Chapter 5 · Continuity and Differentiability

Exercise 5.6 · Q2: $\frac{dy}{dx}$ when $x=a\cos\theta$ ⁠, $y=b\cos\theta$

Exercise 5.6Parametric differentiation · Section 5.6

The question

If $x$ and $y$ are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find $\dfrac{dy}{dx}$ ⁠.

2. $x = a\cos\theta$ ⁠, $y = b\cos\theta$

Exam answer

\begin{aligned} \frac{dx}{d\theta} &= -a\sin\theta, \\ \frac{dy}{d\theta} &= -b\sin\theta \end{aligned}

\begin{aligned} \frac{dy}{dx} &= \frac{dy/d\theta}{dx/d\theta} = \frac{-b\sin\theta}{-a\sin\theta} \end{aligned}

$\therefore\ \dfrac{dy}{dx} = \dfrac{b}{a}$ $(\theta \ne n\pi)$

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

The one idea this problem is built on

The slope is the ratio of the two $\theta$ -rates,

\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}, \quad \text{where } \frac{dx}{d\theta} \neq 0.

Both rates contain the same factor $-\sin\theta$ ⁠, which cancels and leaves a constant, $\frac{b}{a}$ ⁠. A constant slope means the points lie on a straight line.

Given, to find, and the plan

Given: $x = a\cos\theta$ ⁠, $y = b\cos\theta$ ⁠, with constants $a$ ⁠, $b$ and parameter $\theta$ ⁠. We need $a \neq 0$ ⁠: if $a = 0$ ⁠, then $x = 0$ for every $\theta$ and $x$ never changes, so $\frac{dx}{d\theta}$ is always zero and the formula can never be used.
To find: $\dfrac{dy}{dx}$ ⁠, without eliminating $\theta$ ⁠.
Plan: Differentiate $x$ and $y$ with respect to $\theta$ ⁠; find where $\frac{dx}{d\theta} = 0$ ⁠; divide; then interpret the answer with a picture.

Basics you need

1Derivative of cosine

$\frac{d}{d\theta}\cos\theta = -\sin\theta$ ⁠, and a constant factor stays in front (Table 5.3). Basics: derivative rules

2The parametric formula (Section 5.6)

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ ⁠, provided $\frac{dx}{d\theta} \neq 0$ ⁠. Lesson: parametric differentiation

3Where is $\sin\theta = 0$ ⁠?

Exactly at the integer multiples of $\pi$ ⁠: $\theta = n\pi$ ⁠, $n \in \Z$ ⁠. There $\cos\theta = \pm 1$ ⁠. Basics: trigonometric identities

Solution

Differentiate $x$ with respect to $\theta$
$\frac{dx}{d\theta} = a\cdot\frac{d}{d\theta}\cos\theta = -a\sin\theta.$
The parametric formula is built from the two rates with respect to the parameter, so we find each one. The constant $a$ stays in front.

$\frac{d}{d\theta}\cos\theta = -\sin\theta$ (Table 5.3). Basics: derivative rules
Differentiate $y$ with respect to $\theta$
$\frac{dy}{d\theta} = b\cdot\frac{d}{d\theta}\cos\theta = -b\sin\theta.$
Exactly the same computation with $b$ in place of $a$ ⁠.

Constant multiple rule: $\frac{d}{d\theta}(cu) = c\frac{du}{d\theta}$ ⁠.
Find where the formula may be used

We need $\frac{dx}{d\theta} = -a\sin\theta \neq 0$ ⁠. Since $a \neq 0$ ⁠,
$-a\sin\theta \neq 0 \iff \sin\theta \neq 0 \iff \theta \neq n\pi.$
We are about to divide by $-a\sin\theta$ ⁠, so the values where it vanishes must be set aside first. By Basic 3 they are the multiples of $\pi$ ⁠.

$\sin\theta = 0 \iff \theta = n\pi$ ⁠, $n \in \Z$ ⁠.
Divide

For $\theta \neq n\pi$ ⁠,
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-b\sin\theta}{-a\sin\theta} = \frac{b}{a}.$
The common factor $-\sin\theta$ is not zero (Step 3), so it cancels.

The $y$ -rate goes on top. After cancelling, nothing depending on $\theta$ is left: the slope is the same at every point where it is defined.

Parametric formula, Section 5.6. Lesson: parametric differentiation
What a constant slope means

As $\theta$ runs through the real numbers, $\cos\theta$ moves back and forth between $-1$ and $1$ ⁠, so the point $(a\cos\theta,\ b\cos\theta)$ moves back and forth between $(-a, -b)$ and $(a, b)$ ⁠. The slope is $\frac{b}{a}$ everywhere, which is what we expect of a straight line, and indeed these points form the straight segment of slope $\frac{b}{a}$ joining those two ends (the check below confirms it: $y = \frac{b}{a}x$ ⁠). Its ends are reached exactly when $\cos\theta = \pm 1$ ⁠, that is at $\theta = n\pi$ ⁠, the values excluded in Step 3: there the moving point stops for an instant and turns back, so $\frac{dx}{d\theta} = \frac{dy}{d\theta} = 0$ ⁠.

The points for $a = 3$ ⁠, $b = 2$ (equal scales): a segment from $(-3, -2)$ to $(3, 2)$ with slope $\frac{2}{3}$ ⁠. At $\theta = \frac{\pi}{3}$ the point is $\big(\frac{3}{2}, 1\big)$ ⁠.

This is a good example of a slope that does not depend on the parameter at all. If $b = 0$ the answer is $0$ ⁠: all the points lie on the $x$ -axis.

A line has the same slope at every point. Basics: slope at a point

Result

\frac{dy}{dx} = \frac{b}{a} \qquad (\theta \neq n\pi,\ a \neq 0).

(Textbook answer: $\frac{b}{a}$ ⁠.)

Check it with numbers

A short chord. Take $a = 3$ ⁠, $b = 2$ ⁠, $\theta = 1$ ⁠: the point is $(1.6209069,\ 1.0806046)$ ⁠. The points at $\theta = 0.999$ and $\theta = 1.001$ are $(1.623430520,\ 1.082287013)$ and $(1.618381695,\ 1.078921130)$ ⁠, so

\frac{\Delta y}{\Delta x} = \frac{-0.003365883}{-0.005048825} \approx 0.66667,

which is $\frac{b}{a} = \frac{2}{3} \approx 0.6666667$ (here the chord slope is exactly $\frac{2}{3}$ ⁠, because the points lie on a straight line).

Eliminating $\theta$ ⁠, only as a check. Dividing, $\frac{y}{x} = \frac{b\cos\theta}{a\cos\theta} = \frac{b}{a}$ whenever $\cos\theta \neq 0$ ⁠, so $y = \frac{b}{a}x$ (this also holds when $\cos\theta = 0$ ⁠, where $x = y = 0$ ⁠). The derivative of $\frac{b}{a}x$ is $\frac{b}{a}$ ⁠.

Common mistakes

Losing a minus sign. Both rates are negative; their ratio is $+\frac{b}{a}$ ⁠, not $-\frac{b}{a}$ ⁠.
Thinking a constant answer is suspicious. A slope may well not depend on the parameter; here the points lie on a straight line.
Ignoring the condition. The formula needs $\sin\theta \neq 0$ ⁠; state it.
Eliminating $\theta$ as the solution. The question forbids it; $y = \frac{b}{a}x$ belongs only in the check.

What to take away

Differentiate both coordinates with respect to the parameter, then divide.
A common factor of the two rates cancels only where it is not zero, which is exactly where $\frac{dx}{d\theta} \neq 0$ ⁠.
A slope that does not depend on the parameter means the points lie on a straight line.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{d\theta}\cos\theta = -\sin\theta$ ⁠; constant multiple rule	Steps 1, 2	Table 5.3, Section 5.3 (Basics: derivative rules)
$\sin\theta = 0 \iff \theta = n\pi$	Step 3	Class XI, Trigonometric Functions (Basics: trigonometric identities)
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ ⁠, $\frac{dx}{d\theta} \neq 0$	Steps 3, 4	Section 5.6 (Lesson: parametric differentiation)
$-1 \leq \cos\theta \leq 1$ ⁠, with $\cos\theta = \pm1$ at $\theta = n\pi$	Step 5	Class XI, Trigonometric Functions

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Exercise 5.6 · Q2: dydx\frac{dy}{dx}dxdy​ when x=acos⁡θx=a\cos\thetax=acosθ⁠, y=bcos⁡θy=b\cos\thetay=bcosθ

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

What a constant slope means

Every fact used, and where it comes from

Exercise 5.6 · Q2: $\frac{dy}{dx}$ when $x=a\cos\theta$ ⁠, $y=b\cos\theta$

Differentiate $x$ with respect to $\theta$

Differentiate $y$ with respect to $\theta$