Exercise 5.6 · Q8: $\frac{dy}{dx}$ when $x=a\big(\cos t+\log\tan\frac t2\big)$⁠, $y=a\sin t$

Basics you need

Solution

1. Differentiate $\log\tan\frac{t}{2}$⁠: chain rule

   Three layers: outer $\log u$⁠, middle $u = \tan v$⁠, inner $v = \frac{t}{2}$⁠. So

   $$\begin{aligned} \frac{d}{dt}\log\tan\frac{t}{2} &= \frac{1}{\tan\frac{t}{2}}\cdot\sec^2\frac{t}{2}\cdot\frac12 \\ &= \frac{\cos\frac{t}{2}}{\sin\frac{t}{2}}\cdot\frac{1}{\cos^2\frac{t}{2}}\cdot\frac12 \\ &= \frac{1}{2\sin\frac{t}{2}\cos\frac{t}{2}} = \frac{1}{\sin t}. \end{aligned}$$
   Each layer contributes one factor: $\frac{1}{u}$ from the logarithm, $\sec^2 v$ from the tangent, $\frac12$ from $\frac{t}{2}$⁠. Writing everything in sines and cosines of $\frac{t}{2}$⁠, one $\cos\frac{t}{2}$ cancels, and the denominator $2\sin\frac{t}{2}\cos\frac{t}{2}$ is exactly $\sin t$⁠. On our intervals $\sin\frac{t}{2}$ and $\cos\frac{t}{2}$ are both nonzero (their ratio is positive), so every step is legitimate.
   Chain rule (Theorem 4, Section 5.3.1); $\frac{d}{du}\log u = \frac1u$ (Theorem 5, Section 5.4); $\sin t = 2\sin\frac{t}{2}\cos\frac{t}{2}$⁠. Basics: exponentials and logarithms

2. Assemble $\frac{dx}{dt}$

   $$\begin{aligned} \frac{dx}{dt} &= a\left(-\sin t + \frac{1}{\sin t}\right) \\ &= a\cdot\frac{1 - \sin^2 t}{\sin t} = \frac{a\cos^2 t}{\sin t}. \end{aligned}$$
   The derivative of $\cos t$ is $-\sin t$⁠, and Step 1 gives the rest. Combining over the common denominator $\sin t$ (nonzero on our intervals) and using $1 - \sin^2 t = \cos^2 t$ gives a single tidy fraction.
   Sum rule (Section 5.3); $\sin^2 t + \cos^2 t = 1$⁠.

3. Differentiate $y$ with respect to $t$

   $$\frac{dy}{dt} = a\cos t.$$
   The other rate the formula needs; straight from the table.
   $\frac{d}{dt}\sin t = \cos t$ (Table 5.3).

4. Find where the formula may be used

   We need $\frac{a\cos^2 t}{\sin t} \neq 0$⁠. Since $a \neq 0$⁠, this means $\cos t \neq 0$⁠. Inside the intervals $2n\pi \lt t \lt 2n\pi + \pi$⁠, $\cos t = 0$ only at

   $$t = 2n\pi + \frac{\pi}{2}.$$

   So the formula applies at every other $t$ of the domain.

   Before dividing, set aside the values where the denominator $\frac{dx}{dt}$ is zero. Of all the zeros of cosine, only the middle point of each interval lies in the domain.
   $\cos t = 0 \iff t = (2k+1)\frac{\pi}{2}$⁠; in $(2n\pi,\ 2n\pi + \pi)$ only $2n\pi + \frac{\pi}{2}$⁠.

5. Divide

   For $t$ in the domain with $\cos t \neq 0$⁠,

   $$\begin{aligned} \frac{dy}{dx} &= \frac{dy/dt}{dx/dt} = \frac{a\cos t}{a\cos^2 t/\sin t} \\ &= \frac{a\cos t\sin t}{a\cos^2 t} = \frac{\sin t}{\cos t} = \tan t. \end{aligned}$$
   Dividing by the fraction $\frac{a\cos^2 t}{\sin t}$ means multiplying by $\frac{\sin t}{a\cos^2 t}$⁠. The common factor $a\cos t$ is not zero (Step 4), so it cancels.
   Parametric formula, Section 5.6. Lesson: parametric differentiation

6. Picture the curve: a tractrix

   At $t = \frac{\pi}{2}$ both rates are zero: $\frac{a\cos^2 t}{\sin t} = 0$ and $a\cos t = 0$⁠. Since $\tan\frac{\pi}{4} = 1$⁠, the point has $x = a(0 + \log 1) = 0$ and $y = a\sin\frac{\pi}{2} = a$⁠: it is $(0, a)$⁠. Just before, the slope $\tan t$ is very large and positive; just after, very large and negative: the curve has a sharp tip (a cusp) at $(0, a)$⁠, where $\frac{dy}{dx}$ does not exist.

   As $t$ approaches $0$ or $\pi$⁠, $y = a\sin t$ approaches $0$ while $x$ runs off to $\mp\infty$ (because $\log\tan\frac{t}{2} \to \mp\infty$⁠): both branches creep along the $x$-axis.

   That last remark is the famous property of this curve, the tractrix: follow the tangent at the point with parameter $t$ down to the $x$-axis. Since the slope is $\tan t$⁠, the horizontal run is $\frac{y}{\tan t} = \frac{a\sin t}{\tan t} = a\cos t$⁠, so the length of the piece is

   $$\sqrt{(a\cos t)^2 + (a\sin t)^2} = |a|,$$

   the same for every point. (It is the path of a weight dragged on a rope of length $a$ by someone walking along the $x$-axis.)

   The picture explains the excluded value: it is the cusp at the top. The tangent-length property is a pleasing use of the answer $\tan t$⁠: the slope found without ever eliminating $t$ is all we need.
   Slope $= \frac{\text{rise}}{\text{run}}$⁠, so run $= \frac{\text{rise}}{\text{slope}}$⁠. Basics: slope at a point

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