Miscellaneous · Q6: Curve through $(0, \frac{\pi}{4})$ with $\sin x\cos y\,dx + \cos x\sin y\,dy = 0$

Basics you need

Solution

1. Separate the variables

   Near the point $\left(0, \frac{\pi}{4}\right)$⁠, $\cos x > 0$ and $\cos y > 0$⁠. Divide every term by $\cos x\cos y$⁠:

   $$\begin{aligned} \frac{\sin x}{\cos x}\,dx + \frac{\sin y}{\cos y}\,dy &= 0, \\ \tan x\,dx + \tan y\,dy &= 0. \end{aligned}$$
   In the first term $\cos y$ cancels, leaving $\frac{\sin x}{\cos x}$⁠; in the second $\cos x$ cancels, leaving $\frac{\sin y}{\cos y}$⁠. Now each term holds one variable only. The lines where $\cos x = 0$ or $\cos y = 0$ (such as $x = \pm\frac{\pi}{2}$⁠, $y = \frac{\pi}{2}$⁠) are excluded by the division; none of them passes through $\left(0, \frac{\pi}{4}\right)$⁠.
   Section 9.4.1: separate, then integrate. Lesson: variables separable

2. Integrate, with one constant

   $$\begin{aligned} \int\tan x\,dx + \int\tan y\,dy &= \log|C| \\ \log|\sec x| + \log|\sec y| &= \log|C| \\ \log|\sec x\sec y| &= \log|C|, \end{aligned}$$

   so $\sec x\sec y = \pm C$⁠. Renaming $\pm C$ as $C$⁠, the general solution is

   $$\sec x\sec y = C.\quad\ldots(1)$$
   Writing the arbitrary constant as $\log|C|$ (with $C \ne 0$⁠) lets the logarithms combine into one; it is still a single arbitrary constant, the right number for a first order equation. $\log|u| + \log|w| = \log|uw|$⁠.
   Section 7.3.1, standard integral (i): $\int\tan x\,dx = \log|\sec x| + C$⁠. Basics: every method ends with an integral

3. Use the point $\left(0, \frac{\pi}{4}\right)$

   Putting $x = 0$⁠, $y = \frac{\pi}{4}$ in (1):

   $$\sec 0\cdot\sec\frac{\pi}{4} = 1\cdot\sqrt2 = C,$$

   so $C = \sqrt2$⁠.

   "Passing through $\left(0, \frac{\pi}{4}\right)$" means $y = \frac{\pi}{4}$ when $x = 0$⁠. Of all the curves (1), only the one with $C = \sqrt2$ contains this point.
   Section 9.3: a particular solution from the general one. Lesson: Examples 4 to 9

4. The curve, and where it lives

   With $C = \sqrt2$⁠, $\sec x\sec y = \sqrt2$⁠. Since $\sec y = \frac{1}{\cos y}$⁠, this is $\frac{1}{\cos y} = \frac{\sqrt2}{\sec x}$⁠, that is

   $$\cos y = \frac{\sec x}{\sqrt2},$$

   or, multiplying by $\cos x$⁠, $\cos x\cos y = \frac{1}{\sqrt2}$⁠.

   A cosine is at most 1, so $\frac{\sec x}{\sqrt2} \le 1$⁠, which with $\cos x > 0$ means $\cos x \ge \frac{1}{\sqrt2}$⁠, that is $-\frac{\pi}{4} \le x \le \frac{\pi}{4}$⁠. The branch through $\left(0, \frac{\pi}{4}\right)$ is

   $$y = \cos^{-1}\left(\frac{\sec x}{\sqrt2}\right),\qquad -\frac{\pi}{4} < x < \frac{\pi}{4}.$$
   At $x = \pm\frac{\pi}{4}$ the curve reaches $y = 0$⁠, where $\sin y = 0$ and the tangent is vertical; there it stops being a function $y(x)$ with a derivative, so the open interval is the honest domain. The principal value of $\cos^{-1}$ lies in $[0, \pi]$⁠, and $y = \frac{\pi}{4}$ at $x = 0$ is on that branch.
   Principal value branch of $\cos^{-1}$ (Chapter 2, Section 2.2). Lesson: Miscellaneous exercise player

5. Check by substituting back, and see the curve

   Differentiate $\cos x\cos y = \frac{1}{\sqrt2}$ with respect to $x$⁠:

   $$-\sin x\cos y - \cos x\sin y\,\frac{dy}{dx} = 0,$$

   which, multiplied by $-dx$⁠, is $\sin x\cos y\,dx + \cos x\sin y\,dy = 0$⁠: the given equation. And $\cos 0\cos\frac{\pi}{4} = \frac{1}{\sqrt2}$⁠, so the curve passes through the point.

   Getting the equation back proves the answer. The picture shows the whole relation $\cos x\cos y = \frac{1}{\sqrt2}$ is a closed oval; the slope field is tangent to it everywhere, and the given point sits at its top.
   Implicit differentiation (Chapter 5, Section 5.3.2). Basics: a rule for slopes

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