I'm the teaching assistant for a first semester calculus course, and the professor has given the students the following problem:

Find the points on the curve $xy=\sin(x+y)$ that have a vertical tangent line.

Here is a picture of the curve:

enter image description here

My attempt to solve this problem led me to finding points on the given curve which also satisfy $x=\cos(x+y)$, but I can't figure out how to simultaneously solve these equations (without resulting to numerical methods, which the students are not assumed to know). Is there something I'm missing, or has the professor given the students a problem more difficult than he intended?

Edit: I'm still looking for a solution not requiring numerical methods, or proof that no such solution exists.

  • 1
    $\begingroup$ It may be helpful to look at $\cos^2(x+y) + \sin^2(x+y) = 1$. $\endgroup$ – Michael Albanese Sep 25 '13 at 4:49
  • 1
    $\begingroup$ $\frac{dy}{dx} = \frac{xcos(x+y)(1 +\frac{dy}{dx}) - sin(x+y)}{x^2}$ $\endgroup$ – Don Larynx Sep 25 '13 at 4:51
  • $\begingroup$ @MichaelAlbanese: In fact you suggested him to think of $x^2(1+y^2)=1$? $\endgroup$ – mrs Sep 25 '13 at 4:53
  • $\begingroup$ @Calvin: Yes we are. $\endgroup$ – Jared Sep 25 '13 at 4:53
  • $\begingroup$ @BabakS: Sure, why not? $\endgroup$ – Michael Albanese Sep 25 '13 at 5:02

Let's consider $x$ as a function of $y$. We have to find points $\left(x, y\right)$ where $x' = 0$. That yields the equation $x = \cos\left(x + y\right)$. Then, we have a system of two equations: $$ \left\{% \begin{array}{rcl} xy & = & \sin\left(x + y\right) \\ x & = & \cos\left(x + y\right) \end{array}\right. $$

From those equations we get $\left\vert xy \right\vert \leq 1$ and $\left\vert x \right\vert \leq 1$. In addition, we have $x^{2}\left(y^{2} + 1\right) = 1$. The last identity is satisfied by the choice $x = \cos\left(\theta\right)$ and $y = \tan\left(\theta\right)$. That means

$$ \phi \equiv \theta + 2n\pi = \cos\left(\theta\right) + \tan\left(\theta\right)\,, \qquad n \in {\mathbb Z} $$

We have reduced the whole problem to a one variable problem: $$ \cos^{2}\left(\phi\right) - \phi\cos\left(\phi\right) + \sin\left(\phi\right) = 0 \quad\mbox{with}\quad \left\vert% \begin{array}{rcl} \,\,\, x & = & \cos\left(\phi\right) \\ \,\,\, y & = & \tan\left(\phi\right) \end{array}\right. $$

Solutions of this equation require $$ \sin\left(\phi\right) \leq {\phi^{2} \over 4} $$

Below, we can see plots of $\cos^{2}\left(\phi\right) - \phi\cos\left(\phi\right) + \sin\left(\phi\right)$. We have to choose the roots which are consistent with the original equations since we have to pay attention to the solution signs. For example, the first positive root is $\phi \approx 4,50321873398481\ldots$ which yields $x \approx -0.207648303505316\ldots$ and $y \approx 4,710867037490116\ldots$.

enter image description here

  • $\begingroup$ +1 Hm, I still don't see how we're supposed to solve for $\phi$ (assuming that's what is required). Also, why do you state that we require $\sin \phi \leq \frac{\phi^2}{4}$? This is obviously true other that $ 0 < \phi < 2$, but from the first graph there are no solutions in that region. $\endgroup$ – Calvin Lin Sep 25 '13 at 14:07
  • $\begingroup$ Thanks for this answer. Are you implying that you think the solution requires numerical methods? $\endgroup$ – Jared Sep 26 '13 at 4:22
  • $\begingroup$ @Jared Yes. But it's is a very simple one. $\endgroup$ – Felix Marin Sep 26 '13 at 22:57
  • $\begingroup$ @CalvinLin I wrote the "$\large\sin\left(\phi\right)$" condition before I saw the graphic. It came from the second degree equation for $\large\cos\left(\phi\right)$. The benefit of my approach is that the initial two unknowns $\large x$ and $\large y$ problem is reduced to the solution of a $\large{\bf ONE\ VARIABLE}$ problem which is quite simple in numerical terms. $\endgroup$ – Felix Marin Sep 26 '13 at 23:01

(not a complete solution, but I wanted to add a graph)
I suspect that it's a much more difficult question than intended.

By implicit differentiation, $x \frac{dy}{dx} + y = \cos( x+y) ( 1 + \frac{dy}{dx} ) $.
This gives $\frac{dy}{dx} = \frac{-y + \cos(x+y) } { x - \cos (x+y) }$.
We want a vertical tangent, which means that the denominator is 0, so $x = \cos (x+y)$.
Dividing the first equation, we get $ y = \tan (x+y)$.
There are actually infinitely many solutions, (at least) 1 for each $ n\pi < x+y < (n+1) \pi$.

The blue line is bounded between $-1 $ and $1$ and bounces back and forth. The red lines are similar to a tilted $\tan \theta$ graph.

enter image description here


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