Help with Arc Length Problem: Evaluating Integral from x=8 to x=27

[Liger20]

Homework Statement

Hello, I have an arc length problem that I’m stuck on, and I would really appreciate it if someone could help me out. I understand the arc length formula and everything, it’s evaluating the integral produced by it. The author in the book I got this problem from tells the reader that arc length problems can produce tricky integrals, and therefore he shouldn’t have to show the work involved (real helpful). Anyway, the entire problem is irrelevant, so I’m just going to list the part that I don’t understand. Sorry for the crazy notation, but I couldn’t figure out how to do math symbols on the computer. I'm supposed to find the arc length along f(x)=x^(2/3) from x=8 to x=27. The first step listed starts out after everything has been plugged into the formula.




27
INTEGRAL * square root(1+(4/9)x^(-2/3))dx
8



27
= (1/3) INTEGRAL * square root(9+4x^(-2/3))dx
8


27
=(1/3) INTEGRAL*x^(-1/3)* square root(9x^(2/3)+4)dx
8

I understand the first step, but the author completely lost me on steps two and three. If I could find out what happened in those two steps, I would understand the whole problem. I don’t need anything else explained after the third step, because it involves using u-substitution, which I understand fairly well.

Homework Equations

The formula for arc length is:


B
Arc length= INTEGRAL* square root(1+(f ‘(x))^2) dx
A

The Attempt at a Solution

When I first attempted the problem, I didn’t think the integral looked all that difficult. I thought that I could simply take the square root away by raising the inside term to the power of ½ and integrating. But apparently it’s much more subtle than that. I would really appreciate some guidance with this. I hope the notation I used wasn’t confusing, and I can clarify it if anyone wants me to.

 

[snipez90]

Right, so the author is using algebraic manipulations to turn the first expression into something more congenial. To get to step two, the author used the trick of multiplying an expression by 1, but with the aim of getting rid of the fraction (4/9) under the square root. To do so requires multiplying the integrand by the square root of 9 , but of course, doing this alone will change the original expression. But the square root of 9 is just 3, so if we also divide by 3, then sqrt(9)/3 is just 1, and multiplying the first expression by sqrt(9)/3 gives step 2.

To get from step 2 to step 3, the author took the expression under the square root and factored out an x^(-2/3).

 

[Mark44]

For future reference, your integral

27
INTEGRAL * square root(1+(4/9)x^(-2/3))dx
8

would be easier to read using LaTeX tags
<br /> \int_{8}^{27} \sqrt{1 + \frac{4}{9}x^{-2/3}}dx<br />

To see the code I used, just click on the integral expression.

 

[Bohrok]

To the OP:
Would you post how you continue doing the problem? I did the same problem a couple months ago with f(x)=x^2/3 but over [1, 2] and it was probably the longest problems I've ever done. Just want to see if you or anyone else can find an easier way than I did it.

 

[Liger20]

Sure Bohrok, but before I tried to use latex and it didn't work. So I'm going to have to write it out the weird obnoxious way, but this it the only way I can type it. I'm sorry about this, and hopefully I'll be able to use Latex soon. Also, I can’t guarantee maximum clarity since the author of Calculus Workbook for Dummies loves to skip steps. If anyone happens to have that book by the way, this very problem in on page 227. Anyway, here's the rest:

You finish by using u-substitution where u=9x^(2/3)+4

You change the indices of integration by plugging the original indices into the u expression. This gives you:


85
(1/3) INTEGRAL (1/6)u^(½)du
40

=(1/18)[(2/3)u^(3/2)] (from 85 to 40)

=(85sqrt(85)-80sqrt(10))/(27)

=19.65

Tell me if any of this is unclear. Again I'm sorry about the lack of Latex format.

 

[Mark44]

The integral shown in post 3 has the LaTeX code in it. Just click it and a small page should open with the LaTeX code showing. Just copy and paste it into the PF page, and make corrections as needed.

 

[Bohrok]

That rearrangement of the equation sure works out nice where you use just a u-substitution. I made mine much harder with different u and trig substitutions and complicated limits of integration, but I didn't have any steps to go by.
Thanks for posting this.