2
$\begingroup$

Find the positive real number(s) $c$ such that the graphs of $y=x^c$ and $x=y^c$ intersect (somewhere) at an angle $\frac{\pi}{4}$.

A problem from Mark Krusemeyer. Should one be using the symmetry of the graphs along $x=y$ due the functions being inverses of one another? Just a thought.

$\endgroup$

2 Answers 2

8
$\begingroup$

Here's a brute force approach. There's probably a more elegant way of doing it.

The two curves can be written as $y=x^c$ and $y=x^{1/c}$. They always intersect at $x=0$ and $x=1$ (for $c=1$ the curves coincide).

alt text

The derivatives of the two curves are $c x^{c-1}$ and $c^{-1}x^{1/c-1}$ respectively. For $0<c\neq1$ the angle between them at $x=0$ is clearly always $\pi/2$. At $x=1$ we want the angle between them to be $\pi/4$, so $$ \tan^{-1}(c) - \tan^{-1}(1/c) = \pm\frac\pi4 \implies c = \pm 1+\sqrt2 $$

The two solutions just reflect the fact that you can swap the two curves, or equivalently, set $c\to1/c$.

There are also two negative solutions to the above equation which are excluded by the question. These are also valid curves and solutions, which can be obtained by the transformation $x \to 1/x$.

$\endgroup$
5
$\begingroup$

It is possible to use the symmetry as a slight shortcut to Simon's answer: Given the graph and conditions as laid out there, the relevant point of intersection must be at $x=y=1$. Because of the symmetry, the angles formed by the tangent lines to the curves at that point and the line $y=x$ must be $\frac{\pi}{8}$, so the slope of the tangent line to the blue curve (lower for $0< x<1$) at $x=1$ must be $\tan\frac{3\pi}{8}=1+\sqrt{2}$. Now, either $cx^{c-1}|_{x=1}=c=1+\sqrt{2}$ or $c^{-1}x^{1/c-1}|_{x=1}=\frac{1}{c}=1+\sqrt{2}$, so $c=1+\sqrt{2}$ or $c=\frac{1}{1+\sqrt{2}}=-1+\sqrt{2}$.

$\endgroup$
3
  • 1
    $\begingroup$ Your second solution (which is, of course, correct) comes from just swapping the curves, or equivalently $c \to 1/c$. In my solution (now updated) that comes from using $\pm\pi/4$. $\endgroup$
    – Simon
    Jan 12, 2011 at 8:32
  • $\begingroup$ I forgot to say - your solution is superior since the final equation that needs solving is simpler. $\endgroup$
    – Simon
    Jan 12, 2011 at 8:40
  • 1
    $\begingroup$ @Simon: Thanks. Really, though, it's not that much of a difference: $\arctan c-\arctan\frac{1}{c}=\arctan c-\text{arccot } c = \arctan c-(\frac{\pi}{2}-\arctan{c})=2\arctan{c}-\frac{\pi}{2}$, so $\arctan c-\arctan\frac{1}{c}=\frac{\pi}{4}\implies 2\arctan{c}-\frac{\pi}{2}=\frac{\pi}{4}\implies\arctan c=\frac{3\pi}{8}$ and your equation readily turns into mine. $\endgroup$
    – Isaac
    Jan 12, 2011 at 9:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.