WEBVTT mathematics/calculus-ii/murray
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Today we are going to talk about arc length.
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So there is one main formula for arc length that you need to know
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And that is if you are trying to find the length of a curve y=f(x)
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And you are trying to find the length from a to b.
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The way I have drawn it kind of looks like we are looking for the area under the curve.
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That is not what we are looking for today, not the area, but the length of that curve right there.
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The way you work it out is this integral formula
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The integral from x=a to x=b of the sqrt(1+x'(x)) that is the derivate of x² dx.
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Where this formula comes from is it comes from the pythagorean theorem.
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This is the length of the hypotenuse of a triangle where the base has side length 1, the height is f'(x).
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So the length of the diagonal is 1 + f'(x)² and then take square root of all of that.
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That is where the formula comes from.
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It can help you remember that.
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It is a very common mistake for Calculus 2 students to make to integrate the original function.
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You will get an arc length problem, and then what you will try to do is integrate the original function.
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That is very tempting.
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There is nothing inherent in the problem that will tell you you are making a mistake there.
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You may well go ahead an integrate that and get an answer.
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That is a big no no in Calculus because what you actually computed there was the area, not the arc length.
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You use this pythagorean formula with the sqrt and so you do not make this mistake of calculating the area instead.
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Let us try this out with some examples.
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The first example is the length of the curve x²/8 - ln(x) from x = 1 to x = e.
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We are going to use the formula but first we are going to figure out f'(x).
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Derivate of x²/8, derivative of x² is just 2x so this would be x/4.
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Minus the derivative of ln(x) is 1/x.
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So if we square that, f'(x)²
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It is x²/16 + 1/x² - remember that (a+b)² or (a-b)² is a² + b² - 2ab.
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So minus 2(x/4) × 1/x.
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If we combine those, the x's cancel and we just get - 1/2.
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Remember the formulas that we have to look at 1 + f'(x)².
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That is x²/16 + 1/x² now we had minus 1/2 before.
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If we add - 1/2 + 1, that is + 1/2.
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The nice thing about that is it factors again as a perfect square.
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That is (x/4 + 1/x)².
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Because, if you squared this out, you would get,
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Well, x²/16 + 1/x² + 2 × 1/x × x/4 would give you exactly 1/2 again.
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This factors as a perfect square and is a very common feature of arc length problems.
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We will see that again in some of our problems.
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You will probably see that as you work through your homework problems that you are sort of rigged up to make this work out.
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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.
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That is x/4 + 1/x.
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We are going to calculate the arc length, is the integral of that from x=1 to x=e.
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x/4 + 1/x dx, and so x/4 integrates back to x²/8 and 1/x integrates back to ln(x).
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We evaluate that whole thing from x=e down to x=1.
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We get e²/8 + ln(e) - 1/8 - ln(1).
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Of course ln(e) is just e, ln(1) is just 0.
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So, we get e²/8 + 1 - 1/8, so that is + 7/8.
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That is our answer for the length of that curve.
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Just to review what we have to do for this problem,
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We are given a function here, and we do not integrate it directly.
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First we calculate its derivative.
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Then we plug it into this formula, sqrt(f'(x²))
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Then we integrate that and that gives us our answer.
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Let us try another one.
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This time we have to find the length of the curve y = 4x^3/2 + 1.
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Again, we have to look at f'(x), so we have to take the derivative of that.
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The derivative of x^3/2 is 3/2 x^1/2
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So this is 4 × 3/2x^1/2, derivative of 1 is 0.
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This just simplifies into 6x^1/2.
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f'(x)² would be 36, (x^1/2)² would just be x,
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if we add 1 to both sides we get 36x + 1.
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What we are really integrating is just the sqrt(f'(x)²)
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Which would be the sqrt(36x +1).
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We are going to calculate the integral from x=0 to x=1 of the sqrt of 36x + 1 dx.
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The way we want to do this integral, and this is common if you have the square root of something linear,
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This is a common technique, we are going to use u =36x + 1.
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Remember that whenever you use a substitution, you also have to work out du.
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du is 36 dx, and so dx is du/36.
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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.
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The square root of u and then dx is du/36, I will just put the 1/36 on the outside.
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So I get 1/36, now this square root of u is the same as u^1/2
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So the integral of u^1/2 is u^3/2, and then we have to divide by 3/2.
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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.
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This is 1/36, 3/2 in the denominator, I am going to flip that up to be 2/3.
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Then we have u^3/2 but u was 36x + 1.
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36x+1, so (36x+1)^3/2, all that evaluated from x=0 to x=1.
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Now, 1/36 × 2/3 the, 2 and the 36 can cancel.
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That is 1/18, and that is 1/54.
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Now, 36x + 1, if we plug in x=1, that gives us 37^3/2 - x=0, just gives us 1^3/2.
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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.
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That is as far as we can simplify that one.
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Again, the critical step there was looking at the function, y = 4x^3/2 + 1.
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Remembering that for arc length you do not integrate it directly.
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You take its derivative and then you run it through the square root formula.
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Then you integrate the thing that you get there.
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From the on, it is really an integration problem.
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OK, we are going to look at another example problem.
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y = ln(cos(x)) from x=0 to x=π/3.
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Again, we wanted to find the derivative so f'(x).
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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.
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Well the derivative of cos, by the chain rule, is the -sin(x).
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This simplifies down to - tan(x).
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f'(x)² is tan²(x)
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So, (1 + f'(x))² = 1 + tan²(x).
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But, if you remember the pythagorean identity, that is sec²(x)
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So, the square root of (1+f'(x))² is just sec(x).
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That is what we have to integrate.
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We set up the integral from x=0 to x=π/3 of sec(x) dx.
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Now, sec(x) is one of those integrals that we learned how to solve in our section on trigonometric integrals.
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It was one of those tricky ones that you just kind of have to remember.
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The trick was to multiply top and bottom by sec(x) + tan(x).
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That is probably worth just remembering because it is not something that you are that likely to figure out in a pinch.
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When you do that, the integral is the ln(sec(x)) + tan(x)).
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That is what we need to evaluate from x=0 to x=π/3.
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Now, that turns into the ln(sec(x) = 1/cos(x) + tan(x).
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We are evaluating that from x=0 to x=π/3
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The cos(π/3) is 1/2 tan(π/3) is the sin(π/3)/cos(π/3).
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That is sqrt(3)/2 divided by 1/2, so that is sqrt(3).
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That is what we got by evaluating x = π/3.
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Minus 1/cos(0), the cos(0) is 1, + tan(0) is 0.
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This right hand term gives us the ln(1) which goes away to 0.
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This left hand term gives us the ln of 1/1/2 is 2 + s1rt(3).
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I have absolute values signs here but I do not really need them because 2 + sqrt(3) is going to be positive.
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So, that is our answer.
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Ln(2 + sqrt(3)).
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Again, what we did there was look at the function y = ln(cos(x)).
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We did not integrate directly, instead we took its derivative and plugged it into this pythagorean formula and we integrated that.
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From there on, it was an integration problem.