Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trouble taking a Square Root

  1. Jul 23, 2006 #1
    Couldn’t decide if I should put this in the calculus or general math forums but...

    I’m studying for a final that’s coming up this Wednesday and I’ve been looking at some past quizzes with the steps to finding the solutions that my instructor has posted online. Given the problem:

    1. Compute the length of the curve~r(t) = (t^2, 2t, ln(t)), from t = 1 to t = e.

    I understand that I need to take the integral from 1 to e of the sqrt( (dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2) but unfortunately I’m having trouble taking the sqrt.

    I get down to… *EDITED*sqrt( 4t^2 + 4 + ( 1/t^2 ) ) and I get stuck even with the answer and the steps in front of me. In the next step the problem gets simplified so as to have everything over a t^2 and then the sqrt is done and I’m left with the integral from 1 to e of (2t^2 + 1)/t.

    I’m hoping someone could explain to me how sqrt( 4t^2 + 4 + ( 1/t ) ) becomes (2t^2 + 1)/t.

    I’m ashamed I don’t know this but I don’t want to get stuck trying to do algebra on a calculus final.

    Thanks in advance.
     
    Last edited: Jul 23, 2006
  2. jcsd
  3. Jul 23, 2006 #2
    Well,

    [tex]\sqrt{4t^2+4+\frac{1}{t}}[/tex]

    doesn't reduce to

    [tex]\frac{2t^2 +1}{t}.[/tex]

    So it's good that you can't figure that out!

    Your error was forgetting to square [tex]\frac{dz}{dt}[/tex].
     
    Last edited: Jul 23, 2006
  4. Jul 23, 2006 #3
    Ack, I wish that was my problem but I actually just mistyped it up there.

    I still don't see how sqrt( 4t^2 + 4 + ( 1/t^2 ) ) gets reduced to (2t^2 + 1)/t and I'm sure it's some simple step that I'm missing.

    Sorry...
     
  5. Jul 23, 2006 #4
    [tex]\sqrt{4t^2 + 4 + \frac{1}{t^2}} = \sqrt{\frac{4t^4 + 4t^2 + 1}{t^2}} = \frac{1}{t}\sqrt{4t^4+4t^2+1} = \frac{1}{t}\sqrt{(2t^2+1)^2} = \frac{2t^2+1}{t}[/tex] :smile:
    (for [itex]t > 0[/itex])
     
    Last edited: Jul 23, 2006
  6. Jul 23, 2006 #5
    Thank you, I always got stuck at step 2. In fact at one point I was sitting there with sqrt( ( 2t^2 + 1)^2 / t^2 ) ) and was baffled...

    Thanks again, the world makes sense now.
     
  7. Jul 23, 2006 #6
    :rofl:
    happens to everyone sometimes.

    Good :smile:. Don't be afraid to come back with more questions!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Trouble taking a Square Root
  1. Taking Square Roots (Replies: 9)

  2. Square roots (Replies: 13)

  3. Square Roots? (Replies: 7)

  4. The square root (Replies: 7)

Loading...