For more information, please see full course syllabus of College Calculus: Level II

For more information, please see full course syllabus of College Calculus: Level II

## Discussion

## Study Guides

## Practice Questions

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## Table of Contents

## Transcription

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### Parametric Curves

**Main formula:**

Slope *m* =

Arclength =

**Hints and tips:**

You can remember the formula for the slope of the tangent line by thinking that symbolically, the

*dt*’s cancel, leaving you with*dy*⁄*dx*.To find the equation of the tangent line, you also need a point. Use the given value of

*t*into*x*(*t*) and*y*(*t*) to find it. Then you can use the point-slope formula from high school algebra (*y − y*_{0}=*m*(*x − x*_{0})) to find the equation.Sometimes you aren’t given a value of

*t*, but the coordinates (*x*,*y*) instead. Then you must find which value of*t*gives you the correct (*x*(*t*),*y*(*t*)). Make sure you check that both*x*and*y*are correct for your value of*t*.You can remember the arclength formula by recalling that it is derived from the distance formula between two points, which in turn comes from the Pythagorean Theorem.

Don’t make the common algebraic mistake of thinking that reduces to

*a*+*b*! This is extremely wrong, and your teacher will likely be merciless if you do itMany problems in Calculus II classes are “rigged” so that when you expand

*x*′(*t*)² +*y*′(*t*)² , it becomes a perfect square that cancels nicely with the square root.Often this perfect square is achieved by making one of

*x*′(*t*)² and*y*′(*t*)² be something of the form (*a − b*)² =*a*² − 2*ab + b*². Then the other one changes it to*a*² + 2*ab + b*², which you can then factor as (*a + b*)².Another common technique in arclength problems is to make a

*u*-substitution for whatever is under the square root sign. Then (hopefully) you can manipulate the expression outside the square root into being the*du*. However, you might have to do several steps of algebraic manipulation, pulling factors in or out of the square root sign, before this works.You may sometimes be able to use symmetry to find the arclength of part of a curve and then multiply by an appropriate factor to get the total arclength. This can be especially helpful if you just want to examine part of the curve where all the quantities involved are positive.

When it’s feasible, check that your answer makes sense. Unlike area integrals, which can be negative if a curve goes below the

*x*-axis, arclength should always be positive! You might also be able to check that the curve looks about as long as your answer.

### Parametric Curves

- Make a table
t 0 2 π [(3π)/2] 2π x(t) 0 1 0 − 1 0 y(t) 0 1.25 1.77 2.17 2.51

y(t) = t

^{2}

x(t) = t + 2

What is its Cartesian equation?

- Solve for t with x
- x = t + 2
- t = x − 2

y(t) = t

^{2}

y = (x − 2)

^{2}

y(t) = t − 9

x(t) = √t

for t ≥ 0

What is its Cartesian equation?

- Solve for t, we start with x
- x = √t
- t = x
^{2}

y(t) = t − 9

y = x

^{2}− 9

x(t) = [cos(t)/3]

y(t) = [sin(t)/4]

for t ≥ 0

- Isolate x and y
- x = [cos(t)/3]
- 3x = cos(t)
- y = [sin(t)/4]
- 4y = sin(t)

cos

^{2}(t) + sin

^{2}(t) = 1

(3x)

^{2}+ (4y)

^{2}= 1

9x

^{2}+ 16y

^{2}= 1

x(t) = √2 sin(t)

y(t) = 2cos(t) + 3

for t ≥ 0

- Isolate x
- x = √2 sin(t)
- [x/(√2 )] = sin(t)
- Isolate y
- y = 2cos(t) + 3
- y − 3 = 2cos(t)
- [(y − 3)/2] = cos(t)

^{2}(t) + cos

^{2}(t) = 1

([x/(√2 )])

^{2}+ ([(y − 3)/2])

^{2}= 1

[(x

^{2})/2] + [((y − 3)

^{2})/4] = 1

x(t) = t + 5

y(t) = t

^{3}+ 5

- Isolate t with x
- x = t + 5
- x − 5 = t

y = x

^{3}− 15x

^{2}+ 75x − 125 + 125

y = x

^{3}− 15x

^{2}+ 75x

x(t) = 1 + t

y(t) = t

^{2}− 9

- Isolate t with x
- x = 1 + t
- − t = 1 − x
- t = x − 1

y = (x − 1)

^{2}− 9

y = x

^{2}− 2x + 1 − 9

y = x

^{2}− 2x − 8

x(t) = t

^{2}− 3

y(t) = 1 − t

for t − 3

- Isolate t with x
- x = t
^{2}− 3 - x + 3 = t
^{2} - √{x + 3} = t
- Substitute into y(t)
- y = 1 − √{x + 3}
- Graph the equation using shifts
- y = 1 − √{x + 3}

x(t) = arccos([t/4]) − [(π)/2]

y(t) = [t/8] − 3 for − 2π ≤ t ≤ 2π

- Isolate t with x
- x = arccos([t/4]) − [(π)/2]
- x + [(π)/2] = arccos([t/4])
- cos(x + [(π)/2]) = [t/4]
- 4cos(x + [(π)/2]) = t
- Observe half - angle trig identity
- cos(u + [(π)/2]) = − sinu
- 4cos(x + [(π)/2]) = t
- − 4sinx = t
- Substitute into y(t)
- y = [( − 4sin(x))/8] − 3
- y = − [sinx/2] − 3
- Graph using shifts, reflections, and stretching
- y = − [sinx/2] − 3
- The graph y = sinx has been shifted 3 units down to y = − 3, and reflected. It also has been compressed by a factor of 2.
- Remember to graph within the bounds − 2π ≤ t ≤ 2π.

^{[3/2]}from − 10 to 17

- Find y′
- y′ = [3/2](x − 8)
^{1/2}dx - Use Arc Length Formula
- Arc Length = ∫
_{a}^{b}√{1 + ( [dy/dx] )^{2}} dx - = ∫
_{10}^{17}√{1 + ( [3/2](x − 8)^{1/2})^{2}} d x - = ∫
_{10}^{17}√{1 + ( [9/4](x − 8) )} dx - = ∫
_{10}^{17}√{1 + [9/4]x − 18} dx - = ∫
_{10}^{17}√{[9/4]x − 17} dx - Use u substitution
- u = [9/4]x − 17
- du = [9/4]dx
- ∫
_{10}^{17}√{[9/4]x − 17} dx = ∫_{10}^{17}[4/9]√u du - = [4/9]∫
_{10}^{17}√u du - = [4/9][ [2/3]u
^{[3/2]}]_{10}^{17} - = [4/9][ [2/3]( [9/4]x − 17 )
^{[3/2]}]_{10}^{17}

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

[dy/dx] = [secxtanx/secx] = tanx dx

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

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 )

- x = e
^{t}cost, y = e^{t}sint - Find the deratives using product rule
- x′ = e
^{t}cost − e^{t}sint - y′ = e
^{t}sint + e^{t}cost - Apply the parametric Arc Length Formula
- Arc Length = ∫
_{a}^{b}√{( [dx/dt] )^{2}+ ( [dy/dt] )^{2}} dt - = ∫
_{0}^{3}√{( e^{t}cost − e^{t}sint )^{2}+ ( e^{t}sint + e^{t}cost )^{2}} dt - = ∫
_{0}^{3}√{2e^{2t}(cos^{2}t + sin^{2}t)} dt - = ∫
_{0}^{3}√{2e^{2t}} dt - = √2 ∫
_{0}^{3}e^{t}dt - = √2 [ e
^{t}]_{0}^{3}

^{3}− e

^{0}] = √2 [ e

^{3}− 1 ]

- 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]}√{cot^{2}t + 1} dt - Use Pythagoream Identity
- ∫
_{1}^{[(3π)/4]}√{cot^{2}x + 1} dt = ∫_{1}^{[(3π)/4]}√{csc^{2}x} dt - = ∫
_{1}^{[(3π)/4]}csctdt - = [ − ln|csct + cott| ]
_{1}^{[(3π)/4]}

- Find [dr/(dθ)]
- r′ = − 2sinθ
- Apply the polar Arc Length Formula
- Arc Length = ∫
_{a}^{b}√{r^{2}+ ( [dr/(dθ)] )^{2}} dθ - = ∫
_{0}^{5π}√{( 2cosθ )^{2}+ ( − 2sinθ )^{2}} dθ - = ∫
_{0}^{5π}√{4cos^{2}θ+ 4sin^{2}θ} dθ - = ∫
_{0}^{5π}√4 √{cos^{2}θ+ sin^{2}θ} dθ - = 2∫
_{0}^{5π}√{cos^{2}θ+ sin^{2}θ} dθ - = 2∫
_{0}^{5π}√1 dθ - = 2[ θ]
_{0}^{5π}

- Convert using Polar definitions
- x = rcosθ
- = cosθ(cosθ)
- = cos
^{2}θ - 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

^{2θ}

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

_{a}

^{b}√{4e

^{4θ}

^{2}+ e

^{4θ}} dθ = ∫

_{a}

^{b}√{5e

^{4θ}} dθ = √5 ∫

_{a}

^{b}e

^{2θ}dθ

^{2θ}from 0 to 1?

- Apply previous equation with known bounds
- √5 ∫
_{0}^{1}e^{2θ}dθ = [(√5 )/2][ e^{2θ}]_{0}^{1}

^{2}− 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

### Parametric 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
- Important Equations 0:05
- Slope of Tangent Line
- Arc length
- Lecture Example 1 1:40
- Lecture Example 2 4:23
- Lecture Example 3 8:38
- Additional Example 4
- Additional Example 5

### College Calculus 2 Online Course

I. Advanced Integration Techniques | ||
---|---|---|

Integration by Parts | 24:52 | |

Integration of Trigonometric Functions | 25:30 | |

Trigonometric Substitutions | 30:09 | |

Partial Fractions | 41:22 | |

Integration Tables | 20:00 | |

Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule | 22:36 | |

Simpson's Rule | 21:08 | |

Improper Integration | 44:18 | |

II. Applications of Integrals, part 2 | ||

Arclength | 23:20 | |

Surface Area of Revolution | 28:53 | |

Hydrostatic Pressure | 24:37 | |

Center of Mass | 25:39 | |

III. Parametric Functions | ||

Parametric Curves | 22:26 | |

Polar Coordinates | 30:59 | |

IV. Sequences and Series | ||

Sequences | 31:13 | |

Series | 31:46 | |

Integral Test | 23:26 | |

Comparison Test | 22:44 | |

Alternating Series | 25:26 | |

Ratio Test and Root Test | 33:27 | |

Power Series | 38:36 | |

V. Taylor and Maclaurin Series | ||

Taylor Series and Maclaurin Series | 30:18 | |

Taylor Polynomial Applications | 50:50 |

### Transcription: Parametric Curves

*Welcome back and we are trying some more examples of arc lengths.*0000

*And other problems involving parametric equations.*0005

*We have here the length of a curve given by x(t) = cos ^{2}(t) and y(t) = sin^{2}(t) as t goes from 0 to π/2.*0010

*Remember our arc length formula is x' ^{2} + y'^{2}, take the square root of that and integrate it.*0023

*Let us calculate x'.*0029

*X is cos ^{2}(t), so that would be 2 × cos(t) × -sin(t) using the chain rule there,*0034

*The - sign comes from the derivative of cosine.*0044

*y'(t) is 2 × sin(t) × derivative of sin which is cos(t).*0049

*If we square each one of those, x'(t) ^{2} is 4cos^{2}(t) sin^{2}(t).*0059

*y'(t) ^{2} is 4sin^{2}(t) cos^{2}(t).*0070

*If we add those up, x' ^{2} + y'^{2},*0082

*Well those are the same thing.*0091

*We just get the sqrt(8sin ^{2}(t)cos^{2}(t)).*0094

*The sqrt(8) is 2×sqrt(2).*0102

*The square root of sin ^{2} and cos^{2}, since sin and cos are positive when t is between 0 and π/2,*0105

*This is sin(t) × cos(t).*0117

*So, that is what we want to integrate.*0122

*The integral from t=0, to t=π/2.*0125

*I will write the 2×sqrt(2) on the outside.*0130

*sin(t) cos(t) dt.*0138

*This integral is not too bad.*0143

*We can use the substitution u = sin(t).*0145

*The reason that works so well is because du = cos(t) dt.*0149

*So, what we have here is the integral of u du.*0156

*That gives us u ^{2}/2.*0160

*Now to convert things back into t.*0169

*2×sqrt(2) × sin ^{2}(t)/2.*0172

*Evaluated from t=0 to t=π/2.*0180

*Those 2's cancel each other so we get sqrt(2) × sin(π/2) ^{2},*0188

*That is just one.*0195

*Minus sin(0) ^{2}, which is just 0.*0197

*So our arc length is just sqrt(2).*0200

*Again there, the calculus worked nicely.*0206

*What we did was find x' and y' and plugged them into this Pythagorean Formula.*0208

*Then we integrated to get the answer.*0215

*There is actually another way to see how this problem works.*0218

*We can see through this problem geometrically.*0220

*If you actually try graph the path that these equations are describing,*0226

*Notice that cos ^{2}(t) + sin^{2}(t) = 1.*0228

*So, cos ^{2}(t) = x, sin^{2}(t) = y.*0240

*So this path is actually taking place on the line y+x = 1.*0245

*So there is that line,*0255

*And if you plug in t=0, then x=1 and y=0.*0257

*There is t=0.*0264

*At the point (1,0).*0266

*If you plug in t - π/2, then y = 1 and x=0.*0269

*So, there is t=π/2 at the point (0,1).*0277

*So, what this path is really doing is just following a straight line from (1,0) to (0,1).*0286

*Of course the length of that line is sqrt(2).*0293

*That is kind of a geometric check on the calculus we just did.*0297

*Let us try one more example.*0000

*We want to find the length of the curve given by x(t) = 7 + 2t,*0002

*And y(t) = e ^{t} + e^{-t}, where t goes from 0 to 1.*0006

*Again we want to find x' and y', square each one, add them up and take the square root.*0013

*x'(t) = 2, y'(t), y = e ^{t} + e^{-t}, so the derivative of e^{t} is just e^{t}.*0020

*The derivative of e ^{-t} is e^{-t} × the derivative of -t, which is -1.*0037

*So we get -e ^{-t}.*0047

*So x'(t) ^{2} + y'(t)^{2}.*0050

*Well, 2 ^{2} is just 4.*0058

*Now, if we square y'(t), then we are going to follow the formula (a + b) ^{2},*0060

*Is a ^{2} + 2ab + b^{2}.*0074

*Here a is e ^{t}, so (e^{t})^{2} is e^{2t}, b is -e^{-t},*0080

*So this is -2ab, well ab is e ^{t} × e^{-t},*0090

*Then + b ^{2} is e^{-t}, so that is e^{-2t},*0102

*But look at this e ^{t} × e^{-t}, is e^{0} which is 1.*0111

*This is just -2.*0116

*We also have a 4 here, so what we get is e ^{2t} - 2 + 4, is just + 2.*0120

*+ e ^{-2t}.*0131

*The clever thing to do here is to write that 2 as 2e ^{t} × e^{-t} again + e^{-2t}.*0134

*This is (e ^{t} + e^{-t})^{2}.*0145

*The square root of all that, x'(t) ^{2} + y'(t)^{2}.*0155

*The square root cancels that for a fixed square,*0164

*So we get e ^{t} + e^{-t}.*0168

*That is what we want to integrate to find the arc length.*0173

*We take the integral from t=0 to t=1.*0175

*We are getting those bounds from the original problem of e ^{t} + e^{-t} dt.*0181

*That integral is not too bad.*0192

*The integral of e ^{t}, is just e^{t} itself.*0194

*The integral of e ^{-t}, is e^{-t}/the derivative of -t, which is -1.*0197

*That is the same as multiplying by -1.*0207

*So this is e ^{t} - e^{-t}.*0210

*Then we want to evaluate that from t=0 to t=1.*0213

*That is e ^{1} - e^{-1} - e^{0} + e^{0}.*0221

*The e ^{0}'s are both 1 so those cancel each other out.*0232

*This is e - e ^{-1}.*0235

*I will write this as 1/e, and that is our answer for the arc length.*0240

*Again, what we did there, we looked at the x and y that we were given,*0246

*We took the derivative of each one, squared them, added them up, took their square root.*0252

*And integrated to get our answer for the arc length.*0257

*Thanks for watching, this has been educator.com.*0261

*Hi this is educator.com and we are here to talk about parametric curves.*0000

*The idea about parametric curves is that you are given the equation x(t) and y(t) and those define how a point is moving around in the plane.*0007

*The x(t) gives you the x coordinate at a particular time and the y(t) gives you the y coordinate at a particular time.*0019

*There are basically two calculus problems associated with parametric equations.*0026

*One is to find the tangent line to occur at a particular point.*0030

*The way you are going to do that is we will fine the slope of the tangent line by looking at d(y) dt.*0039

*So, just looking at the derivative of the y equation, and dividing by d(x) d(t), which is the derivative of the x equation.*0048

*That will give us the slope of the tangent line and we will also know one point on the tangent line.*0052

*We can use the point slope equation to find the slope of the tangent line.*0059

*The other equation that you use with parametric equations a lot is*0063

*You find the length of a curve and essentially the formula from that comes from the Pythagorean distance formula.*0075

*You look at x'(t ^{2} + y'(t^{2}) and then you find the square root of that.*0076

*That is a unit of arc length representing the length travelled in a very small amount of time.*0084

*Then you integrate that from your starting time to your ending time and that formula represents the total length of the curve.*0089

*Let us try this out with some examples*0097

*The first examples is the equations are x(t) = t+1, y(t) = t ^{2}.*0100

*Maybe I will just graph a couple of points there.*0109

*If t = 0 then x = 1 and y = 0.*0111

*If t = 1 then x = 2 and y = 1.*0118

*If t = 2 then x = 3 and y = 4.*0125

*This point is travelling along a parabolic path here.*0133

*What we are asked to do is find the tangent line at t=1.*0138

*At t=1, remember, x = 2 and y = 1.*0143

*We are trying to find the tangent line at that point right there.*0149

*We need to find the slope but our slope is d(y) dt/d(x) dt.*0156

*Now dy dt, since y = t ^{2} is 2t.*0170

*Dx dt, since x = t+1 is just 1.*0174

*That is 2t, and when we plug in t = 1, that gives us the slope of 2.*0180

*Now we have the slope and we have a point and it is just an algebra problem to find the equation of a line.*0189

*We use the point slope formula, y - y _{0}, which is 1 here, is equal to the slope × x - x_{0} which is 2.*0195

*This is 2x - 4.*0207

*We get the equation of the tangent line, y = 2x-3.*0213

*To recap there, what we are trying to do is to find the tangent line to a curve that is defined by parametric equations.*0220

*What we did was we plugged in the time value we were given, t =1 to get a point on the tangent line.*0230

*Then we used our equation for the slope dy dt/dx dt.*0240

*We figured those out using the equations for x and y that we were given.*0245

*We plugged in the same value of t and we got our slope and then we had a point on our slope,*0250

*And we could use the old point-slope formula to find the equation for our tangent line.*0259

*Let us find another tangent line.*0263

*This time the curve is x(t) = cos(t) y(t) = sin(t).*0265

*You will hopefully recognize that as the equations for a circle because cos ^{2}(t) + sin^{2}(t) = 1.*0272

*Those are the equations that define a point moving around in a circle.*0282

*We want to find the tangent line at the point, (sqrt(3/2), 1/2).*0288

*That is about right there on the circle*0294

*The difference between this one and the previous one is we have not been given a t value.*0297

*We have to figure out what value of t gives us the point (sqrt(3/2), 1/2).*0302

*What value of t if you plug it in to cos(t) sin(t) gives us the sqrt(3/2)?*0308

*The answer is pi/6 because the cos(pi/6) is sqrt(3/2) and the sin(pi/6) is 1/2.*0318

*So, we know that t is pi/6.*0328

*Again, to find the slope, we use dy dt/dx dt.*0330

*Now y = sin(t) so the derivative there is cos(t).*0340

*X = cos(t), the derivative of that is -sin(t).*0345

*Then we plug in the value t = pi/6 to get a number for the slope so the cos(pi/6) = sqrt(3/2)*0354

*The sin(pi/6) = 1/2, and we still have our negative there.*0364

*The two's cancel so we get our slope is -sqrt(3).*0370

*That gives us the slope and we also have a point, so I will plug those into the point slope formula.*0376

*y - 1/2 = -sqrt(3) x - sqrt(3/2).*0380

*Those values are coming from plugging the points in there.*0394

*We can simplify this a bit.*0396

*This is = sqrt(3)x.*0398

*The minuses cancel each other out plus sqrt(3) × sqrt(3) = 3.*0400

*Then we can bring this half over to the other side so we get y = -sqrt(3)x + 3/2 + 1/2, is just 2.*0410

*We get our tangent line to be y = -sqrt(3)x + 2.*0429

*The way that this problem was different from the previous one was that we were not given a value of t.*0435

*We had to look at the point we were given and we had to figure out what the value of t should be.*0440

*From then on we used the same formulas to find out the slope and the equation of the tangent line.*0450

*By the way, this problem you can also check the answer geometrically if you draw that tangent line.*0454

*It forms a 30-60-90 triangle with the x and y axis.*0464

*That is 30 degrees and that is 60 degrees.*0470

*We know what those angles are because the tangent line is perpendicular to the radius of the circle there,*0476

*We also know that the radius of the circle is 1 so we have a smaller 30-60-90 triangle in there.*0484

*With a short side of length 1, so the long side has length 2.*0501

*That confirms that the y intercept of this tangent line = 2.*0506

*That is a little check on our work using some trigonometry and no calculus at all*0511

*Let us try an example of arc length now.*0519

*We are given the curve x(t) = 6 - 2t ^{3}, y(t) = 8 + 3t^{2}.*0520

*We want to find the length of that curve from t = 0 to t = 2.*0530

*Remember our arc length formula says you want to take the integral of sqrt of x'(t) ^{2} + y'(t)^{2} dt.*0536

*x'(t), the derivative of x if x is 6 - 2t ^{3},*0549

*You take its derivative and the 6 just goes away, so the derivative is -6t ^{2}.*0556

*y'(t), if y' is 8 + 3t ^{2}, again the 8 does not have any effect.*0568

*The 3t ^{2} you take the derivative and you get 6t.*0575

*What we want then is sqrt(x') ^{2}, well that is (-6t)^{2}, so that is 36t^{4}*0584

*+ y' ^{2} is 36t^{2}.*0596

*That can simplify because we can pull a 36 out and just get 6,*0600

*we can pull a t ^{2} out and get t.*0610

*Under the radical, what we have left is just t ^{2} + 1.*0615

*The arc length = integral from t=0 to t=2 of 6t × sqrt(t ^{2}+1) dt.*0620

*Now, this integral is not too bad because what we can do is make a little substitution.*0645

*u = t ^{2} + 1.*0650

*Then, du = 2t, dt.*0654

*The reason that works so nicely is that we already have the t and the dt, so we basically have du.*0658

*In fact, 6t dt, is just 3 du.*0666

*Then, we still have the integral from t = 0 to t = 2 of now the sqrt(u) du.*0674

*You can think of the square root as u ^{1/2}.*0683

*To integrate that we get u ^{3/2}/3/2 which is the same as multiplying by 2/3.*0687

*Then we still have that 3 on the outside.*0698

*This is evaluated from t=0 to t=2.*0701

*I have to convert the u's back into t's*0706

*These 3's cancel, so we get 2 × u was (t ^{2} + 1)^{3/2} evaluated from t=0 to t=2.*0709

*This gives us 2 now if we plug in t=2, we get 2 ^{2} + 1, so that is 5^{3/2}.*0727

*- 0 ^{2} + 1, so that is 1^{3/2}.*0737

*We can simplify that to a little bit.*0744

*5 ^{3/2} is the same as 5 × 5^{1/2}.*0746

*1 ^{3/2} is just 1.*0749

*We get our final answer for the arc length there.*0758

*What made this problem work is having this formula that came from the pythagorean theorem.*0762

*We just take the x(t) and the y(t) that we are given and we plug them into this formula which involves computing a couple derivatives,*0768

*Simplifying a little bit if we can, then we integrate and we get out answer.*0775

*We will try a couple more examples later.*0782

1 answer

Last reply by: Dr. William Murray

Thu Nov 20, 2014 3:20 PM

Post by David Llewellyn on November 19, 2014

How do the table and the graph in the answer to the practice question relate to the actual question? Shouldn't the graph show the relationship between x and y and not the two relationships between x (albeit a different function than in the question) and y and t?

1 answer

Last reply by: Dr. William Murray

Fri Aug 16, 2013 6:21 PM

Post by Timothy Davis on August 14, 2013

Hi Dr. Murray,

Thanks for answering my last question. I thoroughly enjoy your videos. You mentioned that once we find the tangent line, we can use the point slope formula to find the equation of the tangent line. My question is, if we have the slope of the tangent line already defined by dy/dt / dx /dt, why do we need to find the equation of the tangent line? What is the difference between the slope of the tangent line and the equation of the tangent line? You go over this around 1:15 into your lecture.

1 answer

Last reply by: Dr. William Murray

Wed Apr 3, 2013 11:52 AM

Post by omatseye ugen on March 12, 2012

good job . but i thought this video will involve mostly multivar. calculus

1 answer

Last reply by: Dr. William Murray

Wed Apr 3, 2013 11:48 AM

Post by Allison Walsh on April 6, 2011

Awesome!

6 answers

Last reply by: Dr. William Murray

Wed Apr 3, 2013 11:46 AM

Post by Wen Geng on March 7, 2011

Dr. Murray's lectures are clear, quick and concise. Very good, unlike the Chemistry ones.