When moving the slider of variable "m", we can see that the red curve is being bend, and you can clearly see that the lowest part of that red curve follow sine wave (green line)

Instead follow sine wave, how to make that lowest part follow blue line (straight line) instead sine wave while changing "m" variable and still keep original shape? If you know how, please explain as simple as possible how did it work and provide the polar function :(.

I know that this question may asked somewhere, but with limit English, I don't know how to search this type of problem and the name of the curve, and the question may unclear to some...

  • $\begingroup$ Forgot to cite: variable "m" is within range 0 <= m <= 1, for not creating more than 2 petals, sorry for confusion. And the lowest part (should I call it "tipped point") should be within y-axis... $\endgroup$ – Trung0246 Sep 20 '17 at 6:17

The green line is NOT a part of the red curve. It is the drawing of a different function.

You can see on the left side of the figure the equations of three different functions, which are respectively drawn in red, green and blue.

The red curve is the is the drawing in polar coordinates. When you change the value of $m$, of course the chape changes, but not as you imagine. You wrote "The red curve is being bend". This because you make $m$ vary only on a small range. Try on a larger range and see what append. For example $m=1.5$ then $m=5$ on the figures below.

enter image description here

enter image description here

The green curve is not "the lowest part of the red curve" as you wrote. It represents a pure sinusoidal function. The change of $m$ translate it, nothing more.

The blue curve represents a linear function. The change of $m$ translate it, that's all.

Drawing the three functions is only a feature of the software that you use. The question "How to make lowest part of this curve follow straight line instead sine wave?" is a non-sens.


The Cartesian coordinates of the "tipped point" are $\left(x=0\:,\:y=\cos(m\frac{\pi}{2})\right)$. If you want that this point follows a moving straight line (drawn in blue), just change the equation of the straight line to : $$y=\left(-x+\cos(m\frac{\pi}{2}) \right)d$$

enter image description here

Another variant. Only the polar function is changed :

enter image description here

  • $\begingroup$ Sorry for confusion. Variable "m" should only within 0 <= m <= 1, and should only have 2 "tipped point" (or "lowest part") and no more than 2 petals... $\endgroup$ – Trung0246 Sep 20 '17 at 6:20
  • $\begingroup$ This doesn't change my answer. What you call "lowest part" is not sinusoidal and is not represented by the green curve. $\endgroup$ – JJacquelin Sep 20 '17 at 6:32
  • $\begingroup$ The translate of green sine curve to the left represent the movement path of the tipped point when you tried to change variable "m". You may disable the drawing of straight line and should clearly see that the tipped point moved along on the sine wave. And so as the question stated, I want the tipped point move along that blue straight line, not green sine wave. And I know that those 3 are different functions, not related to each others, sine wave along with straight line just served for looking purpose. I hope this comment made more clear now... $\endgroup$ – Trung0246 Sep 20 '17 at 6:51
  • $\begingroup$ I am still not sure to well understand what you want. Look at the addition to my main answer. Hoping this will help you. $\endgroup$ – JJacquelin Sep 20 '17 at 7:51
  • $\begingroup$ Not the best I got but well... Thank for your time, I really need to edit my question or just ask new question in different way...Anyway... If you have times to fix your answer, this time you need to change only red curve to make the tipped point move in old blue straight line instead green sine wave without modify anything else and keep the red curve as polar function... $\endgroup$ – Trung0246 Sep 20 '17 at 14:19

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