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Arc Length

  1. Aug 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the length of
    [tex]\ x =t^{3}[/tex]
    [tex]\ y =t^{2}[/tex]
    0 [tex]\leq[/tex]t[tex]\leq[/tex] 4

    2. Relevant equations
    I would write the formula for the arc length but I don't know how to make a definite integral.

    3. The attempt at a solution
    I have the whole thing set up and I'm ready to integrate but I seem to have forgotten how to integrate [tex]\int\ \sqrt{9t^{4}+2t^{2}}dt[/tex]
  2. jcsd
  3. Aug 27, 2009 #2
    From the integral you set up, first factor t^2, then it will become t by square root. After that, use trigonometric substitution. 3t=sqrt(2)tan(theta), 3dt=sqrt(2)sec^2(theta), then substitute them to the integral
  4. Aug 27, 2009 #3
    Unfortunately I don't follow you, after I factor out the t^2 I get
    [tex]\int\sqrt{t^{2}(9t^{2}+2)} [/tex] after that I don't know what your trying to say.
  5. Aug 27, 2009 #4


    Staff: Mentor

    Bring the t2 factor out of the radical so that the integrand becomes
    [tex]t\sqrt{9t^2 + 2}[/tex].

    Also, it is much simpler to evaluate this integral using an ordinary substitution instead of the more complicated trig substitution that darkmagic suggested.
  6. Aug 27, 2009 #5


    Staff: Mentor

    Also, here is the definite integral. Click on it to see the LaTeX code that I used.
    [tex]\int_{0}^{4} t\sqrt{9t^2 + 2}~dt[/tex]
  7. Aug 27, 2009 #6
    I hate to ask again but I think I rushed through U-substitution to fast. I don't see anything thats the derivative of the otherthing so I'm not sure what to pick for U.
  8. Aug 27, 2009 #7
    Recall the chain rule: If f is a function of g and g is a function of t, and we define h(t) = f(g(t)), then h'(t) = f'(g(t))g'(t).
    What is the derivative of the inside of that radicand in your integral? Can you modify the outside of the integral so that the integrand resembles a chain rule?
  9. Aug 28, 2009 #8


    Staff: Mentor

    Then you should review this technique. For your problem, let u = 9t2 + 2. Then du = ______ dt ?
  10. Aug 28, 2009 #9
    I got the answer to be


    is this right?
  11. Aug 28, 2009 #10


    Staff: Mentor

    It's easy enough to check for yourself. Just differentiate what you have and if it's correct, you should get the integrand you started with.

    I will say, though, that it appears you have made an error in your integration. Also, when you do get the correct antiderivative, you will still need to evaluate it at 4 and at 0 -- you're working with a definite integral, so you should get a number for your arc length.

    Show us how you did the substitution and the work after that, and we'll help you out.
  12. Aug 28, 2009 #11
    ok. I do not see that. Just a u-substitution.
  13. Aug 28, 2009 #12
    Here are my steps

    [tex]\ \int_{0}^{4} t\sqrt{9t^2 + 2}~dt[/tex]

    [tex]\ \frac{1}{18}\int 18t\sqrt{9t^2 + 2}~dt[/tex]

    [tex]\ u=9t^2 + 2[/tex]
    [tex]\ du=18t[/tex]

    [tex]\ \frac{1}{18} \int (u)^{1/2} du [/tex]
    [tex]\ \frac{1}{18} \frac{(9t^{2}+2)^{3/2}}{3/2} [/tex]

    I can't see where I went wrong.
    Last edited: Aug 28, 2009
  14. Aug 28, 2009 #13


    Staff: Mentor

    The last line should be du = 18t dt. If you get in the habit of omitting this differential, it can come back around and bite you in more complicated integrals.
    That's slightly different from what you had in post #9. If you look closely, you should see the difference.

    To continue where you left off, and doing a little simplification

    [tex]\ \int_{0}^{4} t\sqrt{9t^2 + 2}~dt[/tex]
    (using your substitution and then undoing the substitution)
    [tex]\ = \ \frac{1}{27} ~(9t^{2}+2)^{3/2}\vert_0^4[/tex]
    Now, just evaluate this antiderivative at 4 and at 0 and subtract.
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