Vector Analysis: Sketching Hyperbola x^2-y^2=1 with Vector-Valued Function

[mikemichiel]

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.

 

[HallsofIvy]

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 cos^2 x+ sin^2 x= 1. Do you know the equivalent formula for cosh and sinh?

 

[Pinu7]

(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.

 

[mikemichiel]

thanks pinu