Results 1 to 10 of 10

Math Help - Find the Arc Length of a Vector Function

  1. #1
    Member
    Joined
    Oct 2009
    From
    Detroit
    Posts
    176
    Thanks
    5

    Find the Arc Length of a Vector Function

    the given vector and the bounds of integration are;
    r_{0}=\langle\sqrt{2},e^t,e^{-t}\rangle; 0\leq t \leq 1

    differentiation gives;
    r^\prime_0=\langle 0, e^t, -e^{-t}\rangle

    definition of the length of a curve (were | r^\prime_0| refers to the magnitude of the vector r^\prime_0);
    L=\int_{a}^{b} | r^\prime_0|dt

    substituting the magnitude of r^\prime_0 and the bounds of integration into the equation;
    L=\int_{0}^{1} [ 0+ (e^t)^2+ (-e^{-t})^2]^{\frac{1}{2}}dt

    So every problem, of this general form, so far has had a very simple integration where there is ultimately just a constant under the radical. So I think I'm looking for a way to combine the exponential functions into a constant and I can't come up with it so far.

    If this is the wrong strategy please let me know.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    4,037
    Thanks
    745

    Re: Find the Arc Length of a Vector Function

    Hey bkbowser.

    One suggestion is to try making the elements of the vector simple polynomials and then integrate the function that way (or use trig functions).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    From
    Detroit
    Posts
    176
    Thanks
    5

    Re: Find the Arc Length of a Vector Function

    I don't think we've discussed either of those methods in my program.

    By trig functions do you mean the hyperbolic trig functions? We've been avoiding those sections in the text book and it's possible that this problem was assigned by mistake.

    Is it possible to get more information on making the elements of the vector simple polynomials? Some keywords to look up on the internet perhaps?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    4,037
    Thanks
    745

    Re: Find the Arc Length of a Vector Function

    If you just want practice for certain integrals, just either pick up a book or make your own up.

    Personally I feel that once you get the hang of a few, you should just stop and move to the next thing.

    Also if you are being tested exam wise on certain classes of integrals, focus on those since that is what you'll be marked on.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,251
    Thanks
    1795

    Re: Find the Arc Length of a Vector Function

    I recommend you go back and check the statement of the problem. The problem, as you give it, \left(\sqrt{2}, e^t, e^{-t}\right), is extremely difficult while \left(\sqrt{2}t, e^t, e^{-t}\right) is a very easy problem.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2009
    From
    Detroit
    Posts
    176
    Thanks
    5

    Re: Find the Arc Length of a Vector Function

    Quote Originally Posted by HallsofIvy View Post
    I recommend you go back and check the statement of the problem. The problem, as you give it, \left(\sqrt{2}, e^t, e^{-t}\right), is extremely difficult while \left(\sqrt{2}t, e^t, e^{-t}\right) is a very easy problem.
    You are correct. I have copied the problem wrong.

    So it should be

    the given vector and the bounds of integration are;
    r_{0}=\langlet\sqrt{2},e^t,e^{-t}\rangle; 0\leq t \leq 1

    differentiation gives;
    r^\prime_0=\langle \sqrt{2}, e^t, -e^{-t}\rangle

    definition of the length of a curve (were | r^\prime_0| refers to the magnitude of the vector r^\prime_0);
    L=\int_{a}^{b} | r^\prime_0|dt

    substituting the magnitude of r^\prime_0 and the bounds of integration into the equation;
    L=\int_{0}^{1} [ 2+ (e^t)^2+ (-e^{-t})^2]^{\frac{1}{2}}dt

    I'm still stuck here. I've been trying to get the radicand to either look something like (something)^2 or get it to be a constant. I can't seem to do either here.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2009
    From
    Detroit
    Posts
    176
    Thanks
    5

    Re: Find the Arc Length of a Vector Function

    the radicand simplifies like so;
    [2+(e^t)^2+(-e^{-t})^2]=2(cosh(t))^2

    is this correct?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,251
    Thanks
    1795

    Re: Find the Arc Length of a Vector Function

    No, it isn't. cosh(2t)= \frac{e^{2t}+ e^{-2t}}{2} so cosh^2(2t)= \left(\frac{e^{2t}+ e^{-2t}}{2}\right)^2= \frac{e^{4t}+ 2+ e^{-4t}}{4} so that cosh^2(2t)= \frac{e^{4t}+ e^{-4t}{2}+ \frac{2}{4}= cos(4t)+ \frac{1}{2}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Oct 2009
    From
    Detroit
    Posts
    176
    Thanks
    5

    Re: Find the Arc Length of a Vector Function

    Should I be trying to substitute in some hyperbolic trig function in the first place?

    I ask because we're not covering anything involving hyperbolic trig functions in my colleges program, so there's a chance I might not have to know how to do this problem at the moment!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,804
    Thanks
    1576

    Re: Find the Arc Length of a Vector Function

    Quote Originally Posted by bkbowser View Post
    You are correct. I have copied the problem wrong.

    So it should be

    the given vector and the bounds of integration are;
    r_{0}=\langlet\sqrt{2},e^t,e^{-t}\rangle; 0\leq t \leq 1

    differentiation gives;
    r^\prime_0=\langle \sqrt{2}, e^t, -e^{-t}\rangle

    definition of the length of a curve (were | r^\prime_0| refers to the magnitude of the vector r^\prime_0);
    L=\int_{a}^{b} | r^\prime_0|dt

    substituting the magnitude of r^\prime_0 and the bounds of integration into the equation;
    L=\int_{0}^{1} [ 2+ (e^t)^2+ (-e^{-t})^2]^{\frac{1}{2}}dt

    I'm still stuck here. I've been trying to get the radicand to either look something like (something)^2 or get it to be a constant. I can't seem to do either here.
    \displaystyle \begin{align*} \int_0^1{\sqrt{2 + \left( e^t \right)^2 + \left( -e^{-t} \right)^2 }\,dt} &= \int_0^1{ \sqrt{e^{2t} + 2 + \frac{1}{e^{2t}} }\,dt } \\ &= \int_0^1{ \sqrt{ e^{2t} \left( 1 + \frac{2}{e^{2t}} + \frac{1}{e^{4t}} \right) } \, dt } \\ &= \int_0^1{ e^t \, \sqrt{ 1 + \frac{2}{e^{2t}} + \frac{1}{e^{4t}} } \, dt } \end{align*}

    Now let \displaystyle \begin{align*} u = e^t \implies du = e^t\, dt \end{align*} and note that \displaystyle \begin{align*} u(0) = 1 \end{align*} and \displaystyle \begin{align*} u(1) = e \end{align*}, then the integral becomes

    \displaystyle \begin{align*} \int_0^1{ e^t \, \sqrt{ 1 + \frac{2}{e^{2t}} + \frac{1}{e^{4t}} } \, dt } &= \int_1^e{ \sqrt{ 1 + \frac{2}{u^2} + \frac{1}{u^4} }\, du } \\ &= \int_1^e { \sqrt{ \frac{ u^4 + 2u^2 + 1 }{u^4} } \, du } \\ &= \int_1^e{ \sqrt{ \frac{\left( u^2 + 1 \right) ^2 }{ u^4 } } \, du } \\ &= \int_1^e{ \frac{u^2 + 1}{u^2} \, du } \\ &= \int_1^e{ 1 + u^{-2} \, du } \\ &= \left[ u - \frac{u^{-3}}{3} \right]_1^e \\ &= \left[ u - \frac{1}{3u^3} \right]_1^e \\ &= \left( e - \frac{1}{3e^3} \right) - \left( 1 - \frac{1}{3} \right) \\ &= e - \frac{1}{3e^3} - \frac{2}{3} \end{align*}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: September 30th 2010, 12:34 PM
  2. Replies: 1
    Last Post: November 28th 2009, 12:13 PM
  3. Find a Vector given a length
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: October 20th 2009, 08:36 AM
  4. calc2. find the arc length of a function
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 5th 2009, 04:01 PM
  5. Replies: 2
    Last Post: September 9th 2008, 07:43 PM

Search Tags


/mathhelpforum @mathhelpforum