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Arc Length

  1. Jul 20, 2006 #1
    [tex]y^2 = 4(x+4)^3 0<=x<=2 ......y>0[/tex]

    Ok..so do I just simplify it to y=etc... and then find the derivative and solve..I tried this and I get a kinda long derivative..just seems like it should be easier?

  2. jcsd
  3. Jul 20, 2006 #2


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    Not all that "long" is it? the derivative of y2, with respect to x, is [itex]\2y \frac{dy}{dt} [/itex] and the derivative of 4(x+4)3 is 12(x+ 4)2 so differentiating both sides you get
    [tex]2y \frac{dy}{dx}= 12(x+4)^2[/tex]
    So what is [itex]\frac{dy}{dx}[/itex]?
  4. Jul 20, 2006 #3
    oh, i did it completely wrong..i took the root of both sides and hen tried to get the derivative
  5. Jul 20, 2006 #4


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    Whether you solve it implicitly (HallsofIvy's post) or explicitly (taking the rott, etc.) the result is the same: the implicit method gives

    [tex]2y \frac{dy}{dx}= 12(x+4)^2\Rightarrow \frac{dy}{dx}= \frac{6(x+4)^2}{y} = \frac{6(x+4)^2}{2(x+4)^{\frac{3}{2}}}=3(x+4)^{\frac{1}{2}}[/tex]

    since solving the given equation for y gives [tex]y=2(x+4)^{\frac{3}{2}[/tex] and the positive root was chosen since y>0. This is the starting point for the explicit method, just differentiate the last expression

    [tex]y=\frac{d}{dx}\left[ 2(x+4)^{\frac{3}{2}}\right] = 3(x+4)^{\frac{1}{2}}[/tex]

    which is actually easier.
  6. Jul 20, 2006 #5
    oohhh..Ok Thanks, I get it
  7. Jul 20, 2006 #6


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    start with u=x+4, then try a trig substitution
  8. Jul 20, 2006 #7
    cant i just do..[tex]\int_0^2 \sqrt{1+(3(x+4)^1^/^2)^2}[/tex]

    [tex]\int_0^2 \sqrt{37+9x}[/tex]
  9. Jul 20, 2006 #8


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    Yes, absolutely: I was in error.
  10. Jul 20, 2006 #9
    but wha tyou have there..is [tex] \sqrt{3(x+4)}^2 = 9(x+4)^2[/tex]..

    shouldnt it be: [tex] \sqrt{3(x+4)}^2 = 8(x+4)[/tex]..which can be simplified and then its an easy substitution.
  11. Jul 20, 2006 #10
  12. Jul 20, 2006 #11


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    use rather u=37+9x so that du=9dx to get

    [tex]\int_0^2 \sqrt{37+9x} dx = \frac{1}{9}\int_{37}^{55} \sqrt{u}du[/tex]
    Last edited: Jul 20, 2006
  13. Jul 20, 2006 #12
    how about this one:

    [tex]y= \frac{x^5}{5} + \frac{1}{10x^3} 1>=x>=2[/tex]

    I found the derivative to be [tex] \frac{5x^4}{6} - \frac{3}{10x^4}[/tex]..im kinda stuck here..should I just keep going and will I eventually see a substituion?
    Last edited: Jul 20, 2006
  14. Jul 20, 2006 #13
    yes that is what i got..except du=9dx
  15. Jul 20, 2006 #14


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    yep, it was
  16. Jul 20, 2006 #15
    if I have [tex] \int_0^\pi^/^3 sqrt{sec^2x}[/tex]..should I substitute u = sec^2x..or can it just be secx..and find the integral?
    Last edited: Jul 20, 2006
  17. Jul 21, 2006 #16
    Do you mean [tex]\int_0^\frac{\pi}{3}\sqrt{sec^2x} dx[/tex]?

    If yes, note that since secx is positive from 0 to [tex]\pi[/tex]/3, then [tex]\sqrt{sec^2x}[/tex] is equivalent to secx.

    There is a standard formula for evaulating [tex]\int_0^\frac{\pi}{3}secxdx[/tex]. Do you know what it is?
    Last edited: Jul 21, 2006
  18. Jul 21, 2006 #17


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    The derivative of [tex]y= \frac{x^5}{5} + \frac{1}{10x^3}[/tex] is [tex]y^{\prime}= x^4 - \frac{3}{10x^4}[/tex]
  19. Jul 21, 2006 #18
    ohh..sorry its suppose to be [tex]y= \frac{x^5}{6} + \frac{1}{10x^3} 1>=x>=2[/tex]
  20. Jul 21, 2006 #19
    and for pizzasky..no I dont know how..I found some stuff on the internet concerning the integral of secx..I never knew this integral up until now..so im guessing theres another way
  21. Jul 21, 2006 #20
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