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June 23

Heart (symbol)

There is a formula to graph it in the middle of the article. Someone on the talk page would like it transposed to input to a graphics calculator (y= as opposed to =0) I tried on the talk page but probably got it wrong. I also don't know how to do the notation in wp. Could someone check my math on the talk page? We may wish to add the calculator input format to the main page as well.--Canoe1967 (talk) 11:56, 23 June 2012 (UTC)[reply]

From my memories of my old TI (which it's been about 10 years since I used, admittedly), it didn't really do single-parameter functions that double back on themselves like that, at least not very well. The best bet would be to graph it as a parametric function, and the functions for that are given in the next graph over. I'll leave a similar note on the talk page. Getting it into Y=X format is beyond me. 24.154.70.132 (talk) 15:24, 23 June 2012 (UTC)[reply]

Yes, when I have tried similar (even with circles) on my TI-84, it has simply refused to plot it.--Gilderien Chat|List of good deeds 15:28, 23 June 2012 (UTC)[reply]

My TI-85 (which, despite its model number, was based on an older and weaker chip than the Ti-84) had a mode for plotting parametric equations {x(t), y(t)}. Is that capability unavailable on the new calculators? Here's a webpage with instructions for the parametric mode on TI-83/84 series. It looks much more difficult than the old '85, which had a quick and easy entry and dedicated hardware buttons for x, y, and t. Sadly the 85/86 line has been discontinued, squeezed out by more expensive '89 series, all the new "color screen" systems, and the entry-level 84+ calculators. If you can get a used TI-85/86, give it a shot; its weaker processor is offset by its better user interface for chemistry, physics, and statistics problems, or for plotting polar or parametric equations (like the cardioid equation and its heart-shaped variants). Nimur (talk) 15:56, 23 June 2012 (UTC)[reply]

Thank you very much :) --Gilderien Chat|List of good deeds 15:58, 23 June 2012 (UTC)[reply]

Thanks all. I wonder if that user will come back nearer Valentine's Day. I think what he meant was solve the equation for y. I seem to remember an old calculator asking for that format if I wanted to install a new formula into it. I suppose my math won't work, even if I did solve for y correctly? Did anyone try and succeed in graphing one yet? We could upload a picture of it, that may improve the article.--Canoe1967 (talk) 16:55, 23 June 2012 (UTC)[reply]

To solve for y, you just need to take one term to the other side, cube root both sides, put it all back on one side, and then you have a quadratic in y to solve in the usual way. As with any quadratic, it has two solutions, so you would need to plot both of them to get the whole heart. --Tango (talk) 20:55, 23 June 2012 (UTC)[reply]

Easy for you, but it has been a few years for me.--Canoe1967 (talk) 03:36, 24 June 2012 (UTC)[reply]

Sorry, I didn't mean to suggest it was easy (just that it's a small number of steps once you know what the steps are). --Tango (talk) 13:29, 24 June 2012 (UTC)[reply]

No need to apologize. I still remember BEODMAS, It is just been a while is all. No one has corrected my transposition, so it may actually be correct. That would really amaze me if it was. The last tricky math I had to do were power factor corrections which can be quite tricky. They were multiple guess and they only wanted the answers that were cosine .9 angles or less because 1.0 corrections cost more. They were far easier for those of us that used polar/rectangular buttons, but most added the triangles with Pythagoras.--Canoe1967 (talk) 17:31, 24 June 2012 (UTC)[reply]

You mean your answer on the talk page? I don't think that is right. The easiest way to check is to test it for a few simple points. Try x=1 and y=1. That satisfies the equation given in the article, but it doesn't satisfy yours, so yours must be wrong. --Tango (talk) 19:13, 24 June 2012 (UTC)[reply]

Thanks. I will go over and strike it. Did we ever sort whether it can be done with the middle option, or just the parametric one? It may still be nice to have actual images uploaded if someone does succeed. I don't have a graphing calculator.--Canoe1967 (talk) 19:53, 24 June 2012 (UTC)[reply]

As I said, there will be two solutions to the quadratic, so you'll need to plot both curves to get the full shape. --Tango (talk) 23:45, 24 June 2012 (UTC)[reply]