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Homework Help: Another simple question that I don't get

  1. Mar 1, 2008 #1
    [SOLVED] Another simple question that I don't get :(

    1. The problem statement, all variables and given/known data
    Find the Length of the indicated curve

    y=2x-3 x[1-3]


    2. Relevant equations

    Arc Length=[tex]\int[/tex][tex]\sqrt{1+[f'(x)]^{2}dx}[/tex]

    3. The attempt at a solution

    ok so y'=2 of course,

    so it becomes a problem of [tex]\int from 1 to 3 for \sqrt{5}[/tex]

    using substitution, u=5 ,and this is where i get a little confused....

    du=0dx which would make dx=du/0

    the answer becomes 2[tex]\sqrt{5}[/tex]

    I'm sure I'm making it harder than it is, I always do..... :(
     
  2. jcsd
  3. Mar 1, 2008 #2
    and thinking about it, i should probably not be using the formula for arc length for this problem, even though it is in the arc length chapter...since this is a linear function
     
  4. Mar 1, 2008 #3
    using the arc length formula the only thing i can think of is that when i get

    [tex]\sqrt{5}[/tex]dx and dx ends up being 0, it becomes neglible so i only have [tex]\sqrt{5}[/tex] to evaluate from 1 to 3.... but then my question would be, I know subtracting 3[tex]\sqrt{5}[/tex] from 1[tex]\sqrt{5}[/tex] gives me the answer 2[tex]\sqrt{5}[/tex] but there is no X in front of the [tex]\sqrt{5}[/tex]


    :(
     
  5. Mar 1, 2008 #4
    In the formula for arc length, the dx should be OUTSIDE of the radicand. Does this help you?

    Hint: it should!

    What is [tex]\int_1^3\sqrt{5}*dx[/tex]
     
    Last edited: Mar 1, 2008
  6. Mar 1, 2008 #5
    yea sorry, forgot to put the dx on the outside, but still confused.... using substitution and writing the problem like this.....

    u[tex]^{1/2}[/tex] where u=5, du=0dx so dx=du/0 which is ?

    continuing, if it is right u[tex]^{1/2}[/tex]du would become u[tex]^{3/2}[/tex] times 2/3 so I'm doing something wrong with the substituion.....I'm just confusing meself :(

    I can do all the problems except this one b/c when I take the derivative of u it becomes nothing..... so that is where im getting lost
     
    Last edited: Mar 1, 2008
  7. Mar 1, 2008 #6
    What are you doing? Why the need for u sub? Again I ask you what is [tex]\int c*dx[/tex]

    where c is any constant.....any constant like, oh let's say [itex]\sqrt 5[/itex]
     
  8. Mar 1, 2008 #7
    so you're saying it would just become (integral)c=x so x(sqrt 5)


    I'm not seeing the obvious, for some reason i 'm thinking i need substitution....wtf
     
    Last edited: Mar 1, 2008
  9. Mar 1, 2008 #8
    Of course! sqrt5 is just a number, just some constant. Pull it through the integral and proceed as normal. Evaluate what you have from x=3--->x=1

    ---->(3-1)*[itex]\sqrt 5[/itex]



    [tex]\int_{x_0}^x cdx=c\int_{x_o}^x dx=c[x]_{x_o}^x[/tex]



    What is there to substitute? What are you going to replace with u?
     
    Last edited: Mar 1, 2008
  10. Mar 1, 2008 #9
    you with me?
     
  11. Mar 1, 2008 #10
    I believe so, i just think i gotta step away from these books for awhile:yuck:
     
  12. Mar 1, 2008 #11
    We all do! I feel the same way most of the time!
     
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