# Possibly Difficult Graphing Problem

#### naickej4

Joined Jul 12, 2015
206
Hi all. Thank you all. I am finally understanding this. Very interesting. I went through all the answers that was provided and tried them but i am really intrigued though by Tesla23 way of doing it using parametric plots. Will using parametric plots always provide the correct answer? I tried it out using the online mathematica and it works. it looks like a cardioid. Heart shaped. As I am a student that has no lectures(not able to consult any lecturer), so forgive me if I might not know these concepts very well.Thanks again.

#### MrAl

Joined Jun 17, 2014
10,105
Hi,

Did you mean using the complex plane example, because WBahn's first method was parametric too, in 'x'.
Yes it looks like a cardioid, but is it really? See if you can figure out why it is or why it is not really a cardioid.

I couldnt help but try the ellipse because that is my favorite plane curve. My favorite 3d surface is probably the toroid.

#### WBahn

Joined Mar 31, 2012
28,514
Did you mean using the complex plane example, because WBahn's first method was parametric too, in 'x'.
Yes, it was parametric, but it was not a function (it was a relation -- I think I'm using those terms correctly, it's been a LONG time). Tesla23's form was functional, which makes it a lot more useful for most purposes, including plotting.

#### Tesla23

Joined May 10, 2009
527
Hi all. Thank you all. I am finally understanding this. Very interesting. I went through all the answers that was provided and tried them but i am really intrigued though by Tesla23 way of doing it using parametric plots. Will using parametric plots always provide the correct answer?
Yes it will.

I think of it a little differently. When asked how the circle $$x^2 + y^2 = 1$$ will transform, the question is really asking you to transform each point on the circle and see what the resulting curve is. You can often sketch the curve by manually transforming a few representative points, and this is what I did in post#8.

To plot the curve you need some way of describing the points you are transforming, i.e. the points on the circle. WBahn described them in terms of 'x', getting two points for each 'x' value, and then transforming each one. There is nothing wrong with doing this, but in this example my preference would be to use complex numbers.

I said that you could describe the points on the circle as $$z = e^{j\theta}$$ where $$0 \le \theta \le 2\pi$$

in this case it is easy to transform each point, so the new curve is the set of all points:

$$z^2 - 2z = e^{2j\theta} - 2e^{j\theta}$$ where $$0 \le \theta \le 2\pi$$

which leads to the parametric plot.

#### naickej4

Joined Jul 12, 2015
206
Hi,

Did you mean using the complex plane example, because WBahn's first method was parametric too, in 'x'.
Yes it looks like a cardioid, but is it really? See if you can figure out why it is or why it is not really a cardioid.

I couldnt help but try the ellipse because that is my favorite plane curve. My favorite 3d surface is probably the toroid.
Hi Sir, Yes the complex parametric function. I am going try this also thanks. Yes I need to know why it is a Cardioid. Because from studying complex mapping you can have different types of transformations. e.g. Magnification,Rotation,Translation.
thank you.

#### naickej4

Joined Jul 12, 2015
206
Yes it will.

I think of it a little differently. When asked how the circle $$x^2 + y^2 = 1$$ will transform, the question is really asking you to transform each point on the circle and see what the resulting curve is. You can often sketch the curve by manually transforming a few representative points, and this is what I did in post#8.

To plot the curve you need some way of describing the points you are transforming, i.e. the points on the circle. WBahn described them in terms of 'x', getting two points for each 'x' value, and then transforming each one. There is nothing wrong with doing this, but in this example my preference would be to use complex numbers.

I said that you could describe the points on the circle as $$z = e^{j\theta}$$ where $$0 \le \theta \le 2\pi$$

in this case it is easy to transform each point, so the new curve is the set of all points:

$$z^2 - 2z = e^{2j\theta} - 2e^{j\theta}$$ where $$0 \le \theta \le 2\pi$$

which leads to the parametric plot.
Hi Tesla23, forgive me if i might be asking a silly question. Is there an Advanced Maths text book that I can read which will explain to me more on parametric plots the approach that you and also WBahn followed. I understand MR AI's method perfectly now, but I need to know this way of looking at it for future references. I need to expect all types of questions for the exams.
If I take the Unit circle on a certain point (x,y) can I say x= cosθ and y = sinθ thereafter substitute these into the transform Z^2 -2Z
will this work?
because, x^2 + y^2 = 1 and therefore (cos(θ))^2 + (sin(θ))^2 = 1

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#### MrAl

Joined Jun 17, 2014
10,105
Hi Sir, Yes the complex parametric function. I am going try this also thanks. Yes I need to know why it is a Cardioid. Because from studying complex mapping you can have different types of transformations. e.g. Magnification,Rotation,Translation.
thank you.
Hi,

Well really i posed the question for you to research

What i think is that the cardioid is a certain type of curve that is represented by a certain equation, and that any curve that can be generated by that equation is a cardioid. So that in turn means that if we have an unknown equation, if we can manipulated it such that it takes on the form of one of the cardioid equations, then it must be a cardioid, but if not then it must not be a cardioid.
For example, we have a circle with equation x^2+y^2=R^2. Then we have an 'unknown' equation x^2/A+y^2/B=1. The question is, is this second equation a circle? The answer is in general no, but if A=B then it is a circle, but only then.
So we seek the same kind of proof for the pseudo cardioid we have been working with in this thread.
It may be that the definition of a cardioid can be "any curve that looks like a heart", but it may not be because some curves are uniquely defined as such because they have certain properties and if the new curve doesnt have those properties then it is not a good idea to call it 'exactly' the same type of curve.
So you can try to solve this mystery if you like

#### WBahn

Joined Mar 31, 2012
28,514
Hi Sir, Yes the complex parametric function. I am going try this also thanks. Yes I need to know why it is a Cardioid. Because from studying complex mapping you can have different types of transformations. e.g. Magnification,Rotation,Translation.
thank you.
Be sure that you understand that we have been working with a different kind of transform -- namely a transformation of variables (or often just called a "change of variables"). When you magnify, rotate, or translate an equation, the plot changes. But here we have just changed the way that the equation is expressed without changing the plot itself.

#### naickej4

Joined Jul 12, 2015
206
Hi
Hi,

Well really i posed the question for you to research

What i think is that the cardioid is a certain type of curve that is represented by a certain equation, and that any curve that can be generated by that equation is a cardioid. So that in turn means that if we have an unknown equation, if we can manipulated it such that it takes on the form of one of the cardioid equations, then it must be a cardioid, but if not then it must not be a cardioid.
For example, we have a circle with equation x^2+y^2=R^2. Then we have an 'unknown' equation x^2/A+y^2/B=1. The question is, is this second equation a circle? The answer is in general no, but if A=B then it is a circle, but only then.
So we seek the same kind of proof for the pseudo cardioid we have been working with in this thread.
It may be that the definition of a cardioid can be "any curve that looks like a heart", but it may not be because some curves are uniquely defined as such because they have certain properties and if the new curve doesnt have those properties then it is not a good idea to call it 'exactly' the same type of curve.
So you can try to solve this mystery if you like
Hi Sir. Thank you. I am going to try this and will research this more and come back to this thread. I'm also busy with DSP examples. So i will definitely be back to respond.
Thanks