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I excuse myself in advance if the answer to this problem seems blatantly simple, yet I seem to be hitting a brick wall and not getting anywhere, and it seems to me to be a relatively simple problem.

The problem:

Find all curves in the first quadrant so that the following is true: the triangle, bounded by the tangent line to the curves in point T, the x-axis and the line OT has a constant surface area of p (where p is a positive integer, naturally). The point O marks the origin. This must be true for every point T (meaning the choice of T is arbitrary).

Attempt at the solution:

The idea is of course to set up a differential equation, since we have $f'$ as the slope of the tangent line. The equation of the tangent line can be derived as follows:

$y-f(a)=f'(a)(x-a)$,

where I marked $y$ as the tangent, $f$ as the function we are seeking and $a$ the point $x$ in which we determined the tangent to the function.

Now, marking the sides of the triangle as:

$|OX|$ - the side laying of the x-axis

$|OT|$ - the side from the origin to the point T

$|TX|$ - the tangent line to the x-axis

Now, I reasoned to obtain the expressions for these sides as follows:

$|OX|=a-\frac{f(a)}{f'(a)}$ (by setting $y$ to $0$)

$|OT|=\sqrt{(a-0)^2+(f(a)-0)^2}=\sqrt{a^2+f(a)^2}$ (Euclidian distance)

$|TX|=\sqrt{(a-(a-\frac{f(a)}{f'(a)}))^2+(f(a)-0)^2}=\sqrt{(\frac{f(a)}{f'(a)})^2+f(a)^2}=\frac{f(a)\sqrt{1+f'(a)^2}}{f'(a)}$

and the generalise $a->x$, $f(a)->y$ and $f'(a)->y'$, thus obtaining:

$|OX|=x-\frac{y}{y'}=a$

$|OT|=\sqrt{x^2+y^2}=b$

$|TX|=\frac{y\sqrt{1+y'^2}}{y'}=c$

We may then connect the sides of the triangle with its surface by Heron's formula:

$S_{triangle}=p=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}$

However all this seems a bit cumbersome and rather than just plunging myself into all this algebra I spent my time thinking about more elegant ways. I thought about $S=rs$, where $r$ is the radius of the incircle, but then I just translated the problem into finding r, for which I had no idea.

Another idea was trying to express the differential equation through integral application as follows:

$S_{triangle}=p=\int_0^{f(a)}dy\int_{\frac{ay}{f(a)}}^{a+\frac{y-f(a)}{f'(a)}}dx$

I have obtained the above integral like this:

  1. First we integrate from $0$ to $f(a)$ by $y$ to avoid having to integrate twice.
  2. We integrate by $x$ from the line $|OT|$ to the line $|TX|$; the equation for $|OT|$ is simply $y=\frac{f(a)}{a}x$ and I have already written the tangent line above.

So, my question here is, if the above reasonings and methods are correct (or so to speak, which, if any, is correct). I am only concerned here with setting up the correct differential equation; solving that is not a problem. I am particularly uncomfortable at the moment by the substitution $a->x$, $f(a)->y$ and $f'(a)->y'$ as it seems kind of forced and I am wondering if it is the right thing to do.

I am grateful for all the answers and possibly (indeed probably) better solutions. Many thanks!

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    $\begingroup$ This may be wrong, but if i'm getting your question right, I would go like this. The height of the triangle is $f(x)$, the base is $\Big(x-\frac{f(x)}{f'(x)}\Big)$, thus $$2p=f(x)\Big(x-\frac{f(x)}{f'(x)}\Big)$$ $$f'=\frac{f^2}{(xf-2p)}$$ The solution is $$f(x)=\frac{x\pm\sqrt{x^{2}-4cp}}{2c}$$ $\endgroup$ – Kiryl Pesotski Feb 7 '18 at 23:51
  • $\begingroup$ @KirylPesotski You should post that as an answer. $\endgroup$ – amd Feb 8 '18 at 1:46
  • $\begingroup$ Hm, yes, the integral approach yields the same result, which means my thinking was correct (albeit your solution being of course more straight forward, and as usual with me, I haven't thought about it). Now it might not make much of a difference with the triangle, but this means, I can use this integral approach with bodies, that do not have such nice area formulas. On retrospect the thing that bothered me so much and thus prevented me from solving it was the fact that you take $a$ to be $x$ and $f(a)$ to be $y$ in the final differential equation (it still does a bit). Anyhow, thanks! $\endgroup$ – Nejc Kejzar Feb 8 '18 at 7:44
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The height of the triangle is $f(x)$, the base is $\Big(x-\frac{f(x)}{f'(x)}\Big)$, thus $$2p=f(x)\Big(x-\frac{f(x)}{f'(x)}\Big)$$ $$2pf'=xff'-f^2$$ The solution is $$f(x)=\frac{x\pm\sqrt{x^{2}-4cp}}{2c}$$

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Write your curve in the form $$y\mapsto\bigl(g(y),y\bigr)$$ with an unknown function $x=g(y)$, assumed decreasing for the moment. Then we obtain the condition $${1\over2} y\bigl(g(y)-y g'(y)\bigr)=p\ ,\tag{1}$$ which is an inhomogeneous linear differential equation for $g$. The corresponding homogeneous equation $g(y)-y g'(y)=0$ has the solutions $g_{\rm hom}(y)=C y$, and the "variation of the constant" Ansatz then leads to $$g(y)=c_0 y+{p\over y}\ .$$ This is only the general solution of $(1)$. One now has to do some qualitative geometric analysis in order to check to which extent this is the solution of the original problem.

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