1. Limited time only! Sign up for a free 30min personal 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!

Length of A Curve

  1. Aug 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the length of the curve given by the equation:

    2. Relevant equations
    [itex] y= \int_{-pi/2}^x √(cos t)\, dt [/itex] for x between -∏/2 and ∏/2

    3. The attempt at a solution
    y= sqrt (cos x)
    dy/dx= (sin x)/[-2 * sqrt(cos x)]

    So now applying the arc length formula of sqrt (1 + (dy/dx)^(2)), I get:

    [itex] \int_{-pi/2}^{pi/2} sqrt(1 + {(sin^2(x))/(4 cos(x))}) \, dx [/itex]
    [itex] \int_{-pi/2}^{pi/2} sqrt(1 + ({sin x * tan x}/4))\, dx[/itex]

    I don't know how to integrate that, and haven't learnt it either.. Any assistance is much appreciated. Thank you!
    Last edited: Aug 4, 2013
  2. jcsd
  3. Aug 4, 2013 #2
    You have not specified the equation of the curve.
  4. Aug 4, 2013 #3
    This is all the information I have. Where I am stumped is why y= sqrt(cos t) dt with the limits of integration being 0 and 1/2.
  5. Aug 4, 2013 #4
    Is this ## y = \int_{\pi/2}^x \sqrt {\cos t} dt ## the equation of the curve?
  6. Aug 4, 2013 #5
    Yes that is sir.
  7. Aug 4, 2013 #6
    So you have to compute ## ds = \sqrt {1 + (\frac {dy}{dx})^2 } dx ##, and for that you need to find ##\frac {dy}{dx}## first. What is it?
  8. Aug 4, 2013 #7
    I think dy/dx is (sin x)/[-2 * sqrt(cos x)] .
  9. Aug 4, 2013 #8
    ## - \frac {\sin x} {2 \sqrt {\cos x} } ## is the derivative of ## \sqrt {\cos x} ##.

    But you need to differentiate ## y(x) = \int_{\pi/2}^x \sqrt {\cos t} dt ##.
  10. Aug 4, 2013 #9
    That is what I do not understand unfortunately. Y is supposed to be in terms of x, yet it is given in terms of t, and I do not know how to figure out x in terms of t, if I want to substitute. Am I not getting something?
  11. Aug 4, 2013 #10
    y is not given in terms of t. The variable t appears only inside the integral. The integral depends on the variable x - and this is what y depends on.
  12. Aug 4, 2013 #11
    Yes, and this is where I am confused. Since the limits of integration are -pi/2 to x and x is between -pi/2 and pi/2, can you not say that the second limit is pi/2?
  13. Aug 4, 2013 #12
    Given a function f(x), where x is between a and b, can you not say that it is just f(b) - a constant?
  14. Aug 4, 2013 #13
    Yes- and I think of that as similar to the method of improper integrals- can I make this connection?
  15. Aug 4, 2013 #14
    Are you seriously saying that ANY function can be taken to be a constant just because its domain is defined?
  16. Aug 4, 2013 #15
    No. I was just presuming that for this question, because of the lower and upper limits of integration.
  17. Aug 4, 2013 #16
    The upper limit is a variable. What you have on the right is a function of this variable. You need to fin the derivative of that function with respect to this variable.
  18. Aug 4, 2013 #17
    That is exactly where I am stumped. How do I know the relationship between f(b) and sqrt(cos t) or just t for that matter?
  19. Aug 4, 2013 #18
    I am pretty sure you have studied the fundamental theorem of calculus.
  20. Aug 4, 2013 #19
    Yes, I have.

    I think applying that logic, I come up with this:

    ∫sqrt(cos t) dt= -2/3 sin (t)^(3/2) with the limits of integration being -pi/2 and x.
    -2/3 sin (0)^(3/2) - (-2/3)sin(x)^(3/2)
    2/3 sin (x)^(3/2)

    Then is y= 2/3 sin (x)^(3/2)?
    And, the length of the curve is therefore,
    dy/dx= sqrt(sin x)

    and sqrt(1 + sin^2(x))= (1+ sin^2(x)^(1/2)

    If I take the integral of that, I have: 2/3 * (1+ sin^2(x))^(3/2) * [0.5x - 0.5sinx*cosx]
    which is equal to 2/3 * (1 + sin^2(x))^(3/2) * 1/2 [x- sinx*cosx]
    = 1/3 (1+ sin^2(x))^(3/2) * [x- sinx*cosx]
    Am I correct so far or have I already gone wrong?
  21. Aug 4, 2013 #20
    The fundamental theorem of calculus states that if ## F(x) = \int_a^x f(t) dt ##, then ## F'(x) = f(x) ##. This is exactly what you have: you have ## y = F(x) = \int_a^x \sqrt {\cos t} dt ##, so ## f(x) = \sqrt {\cos x}##, so ## \frac {dy} {dx} = \sqrt {\cos x} ## follows automatically and effortlessly. It should have, anyway.

    I am not sure how you got ## \sqrt {\sin x} ##, it is certainly not correct.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted