Solution 6.

See text and/or instructor's solution manual.

Answer.   Let  [Graphics:../Images/JoukowskiTransModHome_gr_293.gif]  be fixed, then the circle   [Graphics:../Images/JoukowskiTransModHome_gr_294.gif]   in the z-plane

is mapped onto the cardioid   [Graphics:../Images/JoukowskiTransModHome_gr_295.gif]   in the w-plane.  

Solution.   The equation of a circle in the [Graphics:../Images/JoukowskiTransModHome_gr_296.gif]-plane with center [Graphics:../Images/JoukowskiTransModHome_gr_297.gif] is given in polar coordinates  [Graphics:../Images/JoukowskiTransModHome_gr_298.gif]  by the formula  

                    [Graphics:../Images/JoukowskiTransModHome_gr_299.gif].  

For the mapping  [Graphics:../Images/JoukowskiTransModHome_gr_300.gif]  we have  [Graphics:../Images/JoukowskiTransModHome_gr_301.gif],  and  [Graphics:../Images/JoukowskiTransModHome_gr_302.gif].  

Use the substitutions  [Graphics:../Images/JoukowskiTransModHome_gr_303.gif]  and  [Graphics:../Images/JoukowskiTransModHome_gr_304.gif] in   [Graphics:../Images/JoukowskiTransModHome_gr_305.gif] to obtain  

                    [Graphics:../Images/JoukowskiTransModHome_gr_306.gif].

Square both sides and get

                    [Graphics:../Images/JoukowskiTransModHome_gr_307.gif].  

Interesting Fact.   

        To obtain the standard form of a cardioid use the trigonometric identity  [Graphics:../Images/JoukowskiTransModHome_gr_308.gif],  and write

                    [Graphics:../Images/JoukowskiTransModHome_gr_309.gif]  
                              
Now make the substitution [Graphics:../Images/JoukowskiTransModHome_gr_310.gif]  and get the standard form of a cardioid:

                    [Graphics:../Images/JoukowskiTransModHome_gr_311.gif].  

        Furthermore, the transformation  [Graphics:../Images/JoukowskiTransModHome_gr_312.gif]  maps the point   [Graphics:../Images/JoukowskiTransModHome_gr_313.gif]   onto the point   [Graphics:../Images/JoukowskiTransModHome_gr_314.gif].

For the function  [Graphics:../Images/JoukowskiTransModHome_gr_315.gif]   we have   [Graphics:../Images/JoukowskiTransModHome_gr_316.gif]   and   [Graphics:../Images/JoukowskiTransModHome_gr_317.gif].  

Since   [Graphics:../Images/JoukowskiTransModHome_gr_318.gif],   the mapping is not conformal at  [Graphics:../Images/JoukowskiTransModHome_gr_319.gif],  but  [Graphics:../Images/JoukowskiTransModHome_gr_320.gif].  

Hence, by Theorem 10.2 in Section 10.1 the mapping  [Graphics:../Images/JoukowskiTransModHome_gr_321.gif]  magnifies angles at  [Graphics:../Images/JoukowskiTransModHome_gr_322.gif]  by the factor  [Graphics:../Images/JoukowskiTransModHome_gr_323.gif].

        The rays tangent to the circle at  [Graphics:../Images/JoukowskiTransModHome_gr_324.gif]  make an angle  [Graphics:../Images/JoukowskiTransModHome_gr_325.gif]  and  [Graphics:../Images/JoukowskiTransModHome_gr_326.gif],  respectively.

The image rays tangent to the two curves forming the cusp at  [Graphics:../Images/JoukowskiTransModHome_gr_327.gif]  make the angles  [Graphics:../Images/JoukowskiTransModHome_gr_328.gif]  and  [Graphics:../Images/JoukowskiTransModHome_gr_329.gif],  respectively.

Since these angles point in the same direction, the cusp in the cardioid forms an angle of  0°.     

 

We are done.   

 

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/JoukowskiTransModHome_gr_330.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_331.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_332.gif]


[Graphics:../Images/JoukowskiTransModHome_gr_333.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_334.gif]


[Graphics:../Images/JoukowskiTransModHome_gr_335.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_336.gif]


[Graphics:../Images/JoukowskiTransModHome_gr_337.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_338.gif]

Aside.  We can look at the equations.  

[Graphics:../Images/JoukowskiTransModHome_gr_339.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_340.gif]


[Graphics:../Images/JoukowskiTransModHome_gr_341.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_342.gif]


[Graphics:../Images/JoukowskiTransModHome_gr_343.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_344.gif]


[Graphics:../Images/JoukowskiTransModHome_gr_345.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_346.gif]


[Graphics:../Images/JoukowskiTransModHome_gr_347.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_348.gif]


We are really done.   

 

Aside.  We can explore some graphs.

 

                    [Graphics:../Images/JoukowskiTransModHome_gr_349.gif]

                    The circle   [Graphics:../Images/JoukowskiTransModHome_gr_350.gif]   with center  [Graphics:../Images/JoukowskiTransModHome_gr_351.gif].

                    Rays tangent to the circle at the origin make an angle  [Graphics:../Images/JoukowskiTransModHome_gr_352.gif]  and  [Graphics:../Images/JoukowskiTransModHome_gr_353.gif].

                    Remark.  The image of these two rays will be a single ray making an angle  [Graphics:../Images/JoukowskiTransModHome_gr_354.gif].

 

 

                    [Graphics:../Images/JoukowskiTransModHome_gr_355.gif]

                    The cardioid    [Graphics:../Images/JoukowskiTransModHome_gr_356.gif].

                    The ray tangent to the two curves forming the cusp at the origin makes an angle  [Graphics:../Images/JoukowskiTransModHome_gr_357.gif].

 

                    [Graphics:../Images/JoukowskiTransModHome_gr_358.gif]          [Graphics:../Images/JoukowskiTransModHome_gr_359.gif]

                    A "zoom in" of the curve   [Graphics:../Images/JoukowskiTransModHome_gr_360.gif].

                    The ray tangent to the two curves forming the cusp at the origin makes an angle  [Graphics:../Images/JoukowskiTransModHome_gr_361.gif].

 

We are really really done.   

 

Aside.  We can re-label the center of the circle and the angle of the tangent line as shown in Figure 11.62.

 

                    [Graphics:../Images/JoukowskiTransModHome_gr_362.gif]

                    The circle   [Graphics:../Images/JoukowskiTransModHome_gr_363.gif]   with center  [Graphics:../Images/JoukowskiTransModHome_gr_364.gif].

                    Rays tangent to the circle at the origin make an angle  [Graphics:../Images/JoukowskiTransModHome_gr_365.gif]  and  [Graphics:../Images/JoukowskiTransModHome_gr_366.gif].

                    Remark.  The image of these two rays will be a single ray making an angle  [Graphics:../Images/JoukowskiTransModHome_gr_367.gif].

 

 

                    [Graphics:../Images/JoukowskiTransModHome_gr_368.gif]

                    The cardioid    [Graphics:../Images/JoukowskiTransModHome_gr_369.gif].

                    The ray tangent to the two curves forming the cusp at the origin makes an angle  [Graphics:../Images/JoukowskiTransModHome_gr_370.gif].

 

                    [Graphics:../Images/JoukowskiTransModHome_gr_371.gif]          [Graphics:../Images/JoukowskiTransModHome_gr_372.gif]

                    A "zoom in" of the curve   [Graphics:../Images/JoukowskiTransModHome_gr_373.gif].

                    The ray tangent to the two curves forming the cusp at the origin makes an angle  [Graphics:../Images/JoukowskiTransModHome_gr_374.gif].

 

Again, An Interesting Fact.   

 

        To obtain the standard form of a cardioid use the trigonometric identity  [Graphics:../Images/JoukowskiTransModHome_gr_375.gif],  and write

                    [Graphics:../Images/JoukowskiTransModHome_gr_376.gif]  
                              
Now make the substitution [Graphics:../Images/JoukowskiTransModHome_gr_377.gif]  and get the standard form of a cardioid:

                    [Graphics:../Images/JoukowskiTransModHome_gr_378.gif].  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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