Parametric Curves & Calculus

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

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

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

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

Eliminating the Parameter

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

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

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

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 $t$.

Parametric Basics

Direction & Speed

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

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

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.

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 $= \sqrt{(x')^2 + (y')^2}$
• Note any domain restrictions when converting to Cartesian

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

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

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

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 $dy/dx$ with respect to $t$, then divide by $dx/dt$.

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

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

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

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

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

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

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!

Setup Practice

Computation

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.

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 $t$):

$\boxed{A = \int_a^b y(t) \cdot x'(t)\,dt}$

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 $|A|$.

Area Practice

Setup the Integral

Area Computation

Key Takeaways — Part 4

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

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

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

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

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 $-2 \le t \le 2$.

(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 $t = 2$.

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

Mixed Practice

FRQ Computation

Workshop Recap

FRQ Strategy for Parametric Problems:

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

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

Parametric Curves & Calculus

Part 7 of 7 — Comprehensive Review

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

Comprehensive Assessment

Quick-Reference Decision Guide

Given a parametric problem, identify what is asked:

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.