Investigation: For what values of $k$ does $y=k\sqrt{x^2-1}$ intersect $y=\ln x$

Question

I'm doing a bit of investigating to try to see where two curves intersect.

More specifically, I want to see for what values of $k$ does $y=k\sqrt{x^2-1}$ intersect $y= \ln x$

When you draw the two curves, we notice that the root curve has a vertical tangent and diverges from there, whereas the log curve has gradient 1 and then diverges (pathetically slowly) from there.

Now, for small values of k (ie: $k=0.2$), we can see that the square root curve intersects the log curve. When $k$ is larger, say $k=2$, then it totally misses the log curve.

How would I find the value of $k$ such that they are tangential (ie: the border between intersecting or not intersecting)? Is there a nice closed form?

My current thoughts are to let this point be say $\alpha$ and then equate their derivatives at $\alpha$ as well as their $y$ coordinates at $\alpha$.

However, I get some equation that I cannot really solve, is there a better way?

Answer 1

In order you have a solution, the function $$f(x)=k\sqrt{x^2-1}-\log(x)$$ must go though a minimum and the value of the function at that point must be negative. So, $$f'(x)=\frac{k x}{\sqrt{x^2-1}}-\frac{1}{x}$$ The minimum occurs when $$\frac{k x}{\sqrt{x^2-1}}=\frac{1}{x}$$ (this reduces to a quadratic in $x^2$ and we need to only consider the positive roots because of $\log(x)$) that is to say when $$x=x_1=\frac{\sqrt{\frac{\sqrt{1-4 k^2}+1}{k^2}}}{\sqrt{2}}$$ or $$x=x_2=\frac{\sqrt{2}}{\sqrt{\sqrt{1-4 k^2}+1}}$$ Now, looking at functions $f(x_1)$ and $f(x_2)$ as functions of $k$, we can discard the second since it is always positive. So, we just need $$f(x_1)=k \sqrt{\frac{\sqrt{1-4 k^2}+1}{2 k^2}-1}-\frac{1}{2} \log \left(\frac{\sqrt{1-4 k^2}+1}{2 k^2}\right) \lt 0$$ Graphing this function shows that, more or less, it is negative for $k \le 0.4$; a more precise calculation (using Newton for example would show $k \le 0.402371$).

Using Taylor expansion, you could easily show that, for small values of $k$, the second solution is close to $1+2k^2$.

However, you must take care that for $k<0.402371$, $f(x)=0$ has three solutions; the first one is trivial ($x=1$) and there are two other. For example :

• for $k=0.4$, you would find $x=1.99159$ and $x=2.49377$
• for $k=0.3$, you would find $x=1.25618$ and $x=6.12169$
• for $k=0.2$, you would find $x=1.09135$ and $x=12.7777$

Using Taylor expansion at $x=1$, you could eaily show that, for small values of $k$, the second solution (after $x=1$) is approximated by $1+2k^2$.

Answer 2

It looks like the problem doesn't have a closed form answer, but I think I have a slightly more elegant way of formulating the problem, anyway.

What you're trying to do is to maximize $k$ subject to the equations $y = k\sqrt{x^2-1}$ and $y = \ln x$ for $x > 0$. Solving for k gets you $k = \frac{y}{\sqrt{x^2 - 1}} = \frac{\ln x}{\sqrt{x^2 - 1}}$, so just find the maximum of that.

But the derivative of that still isn't particularly nice, so I doubt there's a closed form solution. WolframAlpha gives an exact solution using the Lambert W function, though.