Unit 9

Parametric Equations, Polar Coordinates, and Vector-Valued Functions: the one-page sheet

9.1 Parametric Equations and Their Derivatives

For a curve x = x(t) , y = y(t) , the derivatives dx/dt and dy/dt are the horizontal and vertical rates, and their signs give the direction of motion. The slope of the curve is dy/dx = (dy/dt)/(dx/dt) where dx/dt ≠ 0 , evaluated at a time. Horizontal tangents need dy/dt = 0 , vertical tangents dx/dt = 0 , with the other rate nonzero.

The slope of a parametric curve is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ , provided $\frac{dx}{dt} \ne 0$ . It is a function of $t$ : evaluate it at the time, not at the point.
A tangent line needs a point and a slope, both at the same $t$ : $y - y(t_0)$ $= \left.\frac{dy}{dx}\right|_{t=t_0}(x - x(t_0))$ .
Signs give direction: $\frac{dx}{dt} > 0$ moves right, $< 0$ left; $\frac{dy}{dt} > 0$ moves up, $< 0$ down.
Horizontal tangent: $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \ne 0$ . Vertical tangent: $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \ne 0$ . Both zero: look closer.

dx/dt and dy/dt are the horizontal and vertical rates of the moving point so dy/dx = (dy/dt)/(dx/dt) wherever dx/dt ≠ 0 the point moves straight across for an instant: a horizontal tangent the point moves straight up or down for an instant: a vertical tangent

slope of a parametric curve: For a curve $x = x(t)$ , $y = y(t)$ , the slope of the tangent line is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ wherever $\frac{dx}{dt} \ne 0$ (the chain rule solved for $\frac{dy}{dx}$ ). The tangent is horizontal where $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \ne 0$ , and vertical where $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \ne 0$ .

9.2 Second Derivatives of Parametric Equations

The second derivative of a parametric curve is the derivative of dy/dx with respect to x : differentiate dy/dx with respect to t , then divide by dx/dt . Its sign gives the concavity of the curve, and with dy/dx = 0 it classifies a horizontal tangent as a relative minimum or maximum. Never stop after differentiating, and never divide d²y/dt² by d²x/dt² .

Step 1: find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ and simplify it as a function of $t$ .
Step 2: differentiate $\frac{dy}{dx}$ with respect to $t$ . Step 3: divide by $\frac{dx}{dt}$ . Skipping step 3 is the classic slip.
$\frac{d^2y}{dx^2}$ is not $\frac{d^2y/dt^2}{d^2x/dt^2}$ : second derivatives do not divide that way.
Concavity, inflection points and the second derivative test work as in Unit 5, with every derivative found from $t$ .

its rate of change with respect to x needs the same slope rule again d²y/dx² = (d/dt (dy/dx))/(dx/dt) gives the concavity of the curve there: positive is concave up, negative concave down the second derivative test classifies the point as a relative minimum or maximum

second derivative of a parametric curve: For a curve $x = x(t)$ , $y = y(t)$ , the second derivative is the derivative of $\frac{dy}{dx}$ with respect to $x$ : $\frac{d^2y}{dx^2}$ $= \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}$ , where $\frac{dx}{dt} \ne 0$ . Its sign gives the concavity of the curve. It is not $\frac{d^2y/dt^2}{d^2x/dt^2}$ , and not $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ alone.

9.3 Arc Length of a Parametric Curve

The length of a parametric curve traced once for a ≤ t ≤ b is ∫ from a to b of √((x′(t))² + (y′(t))²) dt , the integral of the speed of the moving point. The limits are values of t ; a curve traced more than once is counted more than once. With x = t the formula reduces to the y = f(x) formula, and on calculator questions you write the integral and give three decimal places.

$L$ $= \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2}\,dt$ : square each derivative, add, take the root, integrate in $t$ .
The limits are values of $t$ , not of $x$ . Check that $t$ from $a$ to $b$ traces the curve once; a path traced twice gives twice the length.
For $y = f(x)$ , take $x = t$ : the formula becomes the 8.13 formula $\int_a^b \sqrt{1 + (f'(x))^2}\,dx$ .
Exact answers come from integrands that simplify, often by $\cos^2 t + \sin^2 t = 1$ or a perfect square under the root.

the piece of curve is close to a segment of length √((x′(t))² + (y′(t))²) Δt L = ∫ from a to b of √((x′(t))² + (y′(t))²) dt the integral is the distance traveled, which is the curve's length when the curve is traced once on the exam they are calculator questions: write the setup, then three decimals

arc length of a parametric curve: The length of the curve $x = x(t)$ , $y = y(t)$ for $a \le t \le b$ is $L$ $= \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2}\,dt$ , provided the curve is traced exactly once as $t$ runs from $a$ to $b$ . The integrand is the speed of the moving point, so the same integral gives the distance it travels, even when it retraces part of the curve.

9.4 Vector-Valued Functions and Their Derivatives

A vector-valued function r(t) = ⟨x(t), y(t)⟩ gives the position of a particle as a vector. Its derivative, the velocity vector ⟨x′(t), y′(t)⟩ , is tangent to the path and points in the direction of motion, and its magnitude is the speed. The acceleration vector is the second derivative. Every step is done one component at a time.

A vector-valued function $\mathbf{r}(t) = \left\langle x(t), y(t) \right\rangle$ is parametric equations in one package; every rule works one component at a time.
Velocity $\mathbf{v}(t) = \mathbf{r}'(t)$ and acceleration $\mathbf{a}(t) = \mathbf{r}''(t)$ : differentiate each component, keeping the chain rule's inside factors.
The speed is the magnitude of the velocity, $\sqrt{(x'(t))^2 + (y'(t))^2}$ , never a component alone.
From a table, estimate each component of the velocity with a difference quotient over the interval around the time.

a vector-valued function, describing the same path as the parametric equations the velocity vector v(t) = r′(t) = ⟨x′(t), y′(t)⟩ the velocity vector is tangent to the path and points the way the particle goes the acceleration vector a(t) = v′(t) = ⟨x″(t), y″(t)⟩

vector-valued function: A function that gives a vector for each input, such as the position vector $\mathbf{r}(t) = \left\langle x(t), y(t) \right\rangle$ of a moving particle. It describes the same path as the parametric equations $x = x(t)$ , $y = y(t)$ , and it is differentiated and integrated one component at a time.
velocity vector: The derivative of the position vector, $\mathbf{v}(t)$ $= \mathbf{r}'(t)$ $= \left\langle x'(t), y'(t) \right\rangle$ . It is tangent to the path and points in the direction of motion; its magnitude is the speed. The acceleration vector is the next derivative, $\mathbf{a}(t)$ $= \mathbf{v}'(t)$ $= \left\langle x''(t), y''(t) \right\rangle$ .

9.5 Integrating Vector-Valued Functions

To integrate a vector-valued function, integrate each component; an indefinite integral carries a constant vector C . An initial condition fixes it: the position at time b is the starting position plus the integral of the velocity, r(b) = r(a) +∫ from a to b of v(t) dt . From a table, approximate each component's integral with a sum, then add the start.

Integrate a vector one component at a time: $\int \left\langle f(t), g(t) \right\rangle\,dt$ $= \left\langle F(t), G(t) \right\rangle$ $+ \mathbf{C}$ .
Position from velocity always adds the start: $x(b)$ $= x(a)$ $+ \int_a^b x'(t)\,dt$ and $y(b)$ $= y(a)$ $+ \int_a^b y'(t)\,dt$ .
From acceleration, integrate twice: $\mathbf{v}(0)$ fixes the first constant and $\mathbf{r}(0)$ the second.
$\int_a^b y'(t)\,dt$ is the change in $y$ from $t = a$ to $t = b$ , not $y(b)$ and not a distance.

antidifferentiate each component, adding a constant vector C = ⟨C_1, C_2⟩ it fixes C and picks out the one path the particle follows r(b) = r(a) +∫ from a to b of v(t) dt , one component at a time approximate each component's integral with a Riemann or trapezoidal sum, then add the start

antiderivative of a vector-valued function: An antiderivative of a vector-valued function, found one component at a time: $\int \left\langle f(t), g(t) \right\rangle\,dt$ $= \left\langle F(t), G(t) \right\rangle$ $+ \mathbf{C}$ , with a constant vector $\mathbf{C}$ . An initial condition fixes $\mathbf{C}$ ; for position from velocity, $\mathbf{r}(t)$ $= \mathbf{r}(0)$ $+ \int_0^t \mathbf{v}(s)\,ds$ .

9.6 Motion in the Plane

For a particle in the plane, velocity and acceleration are the first and second derivatives of the position vector, and speed is the magnitude of the velocity. The integral of velocity is the displacement vector, so the position at a time is the start plus that integral; the integral of speed is the total distance traveled. Direction questions read the signs of dx/dt and dy/dt , and calculator answers carry their setup and three decimal places.

Speed is the magnitude of the velocity, $\sqrt{(x'(t))^2 + (y'(t))^2}$ . It is not $\left|\frac{dy}{dx}\right|$ , which measures steepness, not how fast.
Position: $x(b)$ $= x(a)$ $+ \int_a^b x'(t)\,dt$ , and the same for $y$ . Add the initial value, keep the $dt$ .
Distance traveled is $\int_a^b \sqrt{(x'(t))^2 + (y'(t))^2}\,dt$ ; displacement is $\int_a^b \mathbf{v}(t)\,dt$ , a vector.
" $y$ is decreasing" reads the sign of $\frac{dy}{dt}$ ; "moving left" reads $\frac{dx}{dt} < 0$ . A collision needs the same point at the same time.

velocity v(t) = ⟨x′(t), y′(t)⟩ , then acceleration a(t) = ⟨x″(t), y″(t)⟩ the speed √((x′(t))² + (y′(t))²) , a number and never negative the displacement vector r(b) − r(a) ; add the start to get the position the total distance traveled along the path

motion in the plane: A particle in the plane with position $\mathbf{r}(t) = \left\langle x(t), y(t) \right\rangle$ has velocity $\mathbf{v} = \mathbf{r}'$ , acceleration $\mathbf{a} = \mathbf{v}'$ and speed $\sqrt{(x'(t))^2 + (y'(t))^2}$ . The displacement vector over $[a, b]$ is $\int_a^b \mathbf{v}(t)\,dt$ $= \mathbf{r}(b) - \mathbf{r}(a)$ ; the total distance traveled is the integral of the speed, $\int_a^b \sqrt{(x'(t))^2 + (y'(t))^2}\,dt$ .

9.7 Polar Coordinates and Polar Derivatives

Treat r = f(θ) as the parametric curve x = rcos θ , y = rsin θ . The product rule gives dx/dθ and dy/dθ ; their quotient is the slope. Separately, the signs of r and dr/dθ say whether the point moves toward the pole (opposite signs) or away from it (the same sign).

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 $|x| = |r\cos\theta|$ .

it is a parametric curve with parameter θ dx/dθ = dr/dθ cos θ − rsin θ and dy/dθ = dr/dθ sin θ + rcos θ the slope dy/dx = (dy/dθ)/(dx/dθ) the same sign: moving away from the pole; opposite signs: moving toward it

slope of a polar curve: The slope of a polar curve $r = f(\theta)$ : $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ , with $x = r\cos\theta$ and $y = r\sin\theta$ , where $\frac{dx}{d\theta} \ne 0$ . Not to be confused with $\frac{dr}{d\theta}$ , the rate of change of $r$ , the signed distance from the pole.

9.8 Area Inside a Polar Curve

The area of the region bounded by r = f(θ) and the rays θ = α and θ = β is 1/2 ∫ from α to β of r² dθ , the limit of sums of thin wedges 1/2 r² Δθ . Square r , keep the 1/2 , and choose limits that sweep the region exactly once.

$A$ $= \frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta$ : the region bounded by $r = f(\theta)$ and the rays $\theta = \alpha$ , $\theta = \beta$ . Write the $\frac{1}{2}$ and the square in every setup.
Limits: a loop or petal runs between consecutive zeros of $r$ . Check that the region is traced once; a circle such as $r = 4\sin\theta$ is traced once on $[0, \pi]$ , twice on $[0, 2\pi]$ .
No calculator: expand $r^2$ and use $\sin^2 u$ $= \frac{1 - \cos(2u)}{2}$ and $\cos^2 u$ $= \frac{1 + \cos(2u)}{2}$ . Calculator: write the integral, then the value to three decimals.

each wedge is nearly a slice of a circle of radius r the region's area is about ∑ 1/2 r² Δθ , a Riemann sum A = 1/2 ∫ from α to β of r² dθ the region is swept exactly once

polar area: The area of the region bounded by a polar curve $r = f(\theta)$ and the rays $\theta = \alpha$ and $\theta = \beta$ : $A$ $= \frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta$ , the limit of sums of thin wedges of area $\frac{1}{2}r^2\,\Delta\theta$ .

9.9 Area Between Two Polar Curves

For the region outside r = r(θ) and inside r = R(θ) , find the limits where the curves meet (and check the pole), decide which curve is outer, and compute 1/2 ∫ from α to β of (R² − r²)dθ . Square each radius before subtracting.

$A$ $= \frac{1}{2}\int_{\alpha}^{\beta} \left(R(\theta)^2 - r(\theta)^2\right)d\theta$ , outer squared minus inner squared. Never $\frac{1}{2}\int (R - r)^2\,d\theta$ .
Limits come from where the curves meet: solve $f(\theta) = g(\theta)$ , then check the pole separately, since each curve may reach it at a different angle.
Area arithmetic often beats a new integral: the region inside both curves is one curve's area minus the region inside it and outside the other.

the crossing angles, which become the limits you know which curve is outer, R , and which is inner, r its area is 1/2 (R² − r²)Δθ A = 1/2 ∫ from α to β of (R² − r²)dθ

area between polar curves: The area of the region inside a polar curve $r = R(\theta)$ and outside $r = r(\theta)$ : $\frac{1}{2}\int_{\alpha}^{\beta} \left(R(\theta)^2 - r(\theta)^2\right)d\theta$ , with limits where the curves meet. Each radius is squared before subtracting.