Chapter 5 · Continuity and Differentiability

Example 33: $\frac{dy}{dx}$ when $x=a(\theta+\sin\theta)$ ⁠, $y=a(1-\cos\theta)$

ExamplesParametric differentiation · Section 5.6

The question

Find $\dfrac{dy}{dx}$ ⁠, if $x = a(\theta + \sin\theta)$ ⁠, $y = a(1 - \cos\theta)$ ⁠.

Exam answer

Here $x = a(\theta + \sin\theta)$ and $y = a(1 - \cos\theta)$ ⁠.

Differentiating w.r.t. $\theta$ ⁠,

\begin{aligned} \frac{dx}{d\theta} &= a(1 + \cos\theta) \\ \frac{dy}{d\theta} &= a\sin\theta \end{aligned}

\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\sin\theta}{a(1 + \cos\theta)}

Using $\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$ and $1 + \cos\theta = 2\cos^2\frac{\theta}{2}$ ⁠,

\frac{dy}{dx} = \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}

$\therefore\ \dfrac{dy}{dx} = \tan\dfrac{\theta}{2}$ $(\theta \ne (2n + 1)\pi,\ n \in \Z)$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

As in every parametric problem, the slope is the ratio of the two $\theta$ -rates:

\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta},

wherever $\frac{dx}{d\theta} \neq 0$ ⁠. The ratio here is $\frac{\sin\theta}{1+\cos\theta}$ ⁠, and the half-angle formulas collapse it to the single value $\tan\frac{\theta}{2}$ ⁠.

Given, to find, and the plan

Given: $x = a(\theta + \sin\theta)$ ⁠, $y = a(1 - \cos\theta)$ ⁠, with $a$ a nonzero constant and $\theta$ the parameter.
To find: $\dfrac{dy}{dx}$ ⁠.
Plan: Differentiate $x$ and $y$ with respect to $\theta$ ⁠; find where $\frac{dx}{d\theta} = 0$ ⁠; divide; simplify with $\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$ and $1 + \cos\theta = 2\cos^2\frac{\theta}{2}$ ⁠, checking that the factor we cancel is not zero. Finally look at the excluded points on a picture of the curve.

Basics you need

1Derivatives you need

$\frac{d}{d\theta}\theta = 1$ ⁠, $\frac{d}{d\theta}\sin\theta = \cos\theta$ ⁠, $\frac{d}{d\theta}\cos\theta = -\sin\theta$ ⁠, and the derivative of a constant is $0$ (Table 5.3 and the sum rule). Basics: derivative rules

2The parametric formula (Section 5.6)

$\frac{dy}{dx} = \frac{g'(\theta)}{f'(\theta)}$ when $x = f(\theta)$ ⁠, $y = g(\theta)$ and $f'(\theta) \neq 0$ ⁠. Lesson: parametric differentiation

3Half-angle formulas

From the double-angle formulas with $\theta = 2\cdot\frac{\theta}{2}$ ⁠:

\begin{aligned} \sin\theta &= 2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}, \\ 1 + \cos\theta &= 2\cos^2\tfrac{\theta}{2}. \end{aligned}

The second comes from $\cos\theta = 2\cos^2\frac{\theta}{2} - 1$ ⁠. Both hold for every $\theta$ ⁠. Basics: trigonometric identities

4When is $\cos\theta = -1$ ⁠?

Exactly when $\theta$ is an odd multiple of $\pi$ ⁠: $\theta = (2n+1)\pi$ ⁠, $n \in \Z$ ⁠. Then $\frac{\theta}{2} = (2n+1)\frac{\pi}{2}$ ⁠, where $\cos\frac{\theta}{2} = 0$ ⁠. Basics: trigonometric identities

Solution

Differentiate $x$ with respect to $\theta$

$x = a(\theta + \sin\theta)$ is a constant times a sum, so differentiate each term:
$\frac{dx}{d\theta} = a\left(1 + \cos\theta\right).$
The parametric formula needs both $\theta$ -rates. The sum rule lets us differentiate $\theta$ (giving 1) and $\sin\theta$ (giving $\cos\theta$ ⁠) separately, and $a$ stays in front.

Sum rule $(u+v)' = u' + v'$ (Section 5.3). Basics: derivative rules
Differentiate $y$ with respect to $\theta$
$\frac{dy}{d\theta} = a\left(0 - (-\sin\theta)\right) = a\sin\theta.$
The constant 1 has derivative 0, and $-\cos\theta$ has derivative $-(-\sin\theta) = +\sin\theta$ ⁠. The two minus signs cancel, which is why the book writes simply $a\sin\theta$ ⁠.

$\frac{d}{d\theta}\cos\theta = -\sin\theta$ (Table 5.3).
Find where the formula may be used

We need $\frac{dx}{d\theta} \neq 0$ ⁠. Since $a \neq 0$ ⁠,
$\begin{aligned} a(1 + \cos\theta) \neq 0 &\iff \cos\theta \neq -1 \\ &\iff \theta \neq (2n+1)\pi. \end{aligned}$
So from now on $\theta$ is not an odd multiple of $\pi$ ⁠.

Dividing by zero is meaningless, so the excluded values must be found before dividing. We will see in Step 6 what the curve does at these values.

Basic 4: $\cos\theta = -1$ only at odd multiples of $\pi$ ⁠.
Divide

For $\theta \neq (2n+1)\pi$ ⁠,
$\begin{aligned} \frac{dy}{dx} &= \frac{dy/d\theta}{dx/d\theta} = \frac{a\sin\theta}{a(1 + \cos\theta)} \\ &= \frac{\sin\theta}{1 + \cos\theta}. \end{aligned}$
The $y$ -rate goes on top. The factor $a$ cancels because $a \neq 0$ ⁠. This is already a correct answer; Step 5 simplifies it.

Parametric formula, Section 5.6. Lesson: parametric differentiation
Simplify with the half-angle formulas

Rewrite top and bottom using Basic 3:
$\begin{aligned} \frac{\sin\theta}{1 + \cos\theta} &= \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}} \\ &= \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} = \tan\frac{\theta}{2}. \end{aligned}$
Cancelling $2\cos\frac{\theta}{2}$ is allowed: by Step 3, $\theta \neq (2n+1)\pi$ ⁠, so $\frac{\theta}{2} \neq (2n+1)\frac{\pi}{2}$ and $\cos\frac{\theta}{2} \neq 0$ ⁠. Hence
$\frac{dy}{dx} = \tan\frac{\theta}{2} \qquad (\theta \neq (2n+1)\pi).$
Both $\sin\theta$ and $1 + \cos\theta$ contain the factor $\cos\frac{\theta}{2}$ once we write them in half-angles. Spotting this common factor is the whole trick. Notice also that the condition for the original formula (Step 3) is exactly the condition for $\tan\frac{\theta}{2}$ to be defined, so nothing is lost or gained by the simplification.

$\tan u = \frac{\sin u}{\cos u}$ ⁠, defined when $\cos u \neq 0$ ⁠. Basics: trigonometric identities
Picture the curve and the excluded points

At $\theta = (2n+1)\pi$ both rates are zero: $\frac{dx}{d\theta} = a(1 + \cos\theta) = 0$ and $\frac{dy}{d\theta} = a\sin\theta = 0$ ⁠. The point there is $\big(a(2n+1)\pi,\ 2a\big)$ ⁠. As $\theta$ approaches $\pi$ from either side, $\tan\frac{\theta}{2}$ grows without bound in size, so the curve comes into that point steeper and steeper, and leaves it the same way: the point is a sharp tip (a cusp), and $\frac{dy}{dx}$ does not exist there.

The curve for $a = 1$ ⁠, $\theta$ from about $-\pi - 0.8$ to $\pi + 0.8$ ⁠. At $\theta = 0$ it is at its lowest point with slope $\tan 0 = 0$ ⁠; at $\theta = \pm\frac{\pi}{2}$ ⁠, the points $\big(\pm(\frac{\pi}{2}+1), 1\big)$ ⁠, the slope is $\tan(\pm\frac{\pi}{4}) = \pm 1$ ⁠; at $\theta = \pm\pi$ the curve has cusps, where both $\theta$ -rates are zero.

For $-\pi \lt \theta \lt \pi$ the answer has a neat meaning: the slope is $\tan\frac{\theta}{2}$ with $-\frac{\pi}{2} \lt \frac{\theta}{2} \lt \frac{\pi}{2}$ ⁠, so the tangent makes the angle $\frac{\theta}{2}$ with the $x$ -axis.

The figure shows why the condition of Step 3 matters: the excluded values are not a technicality but the sharp tips of the curve. (Putting $\theta = \varphi + \pi$ turns the equations into $x = a\pi + a(\varphi - \sin\varphi)$ ⁠, $y = 2a - a(1 - \cos\varphi)$ ⁠: this is the cycloid, the path of a point on the rim of a rolling wheel, shifted and turned upside down.)

A line with angle of inclination $\alpha$ has slope $\tan\alpha$ (Class XI, Straight Lines). Basics: slope at a point

Result

\frac{dy}{dx} = \frac{\sin\theta}{1+\cos\theta} = \tan\frac{\theta}{2},

for every $\theta \neq (2n+1)\pi$ ⁠, $n \in \Z$ ⁠. At $\theta = (2n+1)\pi$ the curve has a cusp and $\frac{dy}{dx}$ does not exist. (Book's answer: $\tan\frac{\theta}{2}$ ⁠.) As the book's Note says, $\frac{dy}{dx}$ is expressed in terms of the parameter only, without using $x$ and $y$ ⁠.

Check it with numbers

Take $a = 2$ ⁠, $\theta = 1$ (radian). The point is $(2(1+\sin 1),\ 2(1-\cos 1))$ ⁠, approximately $(3.6829420,\ 0.9193954)$ ⁠. Predicted slope: $\tan 0.5 \approx 0.5463025$ ⁠, and $\frac{\sin 1}{1+\cos 1} \approx \frac{0.8414710}{1.5403023} \approx 0.5463025$ ⁠, so the two forms agree.

Chord from $\theta = 0.999$ to $\theta = 1.001$ ⁠: the points are $(3.6798605,\ 0.9177130)$ and $(3.6860217,\ 0.9210789)$ ⁠, so

\frac{\Delta y}{\Delta x} \approx \frac{0.0033659}{0.0061612} \approx 0.54630,

agreeing with $\tan 0.5$ to five decimal places.

Common mistakes

Sign slip in $\frac{dy}{d\theta}$ ⁠. $\frac{d}{d\theta}(1 - \cos\theta) = +\sin\theta$ ⁠. Writing $-\sin\theta$ leads to $-\tan\frac{\theta}{2}$ ⁠.
Using the wrong half-angle formula. $1 + \cos\theta = 2\cos^2\frac{\theta}{2}$ ⁠, while $1 - \cos\theta = 2\sin^2\frac{\theta}{2}$ ⁠. Mixing them up gives $\cot\frac{\theta}{2}$ ⁠.
Cancelling without a condition. $\cos\frac{\theta}{2}$ may be cancelled only because it is not zero when $\theta \neq (2n+1)\pi$ ⁠. State it.
Assuming a slope exists everywhere. At $\theta = \pm\pi$ both rates vanish and the curve has a sharp tip: there is no tangent slope.

What to take away

The parametric method gives $\frac{\sin\theta}{1+\cos\theta}$ at once; the half-angle formulas then reduce it to $\tan\frac{\theta}{2}$ ⁠.
$\frac{\sin\theta}{1+\cos\theta} = \tan\frac{\theta}{2}$ is worth remembering. The same half-angle simplification is needed again in Exercise 5.6, Question 6, where $1 - \cos\theta = 2\sin^2\frac{\theta}{2}$ is the formula to use.
When $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$ are both zero, expect something special on the curve, such as a cusp.

Every fact used, and where it comes from

Fact	Used in	Source
Sum rule; derivatives of $\theta$ ⁠, $\sin\theta$ ⁠, $\cos\theta$ ⁠, constants	Steps 1, 2	Section 5.3, Table 5.3
$\frac{dy}{dx} = \frac{g'(\theta)}{f'(\theta)}$ ⁠, $f'(\theta) \neq 0$	Steps 3, 4	Section 5.6 (Lesson: parametric differentiation)
$\cos\theta = -1 \iff \theta = (2n+1)\pi$	Steps 3, 5	Class XI, Trigonometric Functions
$\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$ ⁠, $1+\cos\theta = 2\cos^2\frac{\theta}{2}$	Step 5	Double-angle formulas, Class XI (Basics: trigonometric identities)
Slope $= \tan(\text{angle of inclination})$	Step 6	Class XI, Straight Lines
$\frac{dy}{dx}$ in terms of the parameter only	Result	Note after Example 33, Section 5.6

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Example 33: dydx\frac{dy}{dx}dxdy​ when x=a(θ+sin⁡θ)x=a(\theta+\sin\theta)x=a(θ+sinθ)⁠, y=a(1−cos⁡θ)y=a(1-\cos\theta)y=a(1−cosθ)

Basics you need

Solution

Differentiate xxx with respect to θ\thetaθ

Differentiate yyy with respect to θ\thetaθ

Find where the formula may be used

Divide

Simplify with the half-angle formulas

Picture the curve and the excluded points

Every fact used, and where it comes from

Example 33: $\frac{dy}{dx}$ when $x=a(\theta+\sin\theta)$ ⁠, $y=a(1-\cos\theta)$

Differentiate $x$ with respect to $\theta$

Differentiate $y$ with respect to $\theta$