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Simplify the following

  1. Oct 3, 2015 #1
    • Member warned about posting without the homework template
    I don't understand why my instructor simplified this problem like he did. I get how we took out the square for the (e^t + e^-t)^2, but what happened to the 2 under the square root.

    Pic of solution is attached
     

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    Last edited: Oct 3, 2015
  2. jcsd
  3. Oct 3, 2015 #2

    Student100

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    Do you know what cosh(t) is?

    Don't delete the template, and show work please.
     
  4. Oct 3, 2015 #3
    I attached a screenshot of the solution. I just need clarification on the simplification done from line 3-4. And I know about hyperbolic trig, but I doubt that that was used there.
     
  5. Oct 3, 2015 #4

    Student100

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    You attached a screenshot, but you deleted the template and didn't show any attempt at the solution. That's not allowed here.

    How are you sure it wasn't, have you tried it?

    I'm almost 100 percent it was used in simplifying what you posted.
     
  6. Oct 3, 2015 #5

    SteamKing

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    Under the radical, the expression (e-t)2 + 2 + (et)2 looks suspiciously like (a2 + 2ab + b2).
    If fact, if you let a = e-t and b = et, you will find the the expression under the radical is the perfect square, namely (a + b)2.

    Note: No cosh needed.
     
  7. Oct 3, 2015 #6
    I forgot most of the hyperbolic trig stuff, but this doesn't look like any of the hyperbolic trig functions. And my attempt at the solution is exactly the same as the attached solutions, except I don't know how the stuff under the square root was simplified.
     

    Attached Files:

  8. Oct 3, 2015 #7

    Student100

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    You can get the simplification from cosh(t) like I did, or you can use SteamKings result which is probably easier.
     
  9. Oct 3, 2015 #8
    oh. ok that makes sense. Thanks. How do you solve those kinds of problems again? Going from (a^2 + 2ab + b^2) to (a + b)^2. How do you exactly do that. This looks familiar, but I haven't done it in forever. I know if you have something like z^2 - 2z you can divide the second term and square it, giving you (z-1)^2. Does my question make sense? I want to make sure I know how to simplify more like those in the future. Can you show me a similar (simpler) example and how to simplify it.
     
  10. Oct 3, 2015 #9

    SteamKing

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    Well, sometimes you do it by trial and error. The key in this problem was that you had et and 1/et, so that when you multiply these two terms together, you'll get a 1. One of the algebra facts you should remember is how to expand (a + b)2 into a2 + 2ab + b2.

    As far as z2 - 2z, that simplifies to z(z - 2). IDK how you would get (z-1)2 out of that, unless you meant to write z2 - 2z + 1.
     
  11. Oct 3, 2015 #10
     
  12. Oct 3, 2015 #11

    SteamKing

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    It's just one of those forms you learn to recognize after a bit of practice. It's sort of like solving trig identities. A lot of those can be solved by remembering the Pythagorean identity and not being afraid to slog thru some algebra. It also helped that the solution was presented in the OP. Sometimes, filling in the blanks in a solution is made simpler, since you are given the starting point and the end point of the calculations. This cuts down on the amount of trial and error to reconstruct what happens in the middle of the calculations.

    If algebra is giving you a hard time in calculus, by all means, brush up on your algebra. Making a couple of mistakes in algebra can keep you from solving even simple calculus problems. Same with trig. Get the basics down cold, before going on to tackle more advanced math.
     
  13. Oct 3, 2015 #12
    You know any good online sources that review that stuff in particular?
     
  14. Oct 3, 2015 #13

    SteamKing

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    No, not really, but I'm sure there are some sites which offer review in those subjects. You can't go wrong by Googling "algebra review" or something along those lines.
     
  15. Oct 3, 2015 #14
    I'm fairly good at most algebra, but quadratic stuff sometimes gets me. I'll go hunting on google for some review then. Thanks for the help!!
     
  16. Oct 4, 2015 #15

    Student100

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    SteamKing has given great advice,

    As far as online resources, Purplemath is decent to look up things quickly, alternatively REA problem solvers has a decent algebra/trig reference book.

    If you want to look at the problem as I did, $$\sqrt{{e^{2t}+2+e^{-2t}}}$$ $$\sqrt{{(e^{t}+e^{-t}})^2}$$ $$2\sqrt{(\frac{e^{t}+e^{-t}}{2})^2}$$ $$2\sqrt{cosh^{2}(t)}$$ Assuming t is real = $$2cosh(t)$$ Which is just what you were trying to show. All I did was add an extra step you don't actually need, unless you wanted to evaluate the integral as ##2\int{cosh(t)dt} = 2sinh(t)+C## which is a bit easier to do quickly.
     
    Last edited: Oct 4, 2015
  17. Oct 4, 2015 #16

    Mark44

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    The OP really continue from the last expression above, without having to convert to hyperbolic cosines. As already mentioned, the quantity inside the radical is a perfect square, so simplifying it reduces to ##e^t + e^{-t}##, which makes for a simple integration in finding the arc length.
     
  18. Oct 4, 2015 #17

    Student100

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    Yeah I know, I have the habit of putting those expressions in terms of hyperbolic. It helped me when I went through class as it came up so often, thought it might be a different way to see the same simplification for the OP. Although it does add unnecessary steps.
     
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