Finding a formula for this curve

  • Thread starter BoomMath
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  • #1
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Hi,

I just registered to this forum, i'm working on the following problem.

In the picture you see a family of linear functions. I need to find a function that is tangent/'just touches' (to) the lines. I immediately thought of a tractrix, but it seems to be a little different. I'd like some help in the right direction. :smile:

Can I solve this using differential equations?

https://www.physicsforums.com/attachment.php?attachmentid=44496&stc=1&d=1330459648
View attachment 44499

thanks in advance
 
Last edited:

Answers and Replies

  • #2
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I understand what you want, but your description of the problem is a little off.

Basically, what you want is that if you draw all the possible lines, then you want an equation of the curve it describes.

Let me solve this for you:
Firstly, let's describe the general line. The general line will connect [itex](a,0)[/itex] and (0,10-a), with a in [0,10]. This line is given by

[tex]y=(1-\frac{10}{a})x+10-a[/tex]

So for each a we have a line [itex]L_a[/itex].

What we want to do now is to fix a point x and find the maximum point that the lines take on.
So, for a certain x, we want to find

[tex]\max\{(1-\frac{10}{a})x+10-a~\vert~a\in [0,10]\}[/tex]

To maximize this, we consider the function

[tex]f(a)=(1-\frac{10}{a})x+10-a[/tex]

of which we want to find the maximum. This can be easily done by calculus.

The derivative of f is

[tex]f^{\prime}(x)=\frac{10x}{a^2}-1[/tex]

We find when this is equal to 0 and we find that

[tex]a=\sqrt{10x}[/tex]

So the maximum value is reached for this a. The actual maximum value is now given by

[tex](1-\frac{10}{\sqrt{10x}})x+10-\sqrt{10x}=x-2\sqrt{10x}+10[/tex]

So the desired function is

[tex]y=x-2\sqrt{10x}+10[/tex]
 
  • #3
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Thank you very much!

I didn't realise the answer was so straightforward, i was trying to fit a parabola on the family of functions, hehe.
 

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