Example 34: $\frac{dy}{dx}$ if $x^{2/3}+y^{2/3}=a^{2/3}$

Basics you need

Solution

1. Choose the parametrisation and check it lies on the curve

   Let $x = a\cos^3\theta$⁠, $y = a\sin^3\theta$⁠. By Basic 1, $\sqrt[3]{x} = \sqrt[3]{a}\cos\theta$ and $\sqrt[3]{y} = \sqrt[3]{a}\sin\theta$⁠, so

   $$\begin{aligned} x^{\frac23} + y^{\frac23} &= a^{\frac23}\cos^2\theta + a^{\frac23}\sin^2\theta \\ &= a^{\frac23}\left(\cos^2\theta + \sin^2\theta\right) \\ &= a^{\frac23}. \end{aligned}$$

   So every point $(a\cos^3\theta, a\sin^3\theta)$ is on the curve.

   We want something whose $\frac23$ power gives $\cos^2\theta$⁠. Since $\frac23 \times 3 = 2$⁠, the cube $\cos^3\theta$ does the job: $\big(\cos^3\theta\big)^{\frac23} = \cos^2\theta$⁠. Then the identity $\cos^2\theta + \sin^2\theta = 1$ does all the work. This is the book's choice too.
   $\sqrt[3]{\cos^3\theta} = \cos\theta$ even when $\cos\theta \lt 0$⁠, because real cube roots keep the sign (Basic 1).

2. Check that the parametrisation gives the whole curve

   Take any point $(x, y)$ on the curve and put $u = \frac{\sqrt[3]{x}}{\sqrt[3]{a}}$⁠, $v = \frac{\sqrt[3]{y}}{\sqrt[3]{a}}$⁠. Dividing the equation by $a^{\frac23}$ gives $u^2 + v^2 = 1$⁠, so by Basic 4 there is a $\theta$ with $u = \cos\theta$⁠, $v = \sin\theta$⁠. Cubing, $x = a\cos^3\theta$ and $y = a\sin^3\theta$⁠.

   Step 1 showed the parametric points lie on the curve; this step shows that no point of the curve is missed. Together they say that the parametric equations describe exactly the given curve, so a slope found from them is the slope of the given curve at every point. (The book shows only the first direction.)
   Cubing undoes the real cube root: $\big(\sqrt[3]{x}\big)^3 = x$⁠.

3. Differentiate $x$ and $y$ with respect to $\theta$

   Chain rule with outer function $u^3$ and inner function $u = \cos\theta$ (then $u = \sin\theta$⁠):

   $$\begin{aligned} \frac{dx}{d\theta} &= a\cdot 3\cos^2\theta\cdot(-\sin\theta) \\ &= -3a\cos^2\theta\sin\theta, \\ \frac{dy}{d\theta} &= a\cdot 3\sin^2\theta\cdot\cos\theta \\ &= 3a\sin^2\theta\cos\theta. \end{aligned}$$
   $\cos^3\theta$ means $(\cos\theta)^3$⁠, a function of a function. The chain rule differentiates the outside (cube) first, keeping the inside, then multiplies by the derivative of the inside. The minus sign in $\frac{dx}{d\theta}$ comes from $\frac{d}{d\theta}\cos\theta = -\sin\theta$⁠.
   Chain rule, Theorem 4 (Section 5.3.1). Lesson: chain rule

4. Find where the formula may be used

   We need $\frac{dx}{d\theta} = -3a\cos^2\theta\sin\theta \neq 0$⁠, that is

   $$\cos\theta \neq 0 \ \text{ and } \ \sin\theta \neq 0.$$

   So $\theta$ must not be a multiple of $\frac{\pi}{2}$⁠. The excluded values give the four points $(a, 0)$⁠, $(0, a)$⁠, $(-a, 0)$⁠, $(0, -a)$⁠, and at each of them $\frac{dy}{d\theta} = 0$ as well.

   These four points are the sharp tips (cusps) of the curve, shown in the figure in Step 6. At $(\pm a, 0)$ an upper and a lower branch meet in a point, so near there $y$ is not a single function of $x$⁠; at $(0, \pm a)$ the curve comes in vertically from both sides. Either way $\frac{dy}{dx}$ does not exist there.
   A product is nonzero exactly when every factor is nonzero; $-3a \neq 0$ since $a \neq 0$⁠.

5. Divide and simplify

   For $\theta$ not a multiple of $\frac{\pi}{2}$⁠,

   $$\begin{aligned} \frac{dy}{dx} &= \frac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta} \\ &= -\frac{\sin\theta}{\cos\theta} = -\tan\theta. \end{aligned}$$
   The common factor $3a\sin\theta\cos\theta$ cancels; it is not zero by Step 4. What is left is $\frac{\sin\theta}{\cos\theta}$ with a minus sign.
   Parametric formula (Basic 3) and $\tan\theta = \frac{\sin\theta}{\cos\theta}$⁠. Lesson: parametric differentiation

6. Write the answer in terms of $x$ and $y$

   Since $\cos\theta \neq 0$⁠, $x \neq 0$⁠, and

   $$\frac{y}{x} = \frac{a\sin^3\theta}{a\cos^3\theta} = \tan^3\theta.$$

   Taking real cube roots (Basic 1), $\tan\theta = \sqrt[3]{\dfrac{y}{x}}$⁠, whatever the sign of $\tan\theta$⁠. Hence

   $$\frac{dy}{dx} = -\tan\theta = -\sqrt[3]{\frac{y}{x}}.$$
   The question is stated in $x$ and $y$⁠, so it is natural to give the answer in $x$ and $y$ as well; this is the book's final form. The cube root has no sign problem, because a real cube root exists for negative numbers too. The figure also shows a lovely consequence of the slope $-\tan\theta$⁠: the tangent line $y - a\sin^3\theta = -\tan\theta\,(x - a\cos^3\theta)$ meets the axes at $A(a\cos\theta, 0)$ and $B(0, a\sin\theta)$⁠, so the piece $AB$ always has length $\sqrt{a^2\cos^2\theta + a^2\sin^2\theta} = a$⁠.
   $\frac{\sin^3\theta}{\cos^3\theta} = \left(\frac{\sin\theta}{\cos\theta}\right)^3 = \tan^3\theta$⁠.

7. Check by implicit differentiation

   Differentiate $x^{\frac23} + y^{\frac23} = a^{\frac23}$ with respect to $x$⁠, for $x \neq 0$⁠, $y \neq 0$ (Basic 5 and the chain rule):

   $$\frac23 x^{-\frac13} + \frac23 y^{-\frac13}\,\frac{dy}{dx} = 0.$$

   Solve for $\frac{dy}{dx}$⁠:

   $$\frac{dy}{dx} = -\frac{x^{-\frac13}}{y^{-\frac13}} = -\frac{y^{\frac13}}{x^{\frac13}} = -\sqrt[3]{\frac{y}{x}}.$$

   The same answer. The conditions $x \neq 0$⁠, $y \neq 0$ exclude exactly the four cusps of Step 4.

   The parametric route avoided negative fractional powers; implicit differentiation meets them head on but reaches the same result. Agreement of two independent methods is a strong check.
   $\frac{d}{dx}\big(y^{\frac23}\big) = \frac23 y^{-\frac13}\frac{dy}{dx}$ by the chain rule. Lesson: implicit differentiation

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