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Trying to calculate arc length

  1. Aug 9, 2009 #1
    Hi,

    Im trying to calculate the arc length of the function [tex]f(x)=x\sqrt{x}[/tex]
    From x=1 to x=7
    But im getting the wrong answer and im not sure why.

    The formula is [tex]\int^{7}_{1}\sqrt{f'(x) + 1}[/tex]

    The derivative of f(x) =[tex]\frac{x}{2\sqrt{x}} + \sqrt{x}[/tex]

    Squaring yields [tex]~~\frac{x}{4} + 2x +1[/tex] which simplifies to:[tex]\frac{9x}{4}+1[/tex]

    Integrating, we get [tex]\int\frac{2(\frac{9x^2}{2}+x)^{3/2}}{27}[/tex]

    Inserting the limits of integration i get [tex]\frac{2*(9*49/2 +7)^{3/2}}{27} - \frac{2*(9/2 +1)^{3/2}}{27} = 253.222[/tex]

    This is incorrect as the aswer should be something over 27.
    What am i doing wrong?
     
    Last edited: Aug 9, 2009
  2. jcsd
  3. Aug 9, 2009 #2

    VietDao29

    User Avatar
    Homework Helper

    Your formula is wrong. It should have read:

    [tex]\int_1 ^ 7 \sqrt{{\color{red}[f'(x)] ^ 2} + 1} {\color{red}dx}[/tex]

    You can apply Multiplication Rule here as what you have done. But it may be shorter to try:

    [tex]f(x) = x \sqrt{x} = x ^ {\frac{3}{2}} \Rightarrow f'(x) = \frac{3}{2} x ^ {\frac{3}{2} - 1} = \frac{3}{2} x ^ {\frac{1}{2}} = \frac{3}{2} \sqrt{x}[/tex].

    Ok. But try to write it more clearly, what does "squaring yields" mean? What are you squaring? And if you are just squaring f'(x), then why does "+ 1" appear there?

    A better way is to write:
    [tex][f'(x)] ^ 2 + 1 = \frac{9}{4} x + 1[/tex]

    Why is there still the 'integral sign' standing in the front?

    -------------

    Your integral will now become:

    [tex]\int_1^7 \sqrt{\frac{9}{4} x + 1} dx[/tex]

    You have integrated it incorrectly. Try again then. I'll give you a hint, you need to use u-substitution to do it. Do it step by step. What are u, du? And what are the new lower and upper limits?

    I'll give you an example then:

    Example:
    Integrate:
    [tex]\int_0^1 \sqrt{2x + 1} dx[/tex]

    Let u = 2x + 1 => du = 2dx => dx = du/2.
    x = 0 => u = 1
    x = 1 => u = 3

    So you new integral is:

    [tex]\frac{1}{2}\int_{1} ^ {3} \sqrt{u}du = \frac{1}{2}\left( \left. \frac{u ^ {\frac{3}{2}}}{\frac{3}{2}} \right|_1 ^ 3 \right) = \frac{1}{2} \times \frac{2}{3} \times \left( \left. u ^ \frac{3}{2} \right|_1 ^ 3 \right)[/tex]
    [tex]= \frac{1}{3} \left( \left. u ^ \frac{3}{2} \right|_1 ^ 3 \right) = \frac{1}{3} \left( \sqrt{3 ^ 3} - 1 \right) = \frac{1}{3} \left( \sqrt{27} - 1 \right)[/tex].

    Ok, let's see if you can complete your problem. :)
     
    Last edited: Aug 9, 2009
  4. Aug 9, 2009 #3
    Thanks for the very helpful/informative post!

    Okay.

    I have [tex]\int\sqrt{\frac{9x}{4}+1}[/tex]

    Rewrite:
    [tex]~(\frac{9x}{4} + 1)^{1/2}[/tex]

    [tex]u=\frac{9x}{4}+1[/tex]
    New limits:
    [tex]Upper:~x = 7 \rightarrow u = \frac{67}{4} ~~\\
    Lower:~ x = 1 \rightarrow u = \frac{13}{4} [/tex]
    [tex]du=\frac{9}{4}dx[/tex]

    [tex]dx=\frac{4}{9}du[/tex]

    [tex]\frac{4}{9}\int \sqrt{u}du[/tex]

    [tex]\frac{4\times 2}{9\times 3}\int u^{3/2}[/tex]

    [tex]\frac{4\times 2}{9\times 3}\sqrt{(\frac{67}{4})^3 - (\frac{13}{4})^3}[/tex]
     
    Last edited: Aug 9, 2009
  5. Aug 9, 2009 #4

    VietDao29

    User Avatar
    Homework Helper

    Well, can you just write it formally. I know you can omit the dx, and du thingy. But, honestly, I think you are pretty new to integrals, so writing it down is no harm, right? It can also help you to know whether you are integrating with respect to x, or to u, or whatever variable.

    Nah, why is the integral sign still there?

    I think you should go back, and re-read (or have a skim through) the chapter on Integrals (and maybe, my example), after integrating, there's no more integral sign. When reading the Integrals chapter again, if you have something unclear, you can post it here. Other people, and me, are more than willing to help you.

    You should grab the basic concepts first, before moving forward. Or you'll find it extremely messy. You must make sure to understand everything about anti-derivative, before moving to definite integrals.

    ------------------

    Well, there are two ways of doing it. The first way is to change the integral to:

    [tex]\frac{8}{27} \int_{\frac{13}{4}}^{\frac{67}{4}} \sqrt{u} du[/tex], and solve it, as normal.

    ------------------

    The second way is to find the anti-derivative of [tex]\int \sqrt{\frac{9}{4}x + 1} dx[/tex] (note that this is an indefinite integral, there is no limit there). Do the u-substitution, then switch back to x. After having the anti-derivative of the function, simply use the formula:

    [tex]\int_a^b f(x) dx = F(b) - F(a)[/tex]

    Using this way, you don't have to change the limits. But you have to switch u back to x.
     
    Last edited: Aug 9, 2009
  6. Aug 9, 2009 #5

    VietDao29

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    Homework Helper

    I saw you editing your post when I'm posting mine.. =.="

    You are wrong at the final step.

    Which should read:

    [tex]\frac{8}{27}\left[ \sqrt{\left( \frac{67}{4} \right) ^ 3} - \sqrt{\left( \frac{13}{4} \right) ^ 3} \right][/tex].

    There are 2 square root signs, instead of one.
     
  7. Aug 9, 2009 #6
    You're right. i will go back and check out the basics again.

    Anyways, thanks for a great explanation :)
     
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