ALO Posted January 12, 2019 Report Share Posted January 12, 2019 Hello, I've been been working on my Math IA, and I suddenly got stuck when I came up with this graph: I know it can't be exponential....oh and also the base equation for this graph is E = kl/r^4 Reply Link to post Share on other sites More sharing options...
kw0573 Posted January 12, 2019 Report Share Posted January 12, 2019 Is the plot linear if you plot E against 1/r^4? Reply Link to post Share on other sites More sharing options...
ALO Posted January 12, 2019 Author Report Share Posted January 12, 2019 no. I get this: And also in the equation I mentioned k is a constant, 9. Is there another way I can make this graph into a linear? Reply Link to post Share on other sites More sharing options...
kw0573 Posted January 12, 2019 Report Share Posted January 12, 2019 You have the data for the plot, right? You plotted y = 1/x^4, but I mean to use values of E against values of 1/r^4 Reply Link to post Share on other sites More sharing options...
ALO Posted January 12, 2019 Author Report Share Posted January 12, 2019 hmm. I actually don't have values for E and r as of yet, as I assume I have to do it later. I need to find out what type of relationship there is between E and r. know Reply Link to post Share on other sites More sharing options...
kw0573 Posted January 12, 2019 Report Share Posted January 12, 2019 Didn't you say E = kl/r^4 so E is proportional to 1/r^4? Reply Link to post Share on other sites More sharing options...
ALO Posted January 12, 2019 Author Report Share Posted January 12, 2019 I don't quite understand how E = kl/r^4 would be proportional to 1/r^4. and if E does = 1/r^4, what does it mean? I am sorry for asking too many questions, it's just that I feel more confused now Reply Link to post Share on other sites More sharing options...
kw0573 Posted January 12, 2019 Report Share Posted January 12, 2019 For the sake of argument, let's call 1 / r^4 = x E = klx, which means E is proportional to x. I am not sure if this is relevant, but if you have the graph, you can obtain individual points (x, y) from the graph using the following online software https://automeris.io/WebPlotDigitizer/ Reply Link to post Share on other sites More sharing options...
ALO Posted January 12, 2019 Author Report Share Posted January 12, 2019 Thank you! Reply Link to post Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.