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Calculus BC Arc Length for Parametric & Polar Curves

Section 4: Applications of Integration: Lecture 2 | 17:40 min

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Lecture Comments (10)

	0 answers Post by ? ? on April 7, 2018 in Q5, cannot simply double the integral from 0,1/2pi . need to plus both 0 to 1/2pi and 1/2pi to pi, and the latter is negative. right answer is 1
	2 answers Last reply by: Xinyuan Xing Tue Apr 28, 2015 10:07 AM Post by Tim Zhang on April 24, 2014 ä¸‹æ¬¡ç›´æŽ¥ç”¨ä¸æ–‡è¯´å§, dude,è‹±è¯å¤ªé€—äº†ã€‚
	0 answers Post by Narin Gopaul on November 14, 2013 I find these lecture very easy to understand but in stuwat calculus book it seems rather difficult
	1 answer Last reply by: ? ? Sat Apr 7, 2018 9:13 PM Post by Alexis Mata Betancourt on September 5, 2013 Careful! sqrt(2-2cos t) = 2 sin(t/2) ! Not 2 sin 2t
	1 answer Last reply by: Ziyue Guo Tue Apr 23, 2013 12:10 AM Post by Jinrong Shi on April 15, 2013 I believe there was a mistake in the first example. When solving the problem, the domain should be the root of 19 to the root of 55 rather that 2 to 6 because the domain should be set in term of u but not x
	0 answers Post by H.y. Kim on March 24, 2013 I think he meant "chain rule" at 12:55

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Arc Length for Parametric & Polar Curves

For normal function,
For parametric function,

Differentiate 2 parametric parts individually
Choosing correct bounds
Apply to formula

Arc Length for Parametric & Polar Curves

Find the length of the arc of y = (x − 8)^[3/2] from − 10 to 17

Find y′
y′ = [3/2](x − 8)^1/2dx
Use Arc Length Formula
Arc Length = ∫_a^b√{1 + ( [dy/dx] )²} dx
= ∫₁₀¹⁷ √{1 + ( [3/2](x − 8)^1/2 )²} d x
= ∫₁₀¹⁷ √{1 + ( [9/4](x − 8) )} dx
= ∫₁₀¹⁷√{1 + [9/4]x − 18} dx
= ∫₁₀¹⁷ √{[9/4]x − 17} dx
Use u substitution
u = [9/4]x − 17
du = [9/4]dx
∫₁₀¹⁷ √{[9/4]x − 17} dx = ∫₁₀¹⁷ [4/9]√u du
= [4/9]∫₁₀¹⁷ √u du
= [4/9][ [2/3]u^[3/2] ]₁₀¹⁷
= [4/9][ [2/3]( [9/4]x − 17 )^[3/2] ]₁₀¹⁷

≈ 25.20

Find the derivative of − lncosx

Use log properties and trig identities to alter
− lncosx = ln( cosx )^{− 1}
= ln[1/cosx]
= lnsecx

Use chain rule
[dy/dx] = [secxtanx/secx] = tanx dx

From 1 to 3, find the length of y = − lncosx

Find the derivative of − lncosx
y′ = tanx
Apply Arc Length equation
∫_a^b√{1 + ( [dy/dx] )²} dx = ∫₁³√{1 + ( tanx )²} dx
= ∫₁³√{1 + tan²x} dx
Use Pythagorean identity
∫₁³√{1 + tan²x} dx = ∫₁³√{sec²x} dx
= ∫₁³ secxdx
Use trig integral identity
∫ secxdx = ln|secx + tanx| + C
∫₁³ secxdx = [ ln|secx + tanx| ]₁³
= ln|sec(3) + tan(3)| − ln|sec(1) + tan(1)|

≈ − 1.08

Find the length of − 1 to 4 of the follow parametric functions:
x = [2/3]t^[3/2], y = 2t + 7

Find the deratives using product rule
x′ = √t
y′ = 2
Apply the parametric Arc Length Formula
Arc Length = ∫_a^b √{( [dx/dt] )² + ( [dy/dt] )²} dt
= ∫_{− 1}⁴ √{( √t )² + ( 2 )²} dt
= ∫_{− 1}⁴ √{t + 4} dt
Use substitution with u = t + 4
du = dt
∫_{− 1}⁴ √{t + 4} dt = ∫_{− 1}⁴ √u du
= [3/2][ t + 4 ]_{− 1}⁴
= [3/2]( 4 + 4 − ( − 1 + 4) )
= [3/2]( 8 + 3 )

= [33/2]

Find the length of 0 to 3 of the follow parametric functions:

x = e^tcost, y = e^tsint
Find the deratives using product rule
x′ = e^tcost − e^tsint
y′ = e^tsint + e^tcost
Apply the parametric Arc Length Formula
Arc Length = ∫_a^b √{( [dx/dt] )² + ( [dy/dt] )²} dt
= ∫₀³ √{( e^tcost − e^tsint )² + ( e^tsint + e^tcost )²} dt
= ∫₀³ √{2e^2t(cos²t + sin²t)} dt
= ∫₀³ √{2e^2t} dt
= √2 ∫₀³ e^tdt
= √2 [ e^t ]₀³

= √2 [ e³ − e⁰ ] = √2 [ e³ − 1 ]

Find the length of 1 to [(3π)/4] of the follow parametric functions:

x = ln|sint|, y = t
Find the deratives using trig integral identity
x′ = cott
y′ = 1
Apply the parametric Arc Length Formula
Arc Length = ∫_a^b √{( [dx/dt] )² + ( [dy/dt] )²} dt
= ∫₁^[(3π)/4] √{( cott )² + ( 1 )²} dt
= ∫₁^[(3π)/4] √{cot²t + 1} dt
Use Pythagoream Identity
∫₁^[(3π)/4] √{cot²x + 1} dt = ∫₁^[(3π)/4] √{csc²x} dt
= ∫₁^[(3π)/4] csctdt
= [ − ln|csct + cott| ]₁^[(3π)/4]

≈ 1.49

Find the arc length of the polar function r = 2cosθ from 0 to 5π

Find [dr/(dθ)]
r′ = − 2sinθ
Apply the polar Arc Length Formula
Arc Length = ∫_a^b √{r² + ( [dr/(dθ)] )²} dθ
= ∫₀^5π √{( 2cosθ )² + ( − 2sinθ )²} dθ
= ∫₀^5π √{4cos²θ+ 4sin²θ} dθ
= ∫₀^5π √4 √{cos²θ+ sin²θ} dθ
= 2∫₀^5π √{cos²θ+ sin²θ} dθ
= 2∫₀^5π √1 dθ
= 2[ θ]₀^5π

= 10π

What is the parametric Arc Length equation of r = cosθ?

Convert using Polar definitions
x = rcosθ
= cosθ(cosθ)
= cos²θ
y = rsinθ
= cosθsin θ
Find [dx/(dθ)] and [dy/(dθ)]
x′ = − 2sinθcosθ
y′ = cos( 2x )
Apply parametric Arc Length Formula (but not necessarily solve)
Arc Length = ∫_a^b √{( [dx/dt] )² + ( [dy/dt] )²} dt
= ∫_a^b √{( [dx/(dθ)] )² + ( [dy/(dθ)] )²} dθ

= ∫_a^b √{( − 2sinθcosθ )² + ( cos( 2x ) )²} dθ

What is the polar Arc Length Equation of r = 2e^2θ

Find [dr/(dθ)]
r′ = e^2θ
Arc Length = ∫_a^b √{r² + ( [dr/(dθ)] )²} dθ

= ∫_a^b √{4e^4θ² + e^4θ} dθ = ∫_a^b √{5e^4θ} dθ = √5 ∫_a^b e^2θdθ

What's the Arc Length for r = e^2θ from 0 to 1?

Apply previous equation with known bounds
√5 ∫₀¹ e^2θdθ = [(√5 )/2][ e^2θ ]₀¹

= [(√5 )/2](e² − 1)

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Answer

Arc Length for Parametric & Polar Curves

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

Arc Length

Example 2: Arc Length for Parametric Curves

Example 3: Arc Length for Parametric Curves

Example 4: Arc Length for Parametric Curves

Example 5: Arc Length for Parametric Curves

Example 6: Arc Length for Polar Curves

Example 7: Arc Length for Polar Curves

Intro 0:00
Arc Length 0:13

Arc Length of a Normal Function
Example

Example 2: Arc Length for Parametric Curves 3:31
Example 3: Arc Length for Parametric Curves 4:23
Example 4: Arc Length for Parametric Curves 8:05
Example 5: Arc Length for Parametric Curves 12:22
Example 6: Arc Length for Polar Curves 15:36
Example 7: Arc Length for Polar Curves 16:03

Mathematics: AP Calculus BC

Section 1: Prerequisites
	Parametric Curves	10:10
	Polar Coordinates	14:54
	Vector Functions	12:05
Section 2: Differentiation
	Parametric Differentiation	13:15
	Polar Differentiation	10:14
	L'Hopital's Rule	6:43
Section 3: Integration
	Integration By Parts	19:04
	Integration By Partial Fractions	22:04
	Improper Integrals	19:01
Section 4: Applications of Integration
	Logistic Growth	19:16
	Arc Length for Parametric & Polar Curves	17:40
	Area for Parametric & Polar Curves	4:33
Section 5: Sequences, Series, & Approximations
	Definition & Convergence	7:10
	Geometric Series	10:59
	Harmonic & P Series	3:58
	Integral Test	11:31
	Comparison Test	7:47
	Limit Comparison Test	9:40
	Ratio Test	11:20
	Alternating Series	9:11
	Absolute Convergence	7:13
	Power Series Convergence	12:52
	Taylor Series	15:11
	Power Series Operations	16:40
	Lagrange Error	7:09
Section 6: Practice Test
	AP Calc BC Practice Test Section 1: Multi Choice Part 1	15:45
	AP Calc BC Practice Test Section 1: Multi Choice Part 2	21:38
	AP Calc BC Practice Test Section 1: Multi Choice Part 3	17:31
	AP Calc BC Practice Test Section 1: Multi Choice Part 4	14:17
	AP Calc BC Practice Test Section 1: Multi Choice Part 5	22:45
	AP Calc BC Practice Test Section 1: Free Response Part 1	9:55
	AP Calc BC Practice Test Section 1: Free Response Part 2	10:08
	AP Calc BC Practice Test Section 1: Free Response Part 3	13:58

Related Books

	Cracking the AP Calculus BC Exam, 2015 Edition Author: Princeton Review ISBN: 804124825 Publisher: Princeton Review Year: 2014 The book includes 3 full length practice tests with detailed explanations, a review of all the key concepts, and targeted strateeies to ace th exam for your highest score.
	Cliffs AP: Calculus AB & BC, 3rd Edition Authors: Dale W Johnson, Kerry J King ISBN: 764586831 Publisher: Cliffs Notes Year: 2001 The book has a topic-by topic breakdown and lots of problem approach suggestions for both free response and multiple choice calculus questions. The book also includes two full length tests.

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John Zhu

Arc Length for Parametric & Polar Curves

Slide Duration:

Table of Contents

Section 1: Prerequisites

Parametric Curves

10m 10s

Intro