Exercise 13.  For the temperature  [Graphics:Images/TemperaturesModHome_gr_625.gif]  in the upper half-disk  [Graphics:Images/TemperaturesModHome_gr_626.gif],  

show that the isothermals   [Graphics:Images/TemperaturesModHome_gr_627.gif],   for  [Graphics:Images/TemperaturesModHome_gr_628.gif],   are portions of circles that pass through the points  [Graphics:Images/TemperaturesModHome_gr_629.gif]  and  [Graphics:Images/TemperaturesModHome_gr_630.gif],  as illustrated in Figure 11.33.  

Solution 13.

See text and/or instructor's solution manual.

Solution.   Start with the equation   [Graphics:../Images/TemperaturesModHome_gr_631.gif]   then get  

                    [Graphics:../Images/TemperaturesModHome_gr_632.gif]    

Now substitute  [Graphics:../Images/TemperaturesModHome_gr_633.gif]  and obtain

                    [Graphics:../Images/TemperaturesModHome_gr_634.gif]   

Therefore, the isothermals  [Graphics:../Images/TemperaturesModHome_gr_635.gif]  are portions of circles that pass through the points  [Graphics:../Images/TemperaturesModHome_gr_636.gif]  and  [Graphics:../Images/TemperaturesModHome_gr_637.gif].  

 

We are done.   

 

Aside.  We can let Mathematica double check our work.

In the equation  [Graphics:../Images/TemperaturesModHome_gr_638.gif]   replace   [Graphics:../Images/TemperaturesModHome_gr_639.gif].

[Graphics:../Images/TemperaturesModHome_gr_640.gif]

[Graphics:../Images/TemperaturesModHome_gr_641.gif]

Now solve for  Q.

[Graphics:../Images/TemperaturesModHome_gr_642.gif]

[Graphics:../Images/TemperaturesModHome_gr_643.gif]

Now replace   [Graphics:../Images/TemperaturesModHome_gr_644.gif]   and  [Graphics:../Images/TemperaturesModHome_gr_645.gif]

[Graphics:../Images/TemperaturesModHome_gr_646.gif]

[Graphics:../Images/TemperaturesModHome_gr_647.gif]


[Graphics:../Images/TemperaturesModHome_gr_648.gif]

[Graphics:../Images/TemperaturesModHome_gr_649.gif]


[Graphics:../Images/TemperaturesModHome_gr_650.gif]

[Graphics:../Images/TemperaturesModHome_gr_651.gif]


We are really done.   

 

Aside.   We can graph the function   [Graphics:../Images/TemperaturesModHome_gr_652.gif].   

                     [Graphics:../Images/TemperaturesModHome_gr_653.gif]

                    A contour graph of the function   [Graphics:../Images/TemperaturesModHome_gr_654.gif],

                    where  [Graphics:../Images/TemperaturesModHome_gr_655.gif]   for   [Graphics:../Images/TemperaturesModHome_gr_656.gif].

 

We are really really done.   

 

                     [Graphics:../Images/TemperaturesModHome_gr_657.gif]

                    A graph of the function   [Graphics:../Images/TemperaturesModHome_gr_658.gif].

                     [Graphics:../Images/TemperaturesModHome_gr_659.gif]

                    A graph of the function   [Graphics:../Images/TemperaturesModHome_gr_660.gif].

                     [Graphics:../Images/TemperaturesModHome_gr_661.gif]

                     Another way to view   [Graphics:../Images/TemperaturesModHome_gr_662.gif].

                     [Graphics:../Images/TemperaturesModHome_gr_663.gif]

                    A graph of the function   [Graphics:../Images/TemperaturesModHome_gr_664.gif].

 

















 

This solution is complements of the authors.

 



































 

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