Parametric equations, polar coordinates and vector-valued functions

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1. Derivatives of parametric curves

Warm-up

Three quick questions from earlier lessons.

Discover

A point moves with $x = t^2$ , $y = t^3$ . At $t = 1$ , $x$ grows at 2 units per second and $y$ at 3 units per second. What is the slope of its path there?

$\tfrac{3}{2}$
$\tfrac{2}{3}$

Key idea

Divide the rates.

For

x = x(t)

y = y(t)

\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}

where

\dfrac{dx}{dt} \ne 0

. Horizontal tangent:

\dfrac{dy}{dt} = 0

\dfrac{dx}{dt} \ne 0

. Vertical tangent:

\dfrac{dx}{dt} = 0

\dfrac{dy}{dt} \ne 0

Worked example

$x = t^2 + 1$ , $y = t^3 - 3t$ . Slope at $t = 2$ .

\dfrac{dx}{dt} = 2t

\dfrac{dy}{dt} = 3t^2 - 3

Differentiate each.

\dfrac{dy}{dx} = \dfrac{3t^2 - 3}{2t}

Divide.

= \dfrac{9}{4}

t = 2.

Your turn, guided

Your turn: same curve, where is the tangent horizontal ( $t > 0$ )?

t = 1

3t² − 3 = 0, and dx/dt = 2 ≠ 0 there.

Practice ★

$x = t^2$ , $y = t^3 + 9t$ . Find $\dfrac{dy}{dx}$ at $t = 3$ .
$x = t^2$ , $y = t^3 + 4t$ . Find $\dfrac{dy}{dx}$ at $t = 1$ .
$x = t^2$ , $y = t^3 + 6t$ . Find $\dfrac{dy}{dx}$ at $t = 3$ .

Watch out!

Practice ★★

$x = 3\cos t$ , $y = 3\sin t$ . Find $\dfrac{dy}{dx}$ at $t = \dfrac{\pi}{ 3 }$ . Give an exact value.
$x = 8\cos t$ , $y = 8\sin t$ . Find $\dfrac{dy}{dx}$ at $t = \dfrac{\pi}{ 6 }$ . Give an exact value.

$x = t^2$ , $y = t^3 - 27t$ . For which $t > 0$ is the tangent horizontal?

Boss problem. $x = t^2 + 1$ , $y = 2t^3$ . Study the curve at $t = 1$ , the point $(2, 2)$ .

Exit ticket

Three questions to check you have got it.

$x = t^2$ , $y = t^3 + 9t$ . Find $\dfrac{dy}{dx}$ at $t = 3$ .
$x = t^2$ , $y = t^3 + 4t$ . Find $\dfrac{dy}{dx}$ at $t = 1$ .
$x = t^2$ , $y = t^3 + 6t$ . Find $\dfrac{dy}{dx}$ at $t = 3$ .

Summary card

Parametric slopes

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ .
Horizontal: $y' = 0$ . Vertical: $x' = 0$ .

Watch out: $y$ 's rate on top.

2. Second derivatives of parametric curves

Warm-up

Three quick questions from earlier lessons.

Discover

For $x = t^2$ , $y = t^3$ ( $t > 0$ ), $\dfrac{dy}{dx} = \dfrac{3t}{2}$ . Is $\dfrac{d^2y}{dx^2}$ just $\dfrac{d}{dt}\left(\dfrac{3t}{2}\right) = \dfrac{3}{2}$ ?

No, it must be divided by $\tfrac{dx}{dt}$
Yes

Key idea

Differentiate the slope with respect to $t$ , then convert to $x$ .

\dfrac{d^2y}{dx^2} = \dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}

. Its sign gives the concavity of the path.

Worked example

$x = t^2$ , $y = t^3 - 3t$ , $t > 0$ .

\dfrac{dy}{dx} = \dfrac{3t^2 - 3}{2t} = \tfrac{3}{2}t - \tfrac{3}{2t}

First derivative.

\tfrac{3}{2} + \tfrac{3}{2t^2}

d/dt of it.

\dfrac{d^2y}{dx^2} = \dfrac{3}{4t} + \dfrac{3}{4t^3} > 0

: concave up

Divide by 2t.

Your turn, guided

Your turn: $x = t$ , $y = t^3$ . (Here $x = t$ , so it is the ordinary second derivative.)

\dfrac{d^2y}{dx^2} = 6t

d/dt(3t²) ÷ 1.

Practice ★

$x = t^2$ , $y = 8t^3$ for $t > 0$ . Find $\dfrac{d^2y}{dx^2}$ in terms of $t$ .
$x = t^2$ , $y = 1t^3$ for $t > 0$ . Find $\dfrac{d^2y}{dx^2}$ in terms of $t$ .

$x = t^2$ , $y = 9t^3$ . Find $\dfrac{d^2y}{dx^2}$ at $t = 3$ .
$x = t^2$ , $y = 7t^3$ . Find $\dfrac{d^2y}{dx^2}$ at $t = 3$ .

Watch out!

Practice ★★

$x = e^t$ , $y = e^{-3t}$ . Find $\dfrac{d^2y}{dx^2}$ in terms of $t$ .
$x = e^t$ , $y = e^{-4t}$ . Find $\dfrac{d^2y}{dx^2}$ in terms of $t$ .

Exit ticket

Three questions to check you have got it.

$x = t^2$ , $y = 8t^3$ for $t > 0$ . Find $\dfrac{d^2y}{dx^2}$ in terms of $t$ .
$x = t^2$ , $y = 1t^3$ for $t > 0$ . Find $\dfrac{d^2y}{dx^2}$ in terms of $t$ .
$x = t^2$ , $y = 5t^3$ for $t > 0$ . Find $\dfrac{d^2y}{dx^2}$ in terms of $t$ .

Summary card

Second derivative

$\dfrac{d^2y}{dx^2} = \dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right) \Big/ \dfrac{dx}{dt}$ .

Watch out: the division by $dx/dt$ is needed again.

3. Arc length of parametric curves

Warm-up

Three quick questions from earlier lessons.

Discover

In a tiny time $dt$ , a point moves $dx$ across and $dy$ up. How far does it travel?

$\sqrt{dx^2 + dy^2}$
$dx + dy$

Key idea

Add up tiny hypotenuses.

Arc length of

x(t)

y(t)

for

a \le t \le b

L = \displaystyle\int_a^b \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2}\,dt

(if the curve is traced once).

Worked example

$x = 3t$ , $y = 4t$ , $0 \le t \le 2$ .

x' = 3

y' = 4

Rates.

L = \displaystyle\int_0^2 5\,dt

√(9 + 16) = 5.

= 10

Check: from (0, 0) to (6, 8), distance 10.

Your turn, guided

Your turn: the circle $x = 2\cos t$ , $y = 2\sin t$ , $0 \le t \le 2\pi$ .

L = 4\pi

√(4 sin² + 4 cos²) = 2; ∫₀^{2π} 2 dt.

Practice ★

Find the length of the curve $x = 3\cos t$ , $y = 3\sin t$ , $0 \le t \le 2\pi$ .
Find the length of the curve $x = 5\cos t$ , $y = 5\sin t$ , $0 \le t \le 2\pi$ .

Find the length of the path $x = 3t$ , $y = 6t$ , $0 \le t \le 1$ . Give an exact value.
Find the length of the path $x = 4t$ , $y = 4t$ , $0 \le t \le 2$ . Give an exact value.

Watch out!

Practice ★★

Find the length of $x = t$ , $y = \tfrac{2}{3}t^{3/2}$ for $0 \le t \le 3$ .
Find the length of $x = t$ , $y = \tfrac{2}{3}t^{3/2}$ for $0 \le t \le 15$ .

Exit ticket

Three questions to check you have got it.

Find the length of the path $x = 3t$ , $y = 6t$ , $0 \le t \le 1$ . Give an exact value.
Find the length of the path $x = 4t$ , $y = 4t$ , $0 \le t \le 2$ . Give an exact value.
Find the length of the path $x = 2t$ , $y = 5t$ , $0 \le t \le 2$ . Give an exact value.

Summary card

Arc length

$L = \displaystyle\int_a^b \sqrt{x'(t)^2 + y'(t)^2}\,dt$ .

Watch out: Pythagoras, not a sum.

4. Vector-valued functions: velocity, speed, acceleration

Warm-up

Three quick questions from earlier lessons.

Discover

$\vec r(t) = \langle t^2, 3t \rangle$ gives a position at each time. Its velocity is $\vec v(t) = \langle 2t, 3 \rangle$ . At $t = 2$ , $\vec v = \langle 4, 3 \rangle$ . What is the speed?

Key idea

Differentiate and integrate componentwise.

\vec r(t) = \langle x(t), y(t) \rangle

: velocity

\vec v = \vec r\,'

, acceleration

\vec a = \vec v\,'

, speed

|\vec v| = \sqrt{x'^2 + y'^2}

. Position from velocity:

\vec r(t) = \vec r(0) + \displaystyle\int_0^t \vec v(s)\,ds

Worked example

$\vec r(t) = \langle \cos t, \sin t \rangle$ .

\vec v(t) = \langle -\sin t, \cos t \rangle

Componentwise.

Speed

= 1

sin² + cos² = 1.

\vec a(t) = \langle -\cos t, -\sin t \rangle

Points towards the center.

Your turn, guided

Your turn: $\vec v(t) = \langle 3t^2, 2 \rangle$ , $\vec r(0) = \langle 0, 5 \rangle$ . Find $y(1)$ .

y(1) = 7

5 + ∫₀¹ 2 dt.

Practice ★

A particle moves with position $\vec r(t) = \langle t^2, 6t \rangle$ . Find its speed at $t = 2$ . Give an exact value.
A particle moves with position $\vec r(t) = \langle t^2, 5t \rangle$ . Find its speed at $t = 4$ . Give an exact value.
A particle moves with position $\vec r(t) = \langle t^2, 3t \rangle$ . Find its speed at $t = 1$ . Give an exact value.

Watch out!

Practice ★★

$\vec r(t) = \langle 6t^2, 8\sin t \rangle$ . Find the magnitude of the acceleration at $t = 0$ .
$\vec r(t) = \langle 8t^2, 5\sin t \rangle$ . Find the magnitude of the acceleration at $t = 0$ .

$\vec v(t) = \langle 2t, 3 \rangle$ and $\vec r(0) = \langle 1, 0 \rangle$ . Find the $x$ -coordinate of the position at $t = 2$ .
$\vec v(t) = \langle 2t, 6 \rangle$ and $\vec r(0) = \langle 1, 0 \rangle$ . Find the $x$ -coordinate of the position at $t = 5$ .

Exit ticket

Three questions to check you have got it.

A particle moves with position $\vec r(t) = \langle t^2, 6t \rangle$ . Find its speed at $t = 2$ . Give an exact value.
A particle moves with position $\vec r(t) = \langle t^2, 5t \rangle$ . Find its speed at $t = 4$ . Give an exact value.
A particle moves with position $\vec r(t) = \langle t^2, 3t \rangle$ . Find its speed at $t = 1$ . Give an exact value.

Summary card

Vector motion

$\vec v = \vec r\,'$ , $\vec a = \vec v\,'$ , speed $= |\vec v|$ .
Position: start + $\int \vec v$ .

Watch out: speed is a length, not a sum of components.

5. Slopes of polar curves

Warm-up

Three quick questions from earlier lessons.

Discover

On the circle $r = 2$ , $\dfrac{dr}{d\theta} = 0$ everywhere. Does that make every tangent horizontal?

Key idea

Turn the polar curve into a parametric one, with $\theta$ as the parameter.

x = r\cos\theta

y = r\sin\theta

, so

\dfrac{dy}{dx} = \dfrac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}

, where

r' = \dfrac{dr}{d\theta}

Worked example

$r = 1 + \sin\theta$ at $\theta = 0$ .

r = 1

r' = 1

r = 1, r′ = cos 0 = 1.

\dfrac{dy}{dx} = \dfrac{1}{1}

Top: 1·0 + 1·1; bottom: 1·1 − 1·0.

= 1

The slope.

Your turn, guided

Your turn: $r = 3$ at $\theta = \dfrac{\pi}{2}$ (top of the circle).

slope

0

r′ = 0: (0 + 3·0)/(0 − 3·1).

Practice ★

Find the slope $\dfrac{dy}{dx}$ of the polar curve $r = 8 + \cos\theta$ at $\theta = \dfrac{\pi}{2}$ .
Find the slope $\dfrac{dy}{dx}$ of the polar curve $r = 5 + \cos\theta$ at $\theta = \dfrac{\pi}{2}$ .

Where does the polar curve $r = f(\theta)$ , for instance $r = 1 + \sin\theta$ , have a horizontal tangent?
1. Where $\dfrac{dy}{d\theta} = 0$ and $\dfrac{dx}{d\theta} \ne 0$
2. Where $\dfrac{dr}{d\theta} = 0$
3. Where $\dfrac{dx}{d\theta} = 0$

Watch out!

Practice ★★

Find the slope of the spiral $r = 4\theta$ at $\theta = \dfrac{\pi}{2}$ .
Find the slope of the spiral $r = 8\theta$ at $\theta = \dfrac{\pi}{2}$ .

Exit ticket

Three questions to check you have got it.

Find the slope $\dfrac{dy}{dx}$ of the polar curve $r = 8 + \cos\theta$ at $\theta = \dfrac{\pi}{2}$ .
Find the slope $\dfrac{dy}{dx}$ of the polar curve $r = 5 + \cos\theta$ at $\theta = \dfrac{\pi}{2}$ .
Find the slope $\dfrac{dy}{dx}$ of the polar curve $r = 2 + \cos\theta$ at $\theta = \dfrac{\pi}{2}$ .

Summary card

Polar slopes

$\dfrac{dy}{dx} = \dfrac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}$ .

Watch out: $dr/d\theta$ is not the slope.

6. Area in polar coordinates

Warm-up

Three quick questions from earlier lessons.

Discover

A thin wedge of a disk of radius $r$ with angle $d\theta$ is almost a sector. A full disk ( $2\pi$ ) has area $\pi r^2$ . What is the area of the wedge?

$\tfrac{1}{2}r^2\,d\theta$
$r\,d\theta$

Key idea

Add up thin sectors.

The area swept by

r = f(\theta)

for

\alpha \le \theta \le \beta

A = \dfrac{1}{2}\displaystyle\int_{\alpha}^{\beta} r^2\,d\theta

Worked example

Area inside $r = 2\sin\theta$ , $0 \le \theta \le \pi$ .

A = \tfrac{1}{2}\displaystyle\int_0^{\pi} 4\sin^2\theta\,d\theta

r² = 4 sin²θ.

= 2 \cdot \tfrac{\pi}{2}

sin² averages ½ over [0, π].

= \pi

A circle of radius 1.

Your turn, guided

Your turn: inside $r = 3$ .

A = 9\pi

½ ∫₀^{2π} 9 dθ.

Practice ★

Use $A = \tfrac{1}{2}\displaystyle\int r^2\,d\theta$ to find the area inside $r = 9$ .
Use $A = \tfrac{1}{2}\displaystyle\int r^2\,d\theta$ to find the area inside $r = 4$ .
Use $A = \tfrac{1}{2}\displaystyle\int r^2\,d\theta$ to find the area inside $r = 6$ .

Watch out!

Practice ★★

Find the area of one petal of the rose $r = 3\cos(2\theta)$ (the petal for $-\tfrac{\pi}{4} \le \theta \le \tfrac{\pi}{4}$ ).

Find the area inside the cardioid $r = 5(1 + \cos\theta)$ .
Find the area inside the cardioid $r = 3(1 + \cos\theta)$ .

Exit ticket

Three questions to check you have got it.

Use $A = \tfrac{1}{2}\displaystyle\int r^2\,d\theta$ to find the area inside $r = 9$ .
Use $A = \tfrac{1}{2}\displaystyle\int r^2\,d\theta$ to find the area inside $r = 4$ .
Use $A = \tfrac{1}{2}\displaystyle\int r^2\,d\theta$ to find the area inside $r = 6$ .

Summary card

Polar area

$A = \tfrac{1}{2}\displaystyle\int_{\alpha}^{\beta} r^2\,d\theta$ .

Watch out: the ½ and the square.

Answers: Parametric equations, polar coordinates and vector-valued functions

1. Derivatives of parametric curves

$6$
$\frac{7}{2}$
$\frac{11}{2}$

$- \frac{\sqrt{3}}{3}$
$- \sqrt{3}$

$\frac{3}{2}$

$6$
$\frac{7}{2}$
$\frac{11}{2}$

2. Second derivatives of parametric curves

$\frac{6}{t}$
$\frac{3}{4 t}$

$\frac{9}{4}$
$\frac{7}{4}$

$12 e^{- 5 t}$
$20 e^{- 6 t}$

$\frac{6}{t}$
$\frac{3}{4 t}$
$\frac{15}{4 t}$

3. Arc length of parametric curves

$6 \pi$
$10 \pi$

$3 \sqrt{5}$
$8 \sqrt{2}$

$\frac{14}{3}$
$42$

$3 \sqrt{5}$
$8 \sqrt{2}$
$2 \sqrt{29}$

4. Vector-valued functions: velocity, speed, acceleration

$2 \sqrt{13}$
$\sqrt{89}$
$\sqrt{13}$

$12$
$16$

$5$
$26$

$2 \sqrt{13}$
$\sqrt{89}$
$\sqrt{13}$

5. Slopes of polar curves

$\frac{1}{8}$
$\frac{1}{5}$

Where $\dfrac{dy}{d\theta} = 0$ and $\dfrac{dx}{d\theta} \ne 0$

$- \frac{2}{\pi}$
$- \frac{2}{\pi}$

$\frac{1}{8}$
$\frac{1}{5}$
$\frac{1}{2}$

6. Area in polar coordinates

$81 \pi$
$16 \pi$
$36 \pi$

$\frac{9 \pi}{8}$

$\frac{75 \pi}{2}$
$\frac{27 \pi}{2}$

$81 \pi$
$16 \pi$
$36 \pi$