Creating lung compliance curves but I'm terrible at the math/graphing

sozme
Messages
19
Reaction score
0

Homework Statement



I am trying to replicate the curves in the attachment, but since I am a physiologist it has been quite a while since I've done graphing. I am simply looking for an easy way to replicate these three curves. This is for a practice exam I am creating for first year medical students. The graphing program I got from sourceforge allows you to create these curves by inserting a function in the form of f(x) = or creating a point series.

Homework Equations



I believe the normal curve is roughly f(x) = √x but, I wish I knew what function to insert to make the line cross the X axis at +2.5 instead of 0. I am clueless how to create the shapes of the other two.

The Attempt at a Solution



Above
 
Physics news on Phys.org
sozme said:

Homework Statement



I am trying to replicate the curves in the attachment, but since I am a physiologist it has been quite a while since I've done graphing. I am simply looking for an easy way to replicate these three curves. This is for a practice exam I am creating for first year medical students. The graphing program I got from sourceforge allows you to create these curves by inserting a function in the form of f(x) = or creating a point series.

Homework Equations



I believe the normal curve is roughly f(x) = √x but, I wish I knew what function to insert to make the line cross the X axis at +2.5 instead of 0. I am clueless how to create the shapes of the other two.

The Attempt at a Solution



Above

I don't see an attachment -- try again?
 
If your curve has the same shape as f(x) =√x, it can be translated to the right by 2.5 units to have an x-intercept at (2.5, 0). That would be y = f(x - 2.5) = √(x - 2.5)
 
  • Like
Likes 1 person
berkeman said:
I don't see an attachment -- try again?

I'm sorry, I think it should be attached now. Thanks for that first answer Mark. Now the other 2 curves should be in the attachment and I need a way to recreate them.
 

Attachments

  • lungcompliance.jpg
    lungcompliance.jpg
    33.8 KB · Views: 359
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top