Exercise 5.6 · Q7: $\frac{dy}{dx}$ when $x=\frac{\sin^3t}{\sqrt{\cos2t}}$⁠, $y=\frac{\cos^3t}{\sqrt{\cos2t}}$

Basics you need

Solution

1. Differentiate $x$⁠: product rule and chain rule

   Write $x = \sin^3 t\cdot(\cos 2t)^{-\frac12}$⁠. By the product rule and Basic 2,

   $$\begin{aligned} \frac{dx}{dt} &= 3\sin^2 t\cos t\,(\cos 2t)^{-\frac12} \\ &\quad + \sin^3 t\sin 2t\,(\cos 2t)^{-\frac32}. \end{aligned}$$

   Put both terms over $(\cos 2t)^{\frac32}$ (multiply the first by $\frac{\cos 2t}{\cos 2t}$⁠):

   $$\frac{dx}{dt} = \frac{N_x}{(\cos 2t)^{\frac32}},$$

   where

   $$N_x = 3\sin^2 t\cos t\cos 2t + \sin^3 t\sin 2t.$$
   Writing $\frac{1}{\sqrt{\cos 2t}}$ as the power $(\cos 2t)^{-\frac12}$ turns a quotient into a product, which is easier to differentiate. A common denominator then lets us study the numerator $N_x$ on its own. Since $\cos 2t \gt 0$⁠, the denominator $(\cos 2t)^{\frac32}$ is a positive number.
   Product rule (Section 5.3); chain rule (Theorem 4, Section 5.3.1). Lesson: chain rule

2. Simplify the numerator $N_x$

   Use $\sin 2t = 2\sin t\cos t$ in the second term and take out $\sin^2 t\cos t$⁠:

   $$\begin{aligned} N_x &= 3\sin^2 t\cos t\cos 2t + 2\sin^4 t\cos t \\ &= \sin^2 t\cos t\,(3\cos 2t + 2\sin^2 t). \end{aligned}$$

   Now $3\cos 2t = 3 - 6\sin^2 t$⁠, so the bracket is $3 - 4\sin^2 t$⁠, and

   $$\begin{aligned} N_x &= \sin t\cos t\,(3\sin t - 4\sin^3 t) \\ &= \sin t\cos t\sin 3t. \end{aligned}$$

   Therefore

   $$\frac{dx}{dt} = \frac{\sin t\cos t\sin 3t}{(\cos 2t)^{\frac32}}.$$
   The aim is a product of simple factors, so that we can see where $\frac{dx}{dt}$ vanishes and what cancels later. Writing everything in $\sin t$ turns the bracket into $3 - 4\sin^2 t$⁠; one more $\sin t$ makes it $3\sin t - 4\sin^3 t$⁠, which is exactly $\sin 3t$⁠.
   $\cos 2t = 1 - 2\sin^2 t$ and $\sin 3t = 3\sin t - 4\sin^3 t$ (Basic 3). Basics: trigonometric identities

3. Differentiate $y$ and simplify the same way

   Write $y = \cos^3 t\cdot(\cos 2t)^{-\frac12}$⁠. With $\frac{d}{dt}\cos^3 t = -3\cos^2 t\sin t$⁠,

   $$\frac{dy}{dt} = \frac{N_y}{(\cos 2t)^{\frac32}},$$

   where

   $$N_y = -3\cos^2 t\sin t\cos 2t + \cos^3 t\sin 2t.$$

   Use $\sin 2t = 2\sin t\cos t$⁠, take out $\cos^2 t\sin t$⁠, and use $3\cos 2t = 6\cos^2 t - 3$⁠:

   $$\begin{aligned} N_y &= \cos^2 t\sin t\,(-3\cos 2t + 2\cos^2 t) \\ &= \cos^2 t\sin t\,(3 - 4\cos^2 t) \\ &= -\sin t\cos t\,(4\cos^3 t - 3\cos t) \\ &= -\sin t\cos t\cos 3t. \end{aligned}$$

   Therefore

   $$\frac{dy}{dt} = \frac{-\sin t\cos t\cos 3t}{(\cos 2t)^{\frac32}}.$$
   Exactly the same plan as for $x$⁠, with the roles of sine and cosine swapped. This time the bracket is written in $\cos t$⁠, and $4\cos^3 t - 3\cos t$ is $\cos 3t$⁠; the minus sign comes from taking $-1$ out of $3 - 4\cos^2 t$⁠.
   $\frac{d}{dt}\cos^3 t = 3\cos^2 t\cdot(-\sin t)$ (chain rule); $\cos 2t = 2\cos^2 t - 1$⁠; $\cos 3t = 4\cos^3 t - 3\cos t$⁠.

4. Find where $\frac{dx}{dt} = 0$

   Remember that $t$ lies where $\cos 2t \gt 0$⁠. The derivative $\frac{dx}{dt}$ is zero exactly when $\sin t$⁠, $\cos t$ or $\sin 3t$ is zero. Check each against the domain:

   • $\cos t = 0$ would give $\cos 2t = 2\cos^2 t - 1 = -1 \lt 0$⁠: never in the domain.
   • $\sin t = 0$ at $t = n\pi$⁠, where $\cos 2t = 1 \gt 0$⁠: these are in the domain.
   • $\sin 3t = 0$ at $t = \frac{k\pi}{3}$⁠, $k \in \Z$⁠. Then $\cos 2t = \cos\frac{2k\pi}{3}$⁠, which is $1$ when $k$ is a multiple of 3 and $-\frac12$ otherwise. So only $t = n\pi$ again is in the domain.

   So $\frac{dx}{dt} \neq 0$ at every $t$ of the domain except $t = n\pi$⁠.

   The formula divides by $\frac{dx}{dt}$⁠, so we must know exactly where it vanishes, and only inside the domain, because outside it there is no curve at all. Checking each factor against $\cos 2t \gt 0$ shows that the only trouble is at $t = n\pi$⁠.
   $\sin 3t = 0 \iff 3t = k\pi$⁠; $\cos\frac{2\pi}{3} = \cos\frac{4\pi}{3} = -\frac12$⁠. Basics: trigonometric identities

5. Divide

   For $t$ in the domain with $t \neq n\pi$⁠,

   $$\begin{aligned} \frac{dy}{dx} &= \frac{dy/dt}{dx/dt} = \frac{-\sin t\cos t\cos 3t}{\sin t\cos t\sin 3t} \\ &= -\frac{\cos 3t}{\sin 3t} = -\cot 3t. \end{aligned}$$
   The common denominator $(\cos 2t)^{\frac32}$ cancels at once, and the common factor $\sin t\cos t$ is not zero by Step 4. What remains is $-\frac{\cos 3t}{\sin 3t}$⁠, which is $-\cot 3t$⁠; it is defined because $\sin 3t \neq 0$ (Step 4).
   Parametric formula, Section 5.6. Lesson: parametric differentiation

6. Picture the curve and the points $t = n\pi$

   At $t = n\pi$ both $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are zero (both contain $\sin t$⁠). The point is $(0, 1)$ for even $n$ and $(0, -1)$ for odd $n$⁠. As $t$ increases through $0$⁠, $x$ goes from negative to positive, while the slope $-\cot 3t$ is very large and positive just before $t = 0$ and very large and negative just after: the curve rises steeply into $(0, 1)$ and falls steeply out of it, a sharp tip (a cusp) where $\frac{dy}{dx}$ does not exist.

   The slope is $0$ where $\cos 3t = 0$ inside the domain. On the upper branch this happens at $t = \pm\frac{\pi}{6}$ (there $\cos 2t = \frac12 \gt 0$⁠): two low points $\big(\pm\frac{\sqrt2}{8},\ \frac{3\sqrt6}{8}\big) \approx (\pm 0.18,\ 0.92)$⁠, just below the cusp. On the lower branch it happens at $t = \pi \pm \frac{\pi}{6}$⁠, and since replacing $t$ by $t + \pi$ changes $(x, y)$ into $(-x, -y)$⁠, these are the two high points $\big(\pm\frac{\sqrt2}{8},\ -\frac{3\sqrt6}{8}\big)$⁠, just above the cusp $(0, -1)$⁠.

   The picture shows what the restriction $t \neq n\pi$ means: those are exactly the two cusps. Other values of $n$ give nothing new, because replacing $t$ by $t + 2\pi$ does not change $x$ or $y$⁠, so all the curve is traced by the two intervals in the caption.
   A cusp is a sharp tip of a curve; no single tangent line fits there. Basics: slope at a point

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