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

### Arclength

**Main formula:**

**Hints and tips:**

To remember this formula, it helps to recall that it comes from the distance formula between two points, which in turn comes from the Pythagorean Theorem.

Remember that you must integrate the square root formula above. A common mistake is to integrate the function itself, not the square root formula. Of course, this would give you the area under the curve and not the arclength.

A similar mistake is to mix this up with formula for surface area of revolution, which looks similar. Be careful which one you are asked for.

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 it!Many problems in Calculus II classes are “rigged” so that when you expand 1 +

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

*f*′(*x*)² be something of the form (*a − b*)² =*a*²− 2*ab + b*². Then the +1 changes it to*a*² + 2*ab + b*², which you can then factor as (*a + b*)².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.

### Arclength

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 Equation 0:04
- Why It Works
- Common Mistake
- Lecture Example 1 2:14
- Lecture Example 2 6:26
- Lecture Example 3 10:49
- 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: Arclength

*We are here to look at a couple more examples of arc length problems.*0000

*The first one I have set up here is y = x ^{4}/8 + 1/4x^{2}.*0005

*So remember with the arc length, you do not integrate it directly.*0012

*That is a common mistake that Calculus 2 students make is just integrating the function they are given.*0015

*When you want to find the arc length, you have got to find the derivative f'.*0019

*So here, the derivative of x ^{4} is 4x^{3}/8 is just x^{3}/2.*0026

*Now, if you think of x ^{2} in the denominator as x^{-2},*0037

*The derivative of that is x ^{-2} x^{-3}.*0042

*That is why I have a negative here - 2x ^{-3},*0048

*Then because of that 4 in the denominator, the two and the four cancel and we get a 2 in the denominator.*0052

*x ^{-3} gives you an x^{3} in the denominator.*0060

*So, that is what we get for the derivative.*0066

*We need to square that, f'(x) ^{2} is x^{6}/4 + 1/4x^{6}, squaring out both terms.*0069

*Then, minus 2ab, so minus 2 × 1/2 × 1/2 is 1/2.*0085

*Then the x ^{3} cancel, and this term is what we get by doing the -2ab of the a - b^{2} formula.*0094

*So, that is where that - 1/2 comes in.*0107

*1 + f'(x) ^{2} is x^{6}/4, still + 1/4x^{6}.*0109

*Now, - 1/2 + 1 gives us + 1/2.*0122

*That is almost the same function that we had before except that the minus has turned into a plus.*0130

*What started out as a-b ^{2}, this is now going to turn into a+b^{2}.*0137

*So this is x ^{3}/2 + 1/2x^{3} quantity squared.*0144

*Again, this is a very common pattern in arc length problems.*0154

*They often kind of rig up these examples so that the answer you get for 1 + f'(x) ^{2} factors nicely into a perfect square.*0158

*It is sometimes not so obvious but if you look for it, it is often there.*0168

*That makes it nice because 1 + f'(x) ^{2}, when we take the square root will just simplify back down to x^{3}/2 + 1/2x^{3}.*0173

*So, the arc length is the integral of that function.*0186

*So that is the integral from x=1 to x=2, getting those values from the stem of the problem.*0191

*Of x ^{3}/2 + 1/2x^{3} dx.*0199

*If we integrate x ^{3} that is x^{4}/4, so this is x^{4}/8.*0210

*Now, the integral of x ^{-3} is x^{-2}/-2, using the power rule.*0217

*So this is -, because the -2 in the denominator, 1/4x ^{2}, that is -2x^{-2}.*0229

*Then there is another 2 coming from the 2 there.*0240

*So, we want a value like this from x=1 to x=2.*0244

*So, what we get is 2 ^{4}/8, that is 16/8, so that is just 2.*0251

*Minus 1/4 × 2 ^{2}, so that is 1/16.*0257

*Minus, now if you plug in x=1, you get 1/8.*0266

*Plus 1/4, so if we think of everything in terms of 16ths there,*0270

*That is 32 - 1 - 2 + 4/16.*0277

*So that simplifies down to 33/16.*0287

*As our final answer for the arc length*0294

*So the point there is that we take the function we are given, find its derivative, run it through this Pythagorean formula,*0297

*And we got lucky here in that the perfect square and the square roots cancel.*0308

*Then integrate the thing you get there.*0315

*Let us do 1 more example here, we want to find the arc length of y=ln(x)/2-x ^{2}/4.*0000

*So we do not integrate the function directly, we look at its derivative.*0010

*f' is, well, the derivative of ln(x) is 1/x so this is 1/2x - the derivative of x ^{2} is 2x.*0016

*So the 2 and 4 cancel, so we get x/2.*0025

*So then we are going to find f'(x) ^{2},*0029

*So that is a ^{2}, so 1/4x^{2},*0035

*+ b ^{2} + x^{2}/4.*0041

*Now, - 2ab, so -2 × 1/2 × 1/2 is 1/2.*0046

*Then the x's cancel, this term right here came from -2ab.*0056

*In the formula for the difference of squares.*0064

*Again, we do 1 + f'(x) ^{2}, so that is 1/4x^{2},*0066

*+ x ^{2}/4, now - 1/2 + 1 gives us + 1/2.*0073

*Again, that converts a - 2ab formula into a + 2ab formula.*0083

*So this factors as a perfect square, 1/2x + x/2 quantity squared.*0093

*So if you look at the square root of 1 + f'(x) ^{2},*0103

*Remember that is what we have to integrate to find the arc length, we get 1/2x + x/2.*0110

*So we set up our arc length, and it is the integral from x=1 to x=e.*0116

*Of, 1/2x + x/2 dx.*0123

*Now that integrates to ln(x)/2,*0132

*X/2 integrates to x ^{2}/4 and we evaluate that whole thing from x=1 to x=e.*0136

*So we get ln(e)/2 + e ^{2}/4 - ln(1/2) - 1/4.*0148

*This simplifies a bit, of course ln(e) is just 1 and the ln(1) is 0.*0161

*So, we get 1/2 + e ^{2}/4 - 1/4.*0168

*So that is 1/4, because that is 1/2 - 1/4, + e ^{2}/4.*0176

*So just to reiterate there, the arc length formula, we take the function we are given,*0187

*We take its derivative, we run it through this Pythagorean Formula, and then you integrate what you get from that.*0196

*That is the end of our arc length lecture.*0204

*Today we are going to talk about arc length.*0000

*So there is one main formula for arc length that you need to know*0005

*And that is if you are trying to find the length of a curve y=f(x)*0008

*And you are trying to find the length from a to b.*0016

*The way I have drawn it kind of looks like we are looking for the area under the curve.*0020

*That is not what we are looking for today, not the area, but the length of that curve right there.*0023

*The way you work it out is this integral formula*0035

*The integral from x=a to x=b of the sqrt(1+x'(x)) that is the derivate of x ^{2} dx.*0039

*Where this formula comes from is it comes from the pythagorean theorem.*0048

*This is the length of the hypotenuse of a triangle where the base has side length 1, the height is f'(x).*0053

*So the length of the diagonal is 1 + f'(x) ^{2} and then take square root of all of that.*0066

*That is where the formula comes from.*0077

*It can help you remember that.*0079

*It is a very common mistake for Calculus 2 students to make to integrate the original function.*0081

*You will get an arc length problem, and then what you will try to do is integrate the original function.*0090

*That is very tempting.*0098

*There is nothing inherent in the problem that will tell you you are making a mistake there.*0101

*You may well go ahead an integrate that and get an answer.*0105

*That is a big no no in Calculus because what you actually computed there was the area, not the arc length.*0110

*You use this pythagorean formula with the sqrt and so you do not make this mistake of calculating the area instead.*0123

*Let us try this out with some examples.*0131

*The first example is the length of the curve x ^{2}/8 - ln(x) from x = 1 to x = e.*0133

*We are going to use the formula but first we are going to figure out f'(x).*0145

*Derivate of x ^{2}/8, derivative of x^{2} is just 2x so this would be x/4.*0149

*Minus the derivative of ln(x) is 1/x.*0156

*So if we square that, f'(x) ^{2}*0160

*It is x ^{2}/16 + 1/x^{2} - remember that (a+b)^{2} or (a-b)^{2} is a^{2} + b^{2} - 2ab.*0170

*So minus 2(x/4) × 1/x.*0181

*If we combine those, the x's cancel and we just get - 1/2.*0190

*Remember the formulas that we have to look at 1 + f'(x) ^{2}.*0196

*That is x ^{2}/16 + 1/x^{2} now we had minus 1/2 before.*0203

*If we add - 1/2 + 1, that is + 1/2.*0212

*The nice thing about that is it factors again as a perfect square.*0219

*That is (x/4 + 1/x) ^{2}.*0225

*Because, if you squared this out, you would get,*0230

*Well, x ^{2}/16 + 1/x^{2} + 2 × 1/x × x/4 would give you exactly 1/2 again.*0234

*This factors as a perfect square and is a very common feature of arc length problems.*0248

*We will see that again in some of our problems.*0252

*You will probably see that as you work through your homework problems that you are sort of rigged up to make this work out.*0255

*Remember that what we are supposed to do is integrate the sqrt(1+f'(x) ^{2}) and that just cancels off the prefect square we had.*0264

*That is x/4 + 1/x.*0276

*We are going to calculate the arc length, is the integral of that from x=1 to x=e.*0280

*x/4 + 1/x dx, and so x/4 integrates back to x ^{2}/8 and 1/x integrates back to ln(x).*0288

*We evaluate that whole thing from x=e down to x=1.*0307

*We get e ^{2}/8 + ln(e) - 1/8 - ln(1).*0316

*Of course ln(e) is just e, ln(1) is just 0.*0327

*So, we get e ^{2}/8 + 1 - 1/8, so that is + 7/8.*0337

*That is our answer for the length of that curve.*0355

*Just to review what we have to do for this problem,*0360

*We are given a function here, and we do not integrate it directly.*0364

*First we calculate its derivative.*0368

*Then we plug it into this formula, sqrt(f'(x ^{2}))*0374

*Then we integrate that and that gives us our answer.*0381

*Let us try another one.*0388

*This time we have to find the length of the curve y = 4x ^{3/2} + 1.*0390

*Again, we have to look at f'(x), so we have to take the derivative of that.*0397

*The derivative of x ^{3/2} is 3/2 x^{1/2}*0401

*So this is 4 × 3/2x ^{1/2}, derivative of 1 is 0.*0406

*This just simplifies into 6x ^{1/2}.*0416

*f'(x) ^{2} would be 36, (x^{1/2})^{2} would just be x,*0421

*if we add 1 to both sides we get 36x + 1.*0430

*What we are really integrating is just the sqrt(f'(x) ^{2})*0435

*Which would be the sqrt(36x +1).*0445

*We are going to calculate the integral from x=0 to x=1 of the sqrt of 36x + 1 dx.*0450

*The way we want to do this integral, and this is common if you have the square root of something linear,*0465

*This is a common technique, we are going to use u =36x + 1.*0470

*Remember that whenever you use a substitution, you also have to work out du.*0476

*du is 36 dx, and so dx is du/36.*0480

*This turns into the integral, I am going to write x=0 to x=1, so that we remember those variables refer to the x values.*0492

*The square root of u and then dx is du/36, I will just put the 1/36 on the outside.*0502

*So I get 1/36, now this square root of u is the same as u ^{1/2}*0509

*So the integral of u ^{1/2} is u^{3/2}, and then we have to divide by 3/2.*0523

*Then we evaluate that from x=0 to x=1, but of course we cannot plug that in right away until we convert back to x's.*0530

*This is 1/36, 3/2 in the denominator, I am going to flip that up to be 2/3.*0541

*Then we have u ^{3/2} but u was 36x + 1.*0548

*36x+1, so (36x+1) ^{3/2}, all that evaluated from x=0 to x=1.*0555

*Now, 1/36 × 2/3 the, 2 and the 36 can cancel.*0571

*That is 1/18, and that is 1/54.*0578

*Now, 36x + 1, if we plug in x=1, that gives us 37 ^{3/2} - x=0, just gives us 1^{3/2}.*0582

*This reduces down to 1/54 × 37 ^{3/2} is the same 37 times the square root of 37 - 1^{3/2} is just 1.*0604

*That is as far as we can simplify that one.*0620

*Again, the critical step there was looking at the function, y = 4x ^{3/2} + 1.*0624

*Remembering that for arc length you do not integrate it directly.*0632

*You take its derivative and then you run it through the square root formula.*0637

*Then you integrate the thing that you get there.*0641

*From the on, it is really an integration problem.*0646

*OK, we are going to look at another example problem.*0652

*y = ln(cos(x)) from x=0 to x=pi/3.*0655

*Again, we wanted to find the derivative so f'(x).*0660

*The derivative of ln is 1/x, so the derivative of ln(cos) is 1/cos(x) × the derivative of cos by the chain rule.*0667

*Well the derivative of cos, by the chain rule, is the -sin(x).*0676

*This simplifies down to - tan(x).*0680

*f'(x) ^{2} is tan^{2}(x)*0685

*So, (1 + f'(x)) ^{2} = 1 + tan^{2}(x).*0691

*But, if you remember the pythagorean identity, that is sec ^{2}(x)*0702

*So, the square root of (1+f'(x)) ^{2} is just sec(x).*0707

*That is what we have to integrate.*0715

*We set up the integral from x=0 to x=pi/3 of sec(x) dx.*0716

*Now, sec(x) is one of those integrals that we learned how to solve in our section on trigonometric integrals.*0728

*It was one of those tricky ones that you just kind of have to remember.*0736

*The trick was to multiply top and bottom by sec(x) + tan(x).*0739

*That is probably worth just remembering because it is not something that you are that likely to figure out in a pinch.*0746

*When you do that, the integral is the ln(sec(x)) + tan(x)).*0756

*That is what we need to evaluate from x=0 to x=pi/3.*0765

*Now, that turns into the ln(sec(x) = 1/cos(x) + tan(x).*0775

*We are evaluating that from x=0 to x=pi/3*0789

*The cos(pi/3) is 1/2 tan(pi/3) is the sin(pi/3)/cos(pi/3).*0794

*That is sqrt(3)/2 divided by 1/2, so that is sqrt(3).*0812

*That is what we got by evaluating x = pi/3.*0815

*Minus 1/cos(0), the cos(0) is 1, + tan(0) is 0.*0822

*This right hand term gives us the ln(1) which goes away to 0.*0837

*This left hand term gives us the ln of 1/1/2 is 2 + s1rt(3).*0843

*I have absolute values signs here but I do not really need them because 2 + sqrt(3) is going to be positive.*0850

*So, that is our answer.*0856

*Ln(2 + sqrt(3)).*0858

*Again, what we did there was look at the function y = ln(cos(x)).*0863

*We did not integrate directly, instead we took its derivative and plugged it into this pythagorean formula and we integrated that.*0868

*From there on, it was an integration problem.*0874

1 answer

Last reply by: Dr. William Murray

Fri Oct 4, 2013 6:53 PM

Post by Narin Gopaul on September 29, 2013

I am sorry i saw the mistake I maked

1 answer

Last reply by: Dr. William Murray

Fri Oct 4, 2013 6:52 PM

Post by Narin Gopaul on September 29, 2013

I think you made a mistake in the first example by saying that u add 1/2 to the function i thought you would -1/2

1 answer

Last reply by: Dr. William Murray

Fri Dec 7, 2012 4:48 PM

Post by Riley Argue on June 3, 2012

Excellent lecture.

You elegantly and simply explained this, thank you.

3 answers

Last reply by: Dr. William Murray

Fri Dec 7, 2012 4:47 PM

Post by Alphonse Mbu on March 7, 2012

why do you use a^2+b^2-2ab. i dont understand what makes that valid