Results 1 to 8 of 8

Thread: Two integrals with radicals.

  1. #1

    Two integrals with radicals.

    I dislike finding arc lengths because so very frequently you end up with a radical you can't easily do away with. Such is my situation in two problems on which I am stuck, and after 1 hour of cogitation, I turn to the Internet to see if you might have a hint on where to go next.

    1.) I am given y = (x^3 / 3) + (1 / 4x) for 1<or=x<or=2 and am asked to "Graph the curve and find its exact length" I've got it graphed, but the length? Not so much. I understand that the formula for finding arc length calls for the derivative of the function, so I dutifully dervive the given and arrive at:

    y' = x^2 - 4x^-2. Spiffy.

    So I set L = Integral from 1-->2 of Root(1 + (x^2 - 4x^-2)^2)dx

    Now what? O.o I don't know how to wrestle with this beast. I don't see a u-substitution that readily presents itself and integration by part would just leave me with a REALLY ugly du that I'd be stuck trying to integrate anyway (yuck). If I expand the (x^2 - 4x^-2)^2 I get Root(1+ x^4-8x^-2 + 16x^-4) dx, which doesn't appear to be any better.

    I'm open to suggestions!



    2.) This time I am given the same instructions as above: Graph the curve and find its length. But here they give me x= (e^t)+(e^-t), y= 5-2t, 0<or=t<or=3

    Again, I do a Yeoman's job deriving x and y. x'=(e^t)-(e^-t) and y' = -2.

    I then have L = integral from 0-->3 of Root((-2)^2 + ((e^t)-(e^-t))^2)dt. I recognize that e^t-e^-t = 2sinh t, but I don't think that getting into hyperbolic trig is actually simplifying things at all. So I'm stuck here again. Again, I don't see a good u-substitution to use that would make things simple. I mean I could do u = t, but that's just swapping out letters and not simplifying my life any. Any other u value seems to leave dangling bits that would be messy. Integration by part would give me just a hideous looking integral. I don't know of an easy way (or at least a rational, logical way) forward on this one either.

    Please help!!!

  2. #2
    Senior Member
    Join Date
    Dec 2014
    Posts
    1,408
    Quote Originally Posted by 1John5vs7 View Post
    I dislike finding arc lengths because so very frequently you end up with a radical you can't easily do away with. Such is my situation in two problems on which I am stuck, and after 1 hour of cogitation, I turn to the Internet to see if you might have a hint on where to go next.

    1.) I am given y = (x^3 / 3) + (1 / 4x) for 1<or=x<or=2 and am asked to "Graph the curve and find its exact length" I've got it graphed, but the length? Not so much. I understand that the formula for finding arc length calls for the derivative of the function, so I dutifully dervive the given and arrive at:

    y' = x^2 - 4x^-2. Spiffy.
    y' is not correct.
    Also, you should usually simplify 1+(y')^2 before trying to use parts or u-subs.
    Last edited by Jomo; 05-11-2015 at 09:24 PM.
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  3. #3
    Senior Member
    Join Date
    Dec 2014
    Posts
    1,408
    Quote Originally Posted by 1John5vs7 View Post
    2.) This time I am given the same instructions as above: Graph the curve and find its length. But here they give me x= (e^t)+(e^-t), y= 5-2t, 0<or=t<or=3

    Again, I do a Yeoman's job deriving x and y. x'=(e^t)-(e^-t) and y' = -2.

    I then have L = integral from 0-->3 of Root((-2)^2 + ((e^t)-(e^-t))^2)dt. I recognize that e^t-e^-t = 2sinh t, but I don't think that getting into hyperbolic trig is actually simplifying things at all. So I'm stuck here again. Again, I don't see a good u-substitution to use that would make things simple. I mean I could do u = t, but that's just swapping out letters and not simplifying my life any. Any other u value seems to leave dangling bits that would be messy. Integration by part would give me just a hideous looking integral. I don't know of an easy way (or at least a rational, logical way) forward on this one either.

    Please help!!!
    4 + 4[sinh(t)]^2 = 4(1+[sinh(t)]^2) = 4[cosh(t)]^2. The rest is simple.

    These arc length formulas are not so bad if you make the correct subs and know your identities.
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  4. #4
    Quote Originally Posted by Jomo View Post
    y' is not correct.
    Also, you should usually simplify 1+(y')^2 before trying to use parts or u-subs.
    Actually I don't see how y' is incorrect.

    x^3 + 1
    3 4x

    That becomes x^2 + (4x^-2).
    Last edited by 1John5vs7; 05-11-2015 at 09:41 PM.

  5. #5
    Senior Member
    Join Date
    Dec 2014
    Posts
    1,408
    Quote Originally Posted by 1John5vs7 View Post
    Actually I don't see how y' is incorrect.

    x^3 + 1
    3 4x

    That becomes x^2 + (4x^-2).
    [1/(4x)]' = [(1/4)(1/x)]'=(1/4)[1/x]'= (1/4)[-1/x^2)= -1/(4x^2) NOT -4/x^2.



    You should know the sqrt of (4 cosh^2(t)) and it is NOT 2cosh(t).
    Last edited by Jomo; 05-11-2015 at 09:49 PM.
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  6. #6
    Quote Originally Posted by Jomo View Post
    4 + 4[sinh(t)]^2 = 4(1+[sinh(t)]^2) = 4[cosh(t)]^2. The rest is simple.

    These arc length formulas are not so bad if you make the correct subs and know your identities.
    Ah, I see what you're getting at there. You end with a perfect square, then you substitute your e values back in at the end, the 2 in the numerator is canceled by the 2 in the denominator and you have e^t + e^-t |0->3. Thank you, this helped me to solve this problem.

    I guess I'm just supposed to memorize ever formula in the back of Stewart's Calculus???? O.o There are over 100 of them though. How do I sort out which ones I need to have memorized and which not? This is the first time I've ever had to use hyperbolic trig functions to solve something.

  7. #7
    Senior Member
    Join Date
    Dec 2014
    Posts
    1,408
    Quote Originally Posted by 1John5vs7 View Post
    Ah, I see what you're getting at there. You end with a perfect square, then you substitute your e values back in at the end, the 2 in the numerator is canceled by the 2 in the denominator and you have e^t + e^-t |0->3. Thank you, this helped me to solve this problem.

    I guess I'm just supposed to memorize ever formula in the back of Stewart's Calculus???? O.o There are over 100 of them though. How do I sort out which ones I need to have memorized and which not? This is the first time I've ever had to use hyperbolic trig functions to solve something.
    You memorize some and easily derive the rest. For example I memorize sin^2 x + cos^2 x=1 . Then I think what will I get if I divide all 3 terms by cos^2 x. So I get (easily!!) that tan^2 x +1 = sec^2 x. I do this division if I ever have tan^2 x and +/- 1 and sec^2 x and +/- 1.

    I am not sure why you are are substituting back the e's.

    So what did you get as the integrand after you took the sqrt?
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  8. #8
    Senior Member
    Join Date
    Dec 2014
    Posts
    1,408
    Quote Originally Posted by 1John5vs7 View Post
    Ah, I see what you're getting at there. You end with a perfect square, then you substitute your e values back in at the end, the 2 in the numerator is canceled by the 2 in the denominator and you have e^t + e^-t |0->3. Thank you, this helped me to solve this problem.

    I guess I'm just supposed to memorize ever formula in the back of Stewart's Calculus???? O.o There are over 100 of them though. How do I sort out which ones I need to have memorized and which not? This is the first time I've ever had to use hyperbolic trig functions to solve something.
    You chose to introduce hyperbolic functions.
    ((-2)^2 + ((e^t)-(e^-t))^2)= 4 + e^(2t) -2 + e^(-2t) = e^(2t) + e^(-2t) + 2 = (e^t + e^-t)^2. The sqrt (e^t + e^-t)^2= |(e^t + e^-t)| = (e^t + e^-t)
    OK?
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

Tags for this Thread

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •