Exercise 5.3 · Q8: $\frac{dy}{dx}$ if $\sin^2x+\cos^2y=1$

Basics you need

Solution

1. Differentiate both sides with respect to $x$

   • $\sin^2 x$⁠: outer $t^2$⁠, inner $\sin x$⁠: $$\frac{d}{dx}(\sin^2 x) = 2\sin x\cos x = \sin 2x.$$
   • $\cos^2 y$⁠: outer $t^2$⁠, middle $\cos$⁠, inner $y(x)$⁠: $$\begin{aligned} \frac{d}{dx}(\cos^2 y) &= 2\cos y\cdot(-\sin y)\,\frac{dy}{dx} \\ &= -\sin 2y\,\frac{dy}{dx}. \end{aligned}$$
   • $\frac{d}{dx}(1) = 0$⁠.

   So

   $$\sin 2x - \sin 2y\,\frac{dy}{dx} = 0.$$
   Each layer of a composition gives one factor. For $\cos^2 y$ the last factor is $\frac{dy}{dx}$⁠, because $y$ changes with $x$⁠; $\sin^2 x$ has no such factor because its innermost variable is $x$ itself.
   Chain rule, Theorem 4, Section 5.3.1; $(\sin t)' = \cos t$⁠, $(\cos t)' = -\sin t$⁠, Table 5.3; $2\sin t\cos t = \sin 2t$⁠. Basics: derivative rules

2. Solve for $\frac{dy}{dx}$

   $\sin 2y\,\frac{dy}{dx} = \sin 2x$⁠, so at points of the curve where $\sin 2y \ne 0$⁠,

   $$\frac{dy}{dx} = \frac{\sin 2x}{\sin 2y}.$$
   We may divide only by a non-zero number. Step 4 shows what happens at the points where $\sin 2y = 0$⁠.
   Collect and divide. Basics: solving for dy/dx

3. What the curve really is

   Replace $\cos^2 y$ by $1 - \sin^2 y$⁠: the relation becomes $\sin^2 x - \sin^2 y = 0$⁠. Now factor the left side. Multiplying out,

   $$\begin{aligned} &\sin(x + y)\,\sin(x - y) \\ &= \sin^2 x\cos^2 y - \cos^2 x\sin^2 y \\ &= \sin^2 x(1 - \sin^2 y) - (1 - \sin^2 x)\sin^2 y \\ &= \sin^2 x - \sin^2 y. \end{aligned}$$

   So the relation holds exactly when $\sin(x + y) = 0$ or $\sin(x - y) = 0$⁠, that is, when

   $$x + y = n\pi \quad\text{or}\quad x - y = n\pi, \quad n \in \mathbf{Z}.$$
   The relation is not a single smooth curve but a grid of straight lines (see the figure). The first line uses $(a + b)(a - b) = a^2 - b^2$ with $a = \sin x\cos y$⁠, $b = \cos x\sin y$⁠.
   $\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$⁠; $\sin t = 0 \iff t = n\pi$ (Class XI). Basics: trigonometric identities

4. The formula on the lines, and where it fails

   On a line $x - y = n\pi$⁠: $y = x - n\pi$⁠, so $\sin 2y = \sin(2x - 2n\pi) = \sin 2x$⁠, and the formula gives $\frac{dy}{dx} = 1$⁠, the slope of that line.

   On a line $x + y = n\pi$⁠: $y = n\pi - x$⁠, so $\sin 2y = \sin(2n\pi - 2x) = -\sin 2x$⁠, and the formula gives $\frac{dy}{dx} = -1$⁠, again the slope of the line.

   Where $\sin 2y = 0$⁠: then $y = \frac{k\pi}{2}$ for an integer $k$⁠, so $\sin^2 y$ is $0$ or $1$⁠, and $\sin^2 x = \sin^2 y$ makes $x$ a multiple of $\frac{\pi}{2}$ of the same kind (even or odd). Both $x + y$ and $x - y$ are then multiples of $\pi$⁠: these are exactly the points where a line of slope $1$ crosses a line of slope $-1$⁠. Near a crossing the curve is not the graph of one function, and Step 1 reads $0 = 0$⁠, which gives no slope.

   Checking the formula against the true shape of the curve confirms it, and explains the condition $\sin 2y \ne 0$⁠: it removes exactly the crossing points.
   $\sin(t + 2n\pi) = \sin t$⁠, $\sin(-t) = -\sin t$ (Class XI). Basics: trigonometric identities

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