Miscellaneous · Q12: Parametric $\frac{dy}{dx}$⁠: $x=10(t-\sin t)$⁠, $y=12(1-\cos t)$

Basics you need

Solution

1. Differentiate $x$ and $y$ with respect to $t$

   $$\begin{aligned} \frac{dx}{dt} &= 10(1 - \cos t), \\ \frac{dy}{dt} &= 12(0 + \sin t) = 12\sin t. \end{aligned}$$
   $\frac{d}{dt}(t - \sin t) = 1 - \cos t$ and $\frac{d}{dt}(1 - \cos t) = 0 - (-\sin t) = \sin t$⁠. The constants 10 and 12 stay in front.
   Table 5.3; sum rule, Section 5.3, rule (1); constant multiples (Class XI). Basics: derivative rules

2. Where is $\frac{dx}{dt} = 0$⁠?

   $\frac{dx}{dt} = 0$ means $\cos t = 1$⁠, i.e. $t = 2n\pi$ for an integer $n$⁠. The only such $t$ in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ is $t = 0$⁠. So for every other $t$ in the interval, $\frac{dx}{dt} > 0$ and the formula of Section 5.6 can be used.

   The book's parametric formula carries the condition "$\frac{dx}{dt} \ne 0$". Checking it is part of the answer: at $t = 0$ we would be dividing by zero.
   $\cos t = 1 \iff t = 2n\pi$ (Class XI). Basics: trigonometric identities

3. Divide: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

   For $t \ne 0$⁠:

   $$\frac{dy}{dx} = \frac{12\sin t}{10(1 - \cos t)} = \frac{6}{5}\cdot\frac{\sin t}{1 - \cos t}.$$
   This is the chain rule $\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}$ solved for $\frac{dy}{dx}$⁠. The fraction $\frac{12}{10}$ reduces to $\frac65$⁠.
   Section 5.6, "$\frac{dy}{dx} = \frac{g'(t)}{f'(t)}$⁠, provided $f'(t) \ne 0$". Lesson: parametric differentiation

4. Simplify with half angles

   $$\begin{aligned} \frac{\sin t}{1 - \cos t} &= \frac{2\sin\frac t2\cos\frac t2}{2\sin^2\frac t2} \\ &= \frac{\cos\frac t2}{\sin\frac t2} = \cot\frac{t}{2}. \end{aligned}$$

   Hence, for $-\frac{\pi}{2} < t < \frac{\pi}{2}$⁠, $t \ne 0$⁠:

   $$\frac{dy}{dx} = \frac{6}{5}\cot\frac{t}{2}.$$
   Cancelling one $2\sin\frac t2$ is allowed because $\sin\frac t2 \ne 0$⁠: for $t \ne 0$ in the interval, $\frac t2$ lies in $\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$ and is not 0. This mirrors Example 33, where $\frac{\sin\theta}{1 + \cos\theta}$ became $\tan\frac{\theta}{2}$⁠.
   $\sin 2A = 2\sin A\cos A$⁠, $\cos 2A = 1 - 2\sin^2 A$ with $A = \frac t2$ (Class XI). Basics: trigonometric identities

5. At $t = 0$ there is no derivative

   At $t = 0$ the point is $(x, y) = (0, 0)$⁠. The slope of the chord from the origin to the point with parameter $t \ne 0$ is

   $$\frac{y - 0}{x - 0} = \frac{12(1 - \cos t)}{10(t - \sin t)}.$$

   For $0 < t < \frac{\pi}{2}$ we have $\tan t > t$⁠, i.e. $\sin t > t\cos t$⁠. Hence $0 < t - \sin t < t - t\cos t = t(1 - \cos t)$⁠, and

   $$\frac{12(1 - \cos t)}{10(t - \sin t)} > \frac{12(1 - \cos t)}{10\,t(1 - \cos t)} = \frac{6}{5t}.$$

   As $t \to 0^+$⁠, $\frac{6}{5t} \to +\infty$⁠, so the chord slopes grow without bound. Replacing $t$ by $-t$ changes the sign of $x$ and keeps $y$⁠, so for $t \to 0^-$ the chord slopes tend to $-\infty$⁠. So $\frac{dy}{dx}$ does not exist at $t = 0$⁠: the curve has a sharp point (a cusp) at the origin, with a vertical tangent.

   The formula $\frac65\cot\frac t2$ also blows up as $t \to 0$⁠, with opposite signs on the two sides; the chord slopes confirm it directly. Here $x(-t) = -x(t)$ because $\sin$ is odd, and $y(-t) = y(t)$ because $\cos$ is even. The key gives $\frac65\cot\frac t2$ without the restriction $t \ne 0$⁠; the restriction is needed.
   The derivative as the limit of chord slopes (Section 5.3); $\sin t < t < \tan t$ for $0 < t < \frac{\pi}{2}$ (Class XI, Limits, used to prove $\frac{\sin t}{t} \to 1$⁠). Basics: what a limit is

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