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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.
     
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