1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Another simple question that I don't get
  1. DE I don't get (Replies: 1)

Loading...