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Vector calculus

  1. Aug 27, 2005 #1
    i having so much trouble with this vector calculus question, someone please help.

    At time t, a moving particle has position r(t)=((e^t) + (e^-t))/2 i + ((e^t) - (e^-t))/2 j.

    a) find the cartesian equation of the path

    I know all you have to do is let the i component to equal x(t) and then make t the subject, and then with the j component let it equal y and sub t into y. but i can't get t to be the subject in the equation.
    someone please give it a go
  2. jcsd
  3. Aug 27, 2005 #2
    i don't get it how can you get that equation from the original equation?
  4. Aug 27, 2005 #3
    This is because the coeffcient of the x-coordinate and the y-coordinate is the same: [tex]\left( {\frac{{e^t + e^{ - t} }}{2}} \right)[/tex]
  5. Aug 27, 2005 #4


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    No, that's incorrect (or the original poster has edited the post since you wrote this). The vector equation is
    [tex]\left(\frac{e^t+ e^{-t}}{2}\right)i+ \left(\frac{e^t- e^{-t}}{2}\right)j[/tex]

    The x and y components are not equal. It should be easy to see that the x-component is cosh(t) and the y-component is sinh(t). Now, use an identity: cosh2(t)- sinh2(t)= 1 so that x2-y2= 1, a hyperbola (in fact, that's why those called "hyperbolic" functions!).

    If was not "easy" :biggrin: to see that [itex]\frac{e^t+ e^{-t}}{2}= cosh(t)[/itex] and that [itex]\frac{e^t- e^{-t}}{2}= sinh(t)[/itex], then it is a little harder: (et+ e-t)2= e2t+ 2+ e-2t. (You see how the et and e-t cancel when you mutiply them?)
    Similarly, (et-et)2= e2t+ 2+ e-2t. Subtract those and you just get 4 (which is why you need "2" in the denominators).
  6. Aug 27, 2005 #5


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    Halls already knew where this was going; that's how he figured out to do it that way. :smile:

    If I had not already known where this was going, the first thing I would do is to recognize that you have two things of this form:

    x = A + B
    y = A - B

    (for your actual problem, A = exp(t) / 2 and B = exp(-t) / 2)

    You might recognize that it's easy to isolate A and B by adding and subtracting the equations respectively. If not, you might recognize this is a system of two linear equations in A and B, which can then be solved by A and B.

    So, in any case, the first thing I would have done is exactly that:

    x + y = 2A
    x - y = 2B


    exp(t) = x + y
    exp(-t) = x - y

    And then, it should be clear how to get rid of t!
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