Exercise 10.  Find the image of the vertical strip   [Graphics:Images/MapTrigonometricFunModHome_gr_706.gif]   under the mapping   [Graphics:Images/MapTrigonometricFunModHome_gr_707.gif].

Solution 10.

Answer.   The image of the vertical strip   [Graphics:../Images/MapTrigonometricFunModHome_gr_708.gif]   under the mapping   [Graphics:../Images/MapTrigonometricFunModHome_gr_709.gif]  

is the right half-plane   [Graphics:../Images/MapTrigonometricFunModHome_gr_710.gif]   slit along the ray   [Graphics:../Images/MapTrigonometricFunModHome_gr_711.gif].

Hint.   Extend the results in Example 10.13.  You can use the information in Figure 10.17.

Short Solution.  The image of the vertical strip  [Graphics:../Images/MapTrigonometricFunModHome_gr_712.gif]  under  [Graphics:../Images/MapTrigonometricFunModHome_gr_713.gif]  is the vertical strip  

[Graphics:../Images/MapTrigonometricFunModHome_gr_714.gif].   Then the image of the vertical strip   [Graphics:../Images/MapTrigonometricFunModHome_gr_715.gif]   under the mapping   

[Graphics:../Images/MapTrigonometricFunModHome_gr_716.gif]  is the right half-plane   [Graphics:../Images/MapTrigonometricFunModHome_gr_717.gif]   slit along the ray   [Graphics:../Images/MapTrigonometricFunModHome_gr_718.gif].   

Therefore, the image of the vertical strip   [Graphics:../Images/MapTrigonometricFunModHome_gr_719.gif]   under the mapping   [Graphics:../Images/MapTrigonometricFunModHome_gr_720.gif]   is

the right half-plane   [Graphics:../Images/MapTrigonometricFunModHome_gr_721.gif]   slit along the ray   [Graphics:../Images/MapTrigonometricFunModHome_gr_722.gif].

We might be done.   

Solution.   Use the trigonometric identity  [Graphics:../Images/MapTrigonometricFunModHome_gr_723.gif].   The mapping   [Graphics:../Images/MapTrigonometricFunModHome_gr_724.gif]   can be written as a composition  

                    [Graphics:../Images/MapTrigonometricFunModHome_gr_725.gif],    

where   

                    [Graphics:../Images/MapTrigonometricFunModHome_gr_726.gif],    and   

                    [Graphics:../Images/MapTrigonometricFunModHome_gr_727.gif].

        The mapping   [Graphics:../Images/MapTrigonometricFunModHome_gr_728.gif]   translates points in the plane to the right by the amount  [Graphics:../Images/MapTrigonometricFunModHome_gr_729.gif].  

        Hence, the image of the vertical strip  [Graphics:../Images/MapTrigonometricFunModHome_gr_730.gif]  under  [Graphics:../Images/MapTrigonometricFunModHome_gr_731.gif]  is the vertical strip  [Graphics:../Images/MapTrigonometricFunModHome_gr_732.gif]  .  

        Next we can use Equation (5-33) to write  

                    [Graphics:../Images/MapTrigonometricFunModHome_gr_733.gif].  

If  [Graphics:../Images/MapTrigonometricFunModHome_gr_734.gif],  then the image of the vertical line  [Graphics:../Images/MapTrigonometricFunModHome_gr_735.gif]  is the curve in the w plane given by the parametric equations  

                    [Graphics:../Images/MapTrigonometricFunModHome_gr_736.gif]    and    [Graphics:../Images/MapTrigonometricFunModHome_gr_737.gif],    for    [Graphics:../Images/MapTrigonometricFunModHome_gr_738.gif].  

The inequalities  [Graphics:../Images/MapTrigonometricFunModHome_gr_739.gif]  imply that the image points lie in the right half-plane  [Graphics:../Images/MapTrigonometricFunModHome_gr_740.gif].  

Eliminate Y from the parametric equations and the result is  [Graphics:../Images/MapTrigonometricFunModHome_gr_741.gif].  

When [Graphics:../Images/MapTrigonometricFunModHome_gr_742.gif] the image curve approaches the v-axis where  [Graphics:../Images/MapTrigonometricFunModHome_gr_743.gif].  

When [Graphics:../Images/MapTrigonometricFunModHome_gr_744.gif] the image curve approaches the ray  [Graphics:../Images/MapTrigonometricFunModHome_gr_745.gif].  

        Hence, the image of the vertical strip  [Graphics:../Images/MapTrigonometricFunModHome_gr_746.gif]  under the mapping  [Graphics:../Images/MapTrigonometricFunModHome_gr_747.gif]  

is the right half-plane  [Graphics:../Images/MapTrigonometricFunModHome_gr_748.gif]  slit along the ray  [Graphics:../Images/MapTrigonometricFunModHome_gr_749.gif].

        Therefore, the image of the vertical strip   [Graphics:../Images/MapTrigonometricFunModHome_gr_750.gif]   under the mapping   [Graphics:../Images/MapTrigonometricFunModHome_gr_751.gif]

is the right half-plane   [Graphics:../Images/MapTrigonometricFunModHome_gr_752.gif]   slit along the ray   [Graphics:../Images/MapTrigonometricFunModHome_gr_753.gif].

We are done.   

Aside.  We can look at some graphs of the mapping  [Graphics:../Images/MapTrigonometricFunModHome_gr_754.gif].

 

[Graphics:../Images/MapTrigonometricFunModHome_gr_755.gif]  [Graphics:../Images/MapTrigonometricFunModHome_gr_756.gif]  [Graphics:../Images/MapTrigonometricFunModHome_gr_757.gif]

 

[Graphics:../Images/MapTrigonometricFunModHome_gr_758.gif]  [Graphics:../Images/MapTrigonometricFunModHome_gr_759.gif]  [Graphics:../Images/MapTrigonometricFunModHome_gr_760.gif]

          The mapping   [Graphics:../Images/MapTrigonometricFunModHome_gr_761.gif],   followed by the mapping   [Graphics:../Images/MapTrigonometricFunModHome_gr_762.gif].

          Observe the points [Graphics:../Images/MapTrigonometricFunModHome_gr_763.gif] and their images  [Graphics:../Images/MapTrigonometricFunModHome_gr_764.gif]  and  [Graphics:../Images/MapTrigonometricFunModHome_gr_765.gif]  

          [Graphics:../Images/MapTrigonometricFunModHome_gr_766.gif]          [Graphics:../Images/MapTrigonometricFunModHome_gr_767.gif]          [Graphics:../Images/MapTrigonometricFunModHome_gr_768.gif]  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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