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Parametric Curves & Calculus

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Parametric Curves & Calculus - Complete Interactive Lesson

Part 1: Parametric Equations

Parametric Curves & Calculus

Part 1 of 7 — Parametric Equations & Graphing

Parametric equations describe a curve using a parameter $t$ , giving both $x$ and $y$ as functions of $t$ . This is essential for modeling motion and curves that fail the vertical line test.

Parametric Equations

$\boxed{x = f(t), \quad y = g(t), \quad a \le t \le b}$

Eliminating the Parameter

To convert from parametric to Cartesian, eliminate $t$ :

Example: $x = 2\cos t$ , $y = 2\sin t$

$\cos$ ,

Parametric Basics

Direction & Speed

The direction (orientation) is determined by increasing $t$ .

Speed along the curve at time $t$ : $\boxed{\text{speed} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}}$

Curve Identification

Speed Calculation

Key Takeaways — Part 1

Parametric equations: $x = f(t)$ , $y = g(t)$
Eliminate parameter using algebra or trig identities
Direction determined by increasing $t$
Speed $= \sqrt{}$

Part 2: Second Derivative

Parametric Curves & Calculus

Part 2 of 7 — Derivatives of Parametric Curves

The chain rule gives us a formula for $dy/dx$ in terms of the parameter $t$ . This is one of the most-tested BC topics.

First Derivative

$\boxed{\frac{dy}{dx} = \frac{dy/dt}{dx/dt} \quad \text{provided } \frac{dx}{dt} \ne 0}$

Part 3: Arc Length (Parametric)

Parametric Curves & Calculus

Part 3 of 7 — Arc Length of Parametric Curves

The arc length formula for parametric curves is a direct extension of the Pythagorean theorem applied to infinitesimal segments.

Arc Length Formula

$\boxed{L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt}$

Part 4: Area Under Parametric Curves

Parametric Curves & Calculus

Part 4 of 7 — Area Under Parametric Curves

The area formula for parametric curves converts the standard $\int y\,dx$ into an integral over the parameter $t$ .

Area Formula

For a curve traced left to right ( $x$ increasing with $t$ ):

Part 5: Eliminating the Parameter

Parametric Curves & Calculus

Part 5 of 7 — Surface Area & Volume of Revolution

When parametric curves are revolved around an axis, we can compute the surface area and volume using modified integral formulas.

Surface Area of Revolution

Revolution about the $x$ -axis: $\boxed{SA = 2\pi \int_a^b |y(t)| \cdot \sqrt{(x')^2 + (y')^2}\,dt}$

Part 6: Practice Workshop

Parametric Curves & Calculus

Part 6 of 7 — Problem-Solving Workshop

Mixed practice covering all parametric curve concepts: graphing, derivatives, arc length, area, and applications.

Workshop Problems

AP FRQ-Style Problem

A particle moves with $x(t) = t^3 - 3t$ , $y(t) = 3t^2 - 9$ for .

Part 7: Final Assessment

Parametric Curves & Calculus

Part 7 of 7 — Comprehensive Review

Master all parametric curve concepts: equations, derivatives, arc length, area, and surface area.

Concept	Key Formula
Slope	$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

The parameter $t$ typically represents time. As $t$ increases, the point $(x(t), y(t))$ traces a curve with a specific direction (orientation).

Common Parametric Curves

Curve	$x(t)$	$y(t)$	Shape
Circle	$a\cos t$	$a\sin t$	Circle radius $a$ , CCW
Ellipse	$a\cos t$	$b\sin t$	Ellipse
Line	$x_0 + at$	$y_0 + bt$
Parabola	$t$	$t^2$	Standard parabola
Cycloid	$t - \sin t$	$1 - \cos t$	Arch shape

Key Fact: A single Cartesian curve can have many different parametric representations. What differs is the speed and direction of traversal.

\cos^2 t + \sin^2 t = 1 \implies \frac{x^2}{4} + \frac{y^2}{4} = 1 \implies x^2 + y^2 = 4

Example: $x = t + 1$ , $y = t^2 - 3$

$t = x - 1$ , so $y = (x-1)^2 - 3$

Example: $x = e^t$ , $y = e^{2t} + 1$

$x = e^t \implies e^{2t} = x^2$ , so $y = x^2 + 1$ (with $x > 0$ )

AP Tip: When eliminating the parameter, state any restrictions on $x$ or $y$ from the domain of $t$ .

For $x = 3\cos t$ , $y = 3\sin t$ : $\text{speed} = \sqrt{(-3\sin t)^2 + (3\cos t)^2} = \sqrt{9} = 3$

The particle moves at constant speed $3$ along the circle. This is uniform circular motion.

= (x^{'})^{2} + (y^{'})^{2}

Note any domain restrictions when converting to Cartesian

Coming Up: Part 2 covers derivatives of parametric curves — $dy/dx$ and $d^2y/dx^2$ .

\frac{d y}{d x} = \frac{d y / d t}{d x / d t} provided \frac{d x}{d t} \neq = 0

This gives the slope of the tangent line to the parametric curve at the point corresponding to parameter $t$ .

Example: $x = t^2 - 1$ , $y = t^3 - 3t$

$\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^2 - 3$

$\frac{dy}{dx} = \frac{3t^2 - 3}{2t} = \frac{3(t^2-1)}{2t}$

At $t = 2$ : $\frac{dy}{dx} = \frac{3(4-1)}{4} = \frac{9}{4}$ at the point $(3, 2)$ .

Key Fact: Horizontal tangent when $dy/dt = 0$ (and $dx/dt \ne 0$ ). Vertical tangent when $dx/dt = 0$ (and $dy/dt \ne 0$ ).

Second Derivative

$\boxed{\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}}$

Critical: This is NOT $\frac{d^2y/dt^2}{d^2x/dt^2}$ ! You must differentiate with respect to , then divide by .

Example (continued): $\frac{dy}{dx} = \frac{3t^2-3}{2t} = \frac{3}{2}t - \frac{3}{2t}$

$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{3}{2} + \frac{3}{2t^2}$

$\frac{d^2y}{dx^2} = \frac{\frac{3}{2} + \frac{3}{2t^2}}{2t} = \frac{3t^2 + 3}{4t^3}$

At $t = 2$ : $\frac{d^2y}{dx^2} = \frac{15}{32}$ (concave up since positive).

Derivative Practice

Tangent Lines

The tangent line at $t = t_0$ : $y - y(t_0) = \frac{dy/dx}\bigg|_{t=t_0} \cdot (x - x(t_0))$

Example: $x = t + \sin t$ , $y = t - \cos t$ at $t = 0$

Point: $(0 + 0, 0 - 1) = (0, -1)$

Slope: $\frac{dy/dx} = \frac{1 + \sin t}{1 + \cos t}\bigg|_{t=0} = \frac{1}{2}$

$\boxed{y + 1 = \frac{1}{2}(x - 0) \implies y = \frac{x}{2} - 1}$

AP Tip: Tangent line problems at specific parameter values are guaranteed on the BC exam. Always find the point AND the slope.

Classify the Tangent

Slope Computation

Key Takeaways — Part 2

Formula	Expression
First derivative	$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
Second derivative	$\frac{d^2y}{dx^2} = \frac{(d/dt)(dy/dx)}{dx/dt}$
Horizontal tangent	$dy/dt = 0$ , $dx/dt \ne 0$
Vertical tangent	$dx/dt = 0$ , $dy/dt \ne 0$

Coming Up: Part 3 covers arc length of parametric curves.

Derivation: A tiny piece of the curve has horizontal change $dx$ and vertical change $dy$ . By the Pythagorean theorem: $ds = \sqrt{dx^2 + dy^2} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$

Note: $\sqrt{(dx/dt)^2 + (dy/dt)^2}$ is the speed of the particle. So:

$\boxed{\text{Arc length} = \int_a^b \text{speed}\,dt}$

Key Fact: Arc length equals the integral of speed — this makes physical sense! Distance = speed × time.

Example: Circle $x = 3\cos t$ , $y = 3\sin t$ , $0 \le t \le 2\pi$

Quantity	Value
$dx/dt$	$-3\sin t$
$dy/dt$	$3 \cos$

This confirms: circumference of circle with radius $3$ is $2\pi(3) = 6\pi$ . \checkmark

Example: $x = t^2$ , $y = t^3$ , $0 \le t \le 1$

$L = \int_0^1 \sqrt{(2t)^2 + (3t^2)^2}\,dt = \int_0^1 \sqrt{4t^2 + 9t^4}\,dt = \int_0^1 t\sqrt{4+9t^2}\,dt$

Let $u = 4 + 9t^2$ : $L = \frac{1}{18}\cdot\frac{2}{3}[u^{3/2}]_4^{13} = \frac{1}{27}(13\sqrt{13} - 8)$

Arc Length Practice

Arc Length vs. Displacement

Concept	Formula	Meaning
Arc length	$\int_a^b \sqrt{(x')^2 + (y')^2}\,dt$	Total distance traveled
Displacement	$\sqrt{(\Delta x)^2 + (\Delta y)^2}$

Arc length $\ge$ displacement, with equality only for straight-line motion.

AP Tip: The AP exam often asks for distance traveled (arc length), not displacement. Read carefully!

Key Takeaways — Part 3

$L = \int_a^b \sqrt{(x')^2 + (y')^2}\,dt = \int_a^b \text{speed}\,dt$

Arc length = integral of speed
Always $\ge$ displacement
For circles: confirms $C = 2\pi r$

Coming Up: Part 4 covers area enclosed by parametric curves.

A = \int_{a}^{b} y (t) \cdot x^{'} (t) d t

This comes from substituting $dx = x'(t)\,dt$ into $A = \int y\,dx$ .

For a curve traced right to left ( $x$ decreasing): $A = -\int_a^b y(t) \cdot x'(t)\,dt$

Example: Area under one arch of the cycloid $x = t - \sin t$ , $y = 1 - \cos t$

One arch: $0 \le t \le 2\pi$ . Since $x' = 1 - \cos t \ge 0$ , the curve moves left to right:

$A = \int_0^{2\pi} (1-\cos t)(1-\cos t)\,dt = \int_0^{2\pi} (1-\cos t)^2\,dt$

$= \int_0^{2\pi} (1 - 2\cos t + \cos^2 t)\,dt = 2\pi - 0 + \pi = 3\pi$

$\boxed{\text{Area under one cycloid arch} = 3\pi}$

Area Enclosed by a Closed Curve

For a closed parametric curve traversed counterclockwise:

$A = -\oint y(t)\,x'(t)\,dt = \oint x(t)\,y'(t)\,dt$

Example: Ellipse $x = a\cos t$ , $y = b\sin t$ , $0 \le t \le 2\pi$

$A = \int_0^{2\pi} b\sin t \cdot (-a\sin t)\,dt = -ab\int_0^{2\pi} \sin^2 t\,dt = -ab \cdot \pi$

Since the curve goes counterclockwise and we get a negative result from $y \cdot x'$ , take the absolute value:

$\boxed{A = \pi ab}$

This confirms the well-known ellipse area formula.

AP Tip: Watch the sign! If the formula gives a negative area, the curve is traced in the opposite direction to what you assumed. Take $|A|$ .

Setup the Integral

Area Computation

Key Takeaways — Part 4

Formula	Use
$A = \int_a^b y(t)x'(t)\,dt$	Area under curve (left to right)
$A = \pi ab$	Ellipse $x = a\cos t$ , $y = b\sin t$
Cycloid arch	$3\pi$ (for unit cycloid)

Coming Up: Part 5 covers surface area of revolution for parametric curves.

SA=2π∫ab∣y(t)∣⋅(x′)2+(y′)2dt

Revolution about the $y$ -axis: $\boxed{SA = 2\pi \int_a^b |x(t)| \cdot \sqrt{(x')^2 + (y')^2}\,dt}$

The factor $2\pi r$ comes from the circumference of revolution, and $ds = \sqrt{(x')^2 + (y')^2}\,dt$ is the arc element.

Example: Sphere from $x = R\cos t$ , $y = R\sin t$ , $0 \le t \le \pi$

Revolving the upper semicircle about the $x$ -axis:

$SA = 2\pi \int_0^{\pi} R\sin t \cdot R\,dt = 2\pi R^2 \int_0^{\pi} \sin t\,dt = 2\pi R^2 \cdot 2 = 4\pi R^2$

$\boxed{SA_{\text{sphere}} = 4\pi R^2 \quad \checkmark}$

Volume of Revolution (Disk/Washer)

About the $x$ -axis (using disks): $V = \pi \int_a^b [y(t)]^2 \cdot x'(t)\,dt$

Example: Sphere volume from semicircle

$x = R\cos t$ , $y = R\sin t$ , $0 \le t \le \pi$ :

$V = \pi \int_0^{\pi} R^2\sin^2 t \cdot (-R\sin t)\,dt = -\pi R^3 \int_0^{\pi} \sin^3 t\,dt$

$= \pi R^3 \cdot \frac{4}{3} = \frac{4}{3}\pi R^3$

$\boxed{V_{\text{sphere}} = \frac{4}{3}\pi R^3 \quad \checkmark}$

AP Tip: Volume problems with parametric curves often appear in BC FRQs. Set up the integral carefully, matching the revolution axis.

Surface Area & Volume

Identify the Setup

Quick Computation

Key Takeaways — Part 5

Quantity	About $x$ -axis	About $y$ -axis
Surface area	$2\pi \int	y
Volume (disk)	$\pi \int y^2 x'\,dt$	$\pi \int x^2 y'\,dt$

Coming Up: Part 6 is a Problem-Solving Workshop with mixed parametric problems.

y (t) = 3 t^{2} - 9

(a) Find all times when the particle has a horizontal tangent.

$dy/dt = 6t = 0 \implies t = 0$ . Check: $dx/dt = 3(0)^2 - 3 = -3 \ne 0$ . \checkmark

Horizontal tangent at $t = 0$ : point $(0, -9)$ .

(b) Find all times when the particle has a vertical tangent.

$dx/dt = 3t^2 - 3 = 3(t-1)(t+1) = 0 \implies t = \pm 1$

At $t = 1$ : $dy/dt = 6 \ne 0$ . Point: $(-2, -6)$ . \checkmark At $t = -1$ : $dy/dt = -6 \ne 0$ . Point: $(2, -6)$ . \checkmark

(c) Find $dy/dx$ at $t = 2$ .

$\frac{dy}{dx} = \frac{6(2)}{3(4)-3} = \frac{12}{9} = \frac{4}{3}$

Workshop Recap

FRQ Strategy for Parametric Problems:

Find derivatives: $dx/dt$ , $dy/dt$ , $dy/dx$
Identify special points: horizontal/vertical tangents
Set up and evaluate integrals: arc length, area
Interpret results in context of motion

Coming Up: Part 7 is the Comprehensive Review of parametric curves.

Comprehensive Assessment

Quick-Reference Decision Guide

Given a parametric problem, identify what is asked:

Asked For	Set Up
Tangent line slope	$dy/dx = (dy/dt)/(dx/dt)$ at given $t$
Horizontal tangent	Solve $dy/dt = 0$ , verify $dx/dt \ne 0$
Vertical tangent	Solve $dx/dt = 0$ , verify $dy/dt \ne 0$
Concavity	Compute $d^2y/dx^2$
Distance traveled	$\int_a^b \sqrt{(x')^2+(y')^2}\,dt$
Area enclosed	$-\oint y\,dx = -\int_a^b y(t)\,x'(t)\,dt$
Surface area (about $x$ -axis)	$2\pi\int y\,\sqrt{(x')^2+(y')^2}\,dt$

AP Key Fact: On the AP exam, distinguish between distance traveled (always positive, involves speed integral) and displacement ( $\Delta x, \Delta y$ separately). They ask both!

Concept Connections

Final Computation

Parametric Curves Complete!

You have mastered:

Parametric equations, elimination, and graphing
First and second derivatives via the chain rule
Arc length and speed computations
Area under and enclosed by parametric curves
Surface area and volume of revolution

AP Exam Note: Parametric/polar/vector questions appear as a dedicated FRQ (usually problem 2 or 3). Practice computing derivatives and integrals quickly since both calculator and non-calculator parts appear.

−ab∫02πsin2tdt=

−πR3∫0πsin3tdt

Parametric Curves & Calculus

Parametric Curves & Calculus - Complete Interactive Lesson

Part 1: Parametric Equations

Parametric Curves & Calculus

Parametric Equations

Eliminating the Parameter

Direction & Speed

Key Takeaways — Part 1

Part 2: Second Derivative

Parametric Curves & Calculus

First Derivative

Part 3: Arc Length (Parametric)

Parametric Curves & Calculus

Arc Length Formula

Part 4: Area Under Parametric Curves

Parametric Curves & Calculus

Area Formula

Part 5: Eliminating the Parameter

Parametric Curves & Calculus

Surface Area of Revolution

Part 6: Practice Workshop

Parametric Curves & Calculus

AP FRQ-Style Problem

Part 7: Final Assessment

Parametric Curves & Calculus

Common Parametric Curves

Example: x=t2−1x = t^2 - 1x=t2−1, y=t3−3ty = t^3 - 3ty=t3−3t

Second Derivative

Example (continued): dydx=3t2−32t=32t−32t\frac{dy}{dx} = \frac{3t^2-3}{2t} = \frac{3}{2}t - \frac{3}{2t}dxdy​=2t

Tangent Lines

Example: x=t+sin⁡tx = t + \sin tx=t+sint, y=t−cos⁡ty = t - \cos ty=t−cost at t=0t = 0t=0

Key Takeaways — Part 2

Example: Circle x=3cos⁡tx = 3\cos tx=3cost, y=3sin⁡ty = 3\sin ty=3sint, 0≤t≤2π0 \le t \le 2\pi0≤t≤2π

Example: x=t2x = t^2x=t2, y=t3y = t^3y=t3, 0≤t≤10 \le t \le 10

Arc Length vs. Displacement

Key Takeaways — Part 3

Example: Area under one arch of the cycloid x=t−sin⁡tx = t - \sin tx=t−sint, y=1−cos⁡ty = 1 - \cos ty=1−cost

Area Enclosed by a Closed Curve

Example: Ellipse x=acos⁡tx = a\cos tx=acost, y=bsin⁡ty = b\sin ty=bsint, 0≤t≤2π0 \le t \le 2\pi0≤t≤

Key Takeaways — Part 4

Example: Sphere from x=Rcos⁡tx = R\cos tx=Rcost, y=Rsin⁡ty = R\sin ty=Rsint, 0≤t≤π0 \le t \le \pi0≤t≤π

Volume of Revolution (Disk/Washer)

Example: Sphere volume from semicircle

Key Takeaways — Part 5

Workshop Recap

Quick-Reference Decision Guide

Parametric Curves Complete!

Example: $x = t^2 - 1$ , $y = t^3 - 3t$

Example (continued): $\frac{dy}{dx} = \frac{3t^2-3}{2t} = \frac{3}{2}t - \frac{3}{2t}$

Example: $x = t + \sin t$ , $y = t - \cos t$ at $t = 0$

Example: Circle $x = 3\cos t$ , $y = 3\sin t$ , $0 \le t \le 2\pi$

Example: $x = t^2$ , $y = t^3$ , $0 \le t \le 1$

Example: Area under one arch of the cycloid $x = t - \sin t$ , $y = 1 - \cos t$

Example: Ellipse $x = a\cos t$ , $y = b\sin t$ , $0 \le t \le 2\pi$

Example: Sphere from $x = R\cos t$ , $y = R\sin t$ , $0 \le t \le \pi$