Exercise 3.  Find the image of the vertical line  [Graphics:Images/MapTrigonometricFunModHome_gr_293.gif]  under the transformation   [Graphics:Images/MapTrigonometricFunModHome_gr_294.gif].  

Solution 3.

Answer.   The image of   [Graphics:../Images/MapTrigonometricFunModHome_gr_295.gif]   under   [Graphics:../Images/MapTrigonometricFunModHome_gr_296.gif]   is the right branch of the hyperbola  [Graphics:../Images/MapTrigonometricFunModHome_gr_297.gif].  

Hint.   Use ideas found in Example 10.13.  

Short Solution.   Substitute   [Graphics:../Images/MapTrigonometricFunModHome_gr_298.gif]   into equation (10-23)  [Graphics:../Images/MapTrigonometricFunModHome_gr_299.gif]   

and get the hyperbola    [Graphics:../Images/MapTrigonometricFunModHome_gr_300.gif]   which reduces to    [Graphics:../Images/MapTrigonometricFunModHome_gr_301.gif].

We might be done.   

Solution.   Using Equation (5-33), we write  

                    [Graphics:../Images/MapTrigonometricFunModHome_gr_302.gif].  

The image of the vertical line  [Graphics:../Images/MapTrigonometricFunModHome_gr_303.gif]  is the curve in the w plane given by the parametric equations  

                    [Graphics:../Images/MapTrigonometricFunModHome_gr_304.gif],    for  [Graphics:../Images/MapTrigonometricFunModHome_gr_305.gif].  

Next, we rewrite these equations as  

                    [Graphics:../Images/MapTrigonometricFunModHome_gr_306.gif].  

Eliminate y from these equations by squaring and using the hyperbolic identity  [Graphics:../Images/MapTrigonometricFunModHome_gr_307.gif].  

The result is the single equation    [Graphics:../Images/MapTrigonometricFunModHome_gr_308.gif].   

Now use   [Graphics:../Images/MapTrigonometricFunModHome_gr_309.gif]   and get   [Graphics:../Images/MapTrigonometricFunModHome_gr_310.gif].  

        Therefore, the image of the vertical line   [Graphics:../Images/MapTrigonometricFunModHome_gr_311.gif]   under the mapping   [Graphics:../Images/MapTrigonometricFunModHome_gr_312.gif]   

is the right branch of the hyperbola   [Graphics:../Images/MapTrigonometricFunModHome_gr_313.gif].  

We are done.   

Aside.  We can look at some graphs of the mapping  [Graphics:../Images/MapTrigonometricFunModHome_gr_314.gif].

 

                    [Graphics:../Images/MapTrigonometricFunModHome_gr_315.gif]          [Graphics:../Images/MapTrigonometricFunModHome_gr_316.gif]

                      The image of  [Graphics:../Images/MapTrigonometricFunModHome_gr_319.gif]  under  [Graphics:../Images/MapTrigonometricFunModHome_gr_320.gif]  is the right branch of the hyperbola  [Graphics:../Images/MapTrigonometricFunModHome_gr_321.gif].  

 

 

 

 

 

                    [Graphics:../Images/MapTrigonometricFunModHome_gr_317.gif]          [Graphics:../Images/MapTrigonometricFunModHome_gr_318.gif]

                    The image of  [Graphics:../Images/MapTrigonometricFunModHome_gr_319.gif]  under  [Graphics:../Images/MapTrigonometricFunModHome_gr_320.gif]  is the right branch of the hyperbola  [Graphics:../Images/MapTrigonometricFunModHome_gr_321.gif].  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) 2008 John H. Mathews, Russell W. Howell