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Lecture Comments (8)

2 answers

Last reply by: Xinyuan Xing
Tue Apr 28, 2015 10:07 AM

Post by Tim Zhang on April 24, 2014

下次直接用中文说吧, dude,英语太逗了。

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

0 answers

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

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 = ∫ab √{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
  • ab √{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 = ∫ab √{( [dx/dt] )2 + ( [dy/dt] )2} dt
  • = ∫ − 14 √{( √t )2 + ( 2 )2} dt
  • = ∫ − 14 √{t + 4} dt
  • Use substitution with u = t + 4
  • du = dt
  • − 14 √{t + 4} dt = ∫ − 14 √u du
  • = [3/2][ t + 4 ] − 14
  • = [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 = etcost, y = etsint
  • Find the deratives using product rule
  • x′ = etcost − etsint
  • y′ = etsint + etcost
  • Apply the parametric Arc Length Formula
  • Arc Length = ∫ab √{( [dx/dt] )2 + ( [dy/dt] )2} dt
  • = ∫03 √{( etcost − etsint )2 + ( etsint + etcost )2} dt
  • = ∫03 √{2e2t(cos2t + sin2t)} dt
  • = ∫03 √{2e2t} dt
  • = √2 ∫03 etdt
  • = √2 [ et ]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 = ∫ab √{( [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 = ∫ab √{r2 + ( [dr/(dθ)] )2} dθ
  • = ∫0 √{( 2cosθ )2 + ( − 2sinθ )2} dθ
  • = ∫0 √{4cos2θ+ 4sin2θ} dθ
  • = ∫0 √4 √{cos2θ+ sin2θ} dθ
  • = 2∫0 √{cos2θ+ sin2θ} dθ
  • = 2∫0 √1 dθ
  • = 2[ θ]0
= 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 = ∫ab √{( [dx/dt] )2 + ( [dy/dt] )2} dt
  • = ∫ab √{( [dx/(dθ)] )2 + ( [dy/(dθ)] )2} dθ
= ∫ab √{( − 2sinθcosθ )2 + ( cos( 2x ) )2} dθ
What is the polar Arc Length Equation of r = 2e
  • Find [dr/(dθ)]
  • r′ = e
  • Arc Length = ∫ab √{r2 + ( [dr/(dθ)] )2} dθ
= ∫ab √{4e2 + e} dθ = ∫ab √{5e} dθ = √5 ∫ab e
What's the Arc Length for r = e from 0 to 1?
  • Apply previous equation with known bounds
  • √5 ∫01 edθ = [(√5 )/2][ e ]01
= [(√5 )/2](e2 − 1)

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

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 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