Polar Coordinates and Derivatives

Part 1 · Hook

Why this matters

A polar curve is drawn by a beam that turns:

\theta

goes up and the point slides along the curve. So it is a parametric curve, with

\theta

as the parameter, and its slope comes from the rule you already know. The new question polar adds is whether the point is moving toward the pole or away from it, and that lives in

\frac{dr}{d\theta}

.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. For a parametric curve $x(t)$ , $y(t)$ , the slope $\frac{dy}{dx}$ is

$\frac{dy/dt}{dx/dt}$ , where $\frac{dx}{dt} \ne 0$
$\frac{dx/dt}{dy/dt}$
$\frac{dy}{dt} \cdot \frac{dx}{dt}$
$\frac{dy}{dt}$

Show the answer

Divide the rate of change of $y$ by that of $x$ (Topic 9.1).

Correct: $\frac{dy/dt}{dx/dt}$ , where $\frac{dx}{dt} \ne 0$ :
$\frac{dx/dt}{dy/dt}$ :
$\frac{dy}{dt} \cdot \frac{dx}{dt}$ :
$\frac{dy}{dt}$ :

2. Which point is the polar point $\left(2, \frac{\pi}{2}\right)$ ?

$(0, 2)$
$(2, 0)$
$(0, -2)$
$\left(2, \frac{\pi}{2}\right)$ in $x$ and $y$

Show the answer

$x = 2\cos\frac{\pi}{2} = 0$ and $y = 2\sin\frac{\pi}{2} = 2$ .

Correct: $(0, 2)$ :
$(2, 0)$ :
$(0, -2)$ :
$\left(2, \frac{\pi}{2}\right)$ in $x$ and $y$ :

3. What is $\frac{d}{d\theta}\left[f(\theta)\sin\theta\right]$ ?

$f'(\theta)\sin\theta + f(\theta)\cos\theta$
$f'(\theta)\cos\theta$
$f'(\theta)\sin\theta - f(\theta)\cos\theta$
$f(\theta)\cos\theta$

Show the answer

Product rule: derivative of the first times the second, plus the first times the derivative of the second.

Correct: $f'(\theta)\sin\theta + f(\theta)\cos\theta$ :
$f'(\theta)\cos\theta$ :
$f'(\theta)\sin\theta - f(\theta)\cos\theta$ :
$f(\theta)\cos\theta$ :

Part 4 · See it

See it first

The cardioid r = 2 + 2 cos θ on a polar grid. At θ = π/2 the point is (0, 2) and the blue tangent line has slope 1: there r = 2, dr/dθ = −2, dx/dθ = −2 and dy/dθ = −2; since dr/dθ is negative while r is positive, the point is moving toward the pole. At θ = π/3 the point is (3/2, 3√3/2), the top of the curve, where the vermillion tangent is horizontal: dy/dθ = 0 and dx/dθ is not 0. — On $r = 2 + 2\cos\theta$ : slope 1 at $\theta = \frac{\pi}{2}$ , and a horizontal tangent at $\theta = \frac{\pi}{3}$ , where $\frac{dy}{d\theta} = 0$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Write the curve as $x = r\cos\theta$ , $y = r\sin\theta$ with $r = f(\theta)$ it is a parametric curve with parameter $\theta$
Differentiate each with the product rule $\frac{dx}{d\theta}$ $= \frac{dr}{d\theta}\cos\theta - r\sin\theta$ and $\frac{dy}{d\theta}$ $= \frac{dr}{d\theta}\sin\theta + r\cos\theta$
Divide, where $\frac{dx}{d\theta} \ne 0$ the slope $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$
Read the signs of $r$ and $\frac{dr}{d\theta}$ togetherthe same sign: moving away from the pole; opposite signs: moving toward it

Part 6 · Key ideas

Key ideas

Slope: $\frac{dy}{dx}$ $= \frac{dr/d\theta\,\sin\theta + r\cos\theta}{dr/d\theta\,\cos\theta - r\sin\theta}$ . Horizontal tangent: $\frac{dy}{d\theta} = 0$ with $\frac{dx}{d\theta} \ne 0$ ; vertical tangent: $\frac{dx}{d\theta} = 0$ with $\frac{dy}{d\theta} \ne 0$ .
$\frac{dr}{d\theta}$ is not the slope. It is the rate of change of the signed distance from the pole.
Distance from the pole is $|r|$ . It increases when $r$ and $\frac{dr}{d\theta}$ have the same sign and decreases when they have opposite signs.
$\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$ say whether the point moves right or left, up or down. Distance from the $y$ -axis is $|x| = |r\cos\theta|$ .

Part 7 · Misconception

A common mistake

The wrong idea: The slope of a polar curve is $\frac{dr}{d\theta}$ .

What actually happens: $\frac{dr}{d\theta}$ tells how fast the distance from the pole changes. The slope needs $x$ and $y$ : $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

1. Find the slope of the tangent line to $r = 3\sin\theta$ at $\theta = \frac{\pi}{6}$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$r = \frac{3}{2}$ and $\frac{dr}{d\theta} = 3\cos\theta = \frac{3\sqrt{3}}{2}$ . Then $\frac{dy}{d\theta}$ $= \frac{dr}{d\theta}\sin\theta$ $+ r\cos\theta$ $= \frac{3\sqrt{3}}{4}$ $+ \frac{3\sqrt{3}}{4}$ $= \frac{3\sqrt{3}}{2}$ and $\frac{dx}{d\theta}$ $= \frac{dr}{d\theta}\cos\theta - r\sin\theta$ $= \frac{9}{4} - \frac{3}{4}$ $= \frac{3}{2}$ , so $\frac{dy}{dx}$ $= \frac{3\sqrt{3}/2}{3/2}$ $= \sqrt{3}$ .

Answer: $\sqrt{3}$ ≈ 1.732

Table

A polar curve given by a table

Values of $r = f(\theta)$ and its derivative for a differentiable function $f$
$\theta$	$r = f(\theta)$	$\frac{dr}{d\theta} = f'(\theta)$
$0$	$2$	$3$
$\frac{\pi}{2}$	$4$	$-2$
$\pi$	$-1$	$-3$
$\frac{3\pi}{2}$	$-2$	$1$

2. At $\theta = \frac{\pi}{2}$ , is the point on the curve moving toward the pole or away from it, as $\theta$ increases?

Away from the pole: $r = 4$ is large
Toward the pole: $r > 0$ and $\frac{dr}{d\theta} < 0$
Away from the pole: $\frac{dx}{d\theta} < 0$
Neither: $\cos\frac{\pi}{2} = 0$ makes $\frac{dr}{d\theta} = 0$

Show the answer

The distance from the pole is $|r| = r$ here, and $\frac{dr}{d\theta} = -2 < 0$ : the distance is decreasing, so the point moves toward the pole.

Away from the pole: $r = 4$ is large: The size of $r$ says where the point is; the sign of $\frac{dr}{d\theta}$ , with the sign of $r$ , says how the distance changes.
Correct: Toward the pole: $r > 0$ and $\frac{dr}{d\theta} < 0$ : Right: a positive distance that is decreasing.
Away from the pole: $\frac{dx}{d\theta} < 0$ : $\frac{dx}{d\theta}$ tracks the horizontal position. The distance from the pole is $|r|$ , read from $r$ and $\frac{dr}{d\theta}$ .
Neither: $\cos\frac{\pi}{2} = 0$ makes $\frac{dr}{d\theta} = 0$ : $\frac{dr}{d\theta} = -2$ is read from the table; it is not multiplied by $\cos\theta$ .

3. At $\theta = \pi$ , is the point on the curve moving toward the pole or away from it, as $\theta$ increases?

Away: $r < 0$ and $\frac{dr}{d\theta} < 0$ , so $|r|$ grows
Toward: $\frac{dr}{d\theta} < 0$ , so the value of $r$ is decreasing
Toward: $r < 0$ puts the point on the opposite ray
Neither: the point is at the pole

Show the answer

The distance from the pole is $|r|$ . With $r = -1$ and $\frac{dr}{d\theta} = -3$ , $r$ is becoming more negative, so $|r|$ is increasing: the point moves away from the pole. The rule: $|r|$ increases when $r$ and $\frac{dr}{d\theta}$ have the same sign.

Correct: Away: $r < 0$ and $\frac{dr}{d\theta} < 0$ , so $|r|$ grows: Right: $r$ and $\frac{dr}{d\theta}$ have the same sign, so the distance $|r|$ increases.
Toward: $\frac{dr}{d\theta} < 0$ , so the value of $r$ is decreasing: A negative $\frac{dr}{d\theta}$ means toward the pole when $r > 0$ . Here $r = -1$ , and $r$ becoming more negative moves it away.
Toward: $r < 0$ puts the point on the opposite ray: A negative $r$ puts the point on the opposite ray; it says nothing about how the distance is changing.
Neither: the point is at the pole: The point is at the pole when $r = 0$ ; here $r = -1$ , one unit away.

4. Find the slope $\frac{dy}{dx}$ of the curve at $\theta = \frac{\pi}{2}$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\frac{dx}{d\theta}$ $= \frac{dr}{d\theta}\cos\theta - r\sin\theta$ $= (-2)(0) - 4(1)$ $= -4$ and $\frac{dy}{d\theta}$ $= \frac{dr}{d\theta}\sin\theta$ $+ r\cos\theta$ $= (-2)(1)$ $+ 4(0)$ $= -2$ , so $\frac{dy}{dx} = \frac{-2}{-4} = \frac{1}{2}$ .

Answer: $\frac{1}{2}$ ≈ 0.500

5. At which points does a polar curve $r = f(\theta)$ have a horizontal tangent line?

Where $\frac{dy}{d\theta} = 0$ and $\frac{dx}{d\theta} \ne 0$
Where $\frac{dr}{d\theta} = 0$
Where $\frac{dx}{d\theta} = 0$ and $\frac{dy}{d\theta} \ne 0$
Where $r = 0$

Show the answer

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ is 0 when the numerator is 0 and the denominator is not.

Correct: Where $\frac{dy}{d\theta} = 0$ and $\frac{dx}{d\theta} \ne 0$ : Right: the slope $\frac{dy/d\theta}{dx/d\theta}$ is 0 there.
Where $\frac{dr}{d\theta} = 0$ : $\frac{dr}{d\theta} = 0$ is where the distance from the pole stops changing, such as the tip of a petal; the tangent there is perpendicular to the ray, not horizontal in general.
Where $\frac{dx}{d\theta} = 0$ and $\frac{dy}{d\theta} \ne 0$ : That makes the slope's denominator 0: a vertical tangent.
Where $r = 0$ : $r = 0$ is the pole. A curve through the pole has tangent along the ray $\theta$ there, horizontal only if that ray is horizontal.

6. For a polar curve $r = f(\theta)$ , $f(1) = 2$ and $f'(1) = -3$ . What does $f'(1) = -3$ mean?

At $\theta = 1$ , the slope of the curve is $-3$
The distance from the pole shrinks at 3 units per radian
At $\theta = 1$ , the point is moving downward at 3 units per radian
At $\theta = 1$ , the point is 3 units from the pole

Show the answer

$f'(\theta) = \frac{dr}{d\theta}$ is the rate of change of $r$ with respect to $\theta$ . With $r = 2 > 0$ , a negative rate means the distance from the pole is decreasing, at 3 units per radian.

At $\theta = 1$ , the slope of the curve is $-3$ : The slope is $\frac{dy}{dx}$ , which needs both $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$ .
Correct: The distance from the pole shrinks at 3 units per radian: Right: $r = 2 > 0$ is the distance, and $\frac{dr}{d\theta} < 0$ means it is shrinking.
At $\theta = 1$ , the point is moving downward at 3 units per radian: Moving down is the sign of $\frac{dy}{d\theta}$ $= f'(1)\sin 1 + f(1)\cos 1$ , not of $f'(1)$ .
At $\theta = 1$ , the point is 3 units from the pole: The distance is $|f(1)| = 2$ ; $f'(1)$ is its rate of change.

Part 9 · Summary

Summary

Treat

r = f(\theta)

as the parametric curve

x = r\cos\theta

,

y = r\sin\theta

. The product rule gives

\frac{dx}{d\theta}

and

\frac{dy}{d\theta}

; their quotient is the slope. Separately, the signs of

r

and

\frac{dr}{d\theta}

say whether the point moves toward the pole (opposite signs) or away from it (the same sign).

Part 10 · Up next

What comes next

Part 11 · Connections