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Vector analysis.

  1. Jul 23, 2009 #1
    1. Sketch the plane curve represented by the vector-valued function r(t)=cosh ti +sinh tj on the interval 0</(trying to say less then or equal to)t</(also less then or equal to)5. Show that the rectangular equation corresponding to r(t) is the hyperbola x^2-y^2=1. Verify your sketch using a graphing utility to graph the hyperbola.

    2. Let P=(coshφ,sinhφ) be the point on the hyperbola corresponding to r(φ) for φ>0. use the formula for the area

    A= 1/2(integral)xdy-ydx
    I don't know where to begin, can someone please guide me in the right direction.
     
  2. jcsd
  3. Jul 23, 2009 #2

    HallsofIvy

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    These look a lot like homework problems. Why were they not posted there?

    These are labeled "1" and "2". Are they 2 separate questions? "Use the formula for area" to do what?

    Presumably you know that [itex]cos^2 x+ sin^2 x= 1[/itex]. Do you know the equivalent formula for cosh and sinh?
     
  4. Jul 23, 2009 #3
    (1)Make x equal to the horizontal component of r(t) and y the vertical component.
    Remember that cosh^2(t)-sinh^2(t)=1.


    (2) Looks like a line integral, which path are you integrating with? If it is is r(t) from above then substitute x with the horizontal part of r(t) and y with the vertical and relate dx and dy to dt.
     
  5. Jul 23, 2009 #4
    thanks pinu
     
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