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x^2+(y-∛(x^2 ) )^2=1

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mollykins | 17:26 Tue 08th Feb 2011 | Science
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is this the equation for a line in the shape of a heart? x^2+(y-∛(x^2 ) )^2=1

I heard it was, but i'm not convinced, I might try and draw it out later, when I have more time, to see though.
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Here you go Molly.

You should acquaint yourself with Wolfram Alpha:

http://www.wolframalp...2%88%9B(X^2+)+)^2%3D1
17:53 Tue 08th Feb 2011
I've never been scared by a question title until now. :o(
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Hi Molly,

Scroll down on link....it might help you. :0)

http://en.wikipedia.o...ki/Heart_%28symbol%29
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Question Author
it doesn't match any of the ones on wiki, but it could just be in a different style.
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*snigger*
you could use MS Mathematics http://www.microsoft....default.aspx#overview to plot it
I just tried that, bibble and couldn't get it to work (admittedly I only tried for about a minute)
Here you go Molly.

You should acquaint yourself with Wolfram Alpha:

http://www.wolframalp...2%88%9B(X^2+)+)^2%3D1
Question Author
Thanks bibble, i'm downloading it now although it's in three bits so i'm slightly confused.
should be possible to set up a spread sheet, input some values and then draw it out, it's not as bad an equation as it looks!
This video is a good introduction to the aims of Wolfram Alpha
http://www.youtube.com/watch?v=60P7717-XOQ
Question Author
Thanks ed, so it is a heart.
Thats brilliant Ed

........Molly, you have your answer thanks to the Ed :0)
In short, yes!
Yes Chuck I found it necessary to rearrange the terms and have 2 equations, one for a positive square root and the other for the negative square root. First time I've tried it for real.
Yip and so does the Ed Molly :)
bit tricky for a spreasheet as you'll get positive and negative parts from the square root that will need to be accounted for.

The figure you're asking about is a cardioid

http://www-history.mc.../Curves/Cardioid.html

Was the answer to a University challenge question last week - bit of a coincidence!

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