cardiod parametric equation problem be solved using different methods?

Yes, there are different methods for solving the Cardiod parametric equation problem. One approach is to use the standard equations x = a(1 - cos(t)) and y = a sin(t), as mentioned earlier. Another method is to use polar coordinates, where r = a(1 + cos(t)) and θ = t. Both methods will result in the same cardiod graph.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
956
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
4K
  • Calculus
Replies
11
Views
3K
  • Calculus and Beyond Homework Help
Replies
4
Views
8K
  • Calculus and Beyond Homework Help
Replies
6
Views
3K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
Back
Top