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Quarter Circles

Date: 12/20/96 at 22:07:38
From: Jason k Cheung
Subject: Functions

How would I make a function that would graph a quarter of a circle in 
the quadrants II and IV?

Date: 12/21/96 at 03:33:24
From: Doctor Pete
Subject: Re: Functions

One possible way to do this is as follows:  We know we can draw a 
semicircle in quadrants I and II with:

     f(x) = +Sqrt[1-x^2]

(i.e., the postive square root of 1 minus x squared).  This gives the 
correct graph in quadrant II, but we would like to "flip" the graph 
about the x-axis for x > 0.  That is, if g(x) is our desired function, 
then g(x) = -f(x) for x > 0, and g(x) = f(x) for x < 0.

To do this, consider the function h(x) = |x|/x, that is, the absolute 
value of x divided by x.  For x > 0, this is 1, and for x < 0, this 
is -1. (Question:  What is h(0)?  This is of interest, because your 
answer to this will affect what g(0) will be.)

Now, I leave it to you to figure out how to apply h(x) to f(x) to get 

-Doctor Pete,  The Math Forum
 Check out our web site!   

Date: 12/24/96 at 15:10:41
From: Jason k Cheung
Subject: Re: Functions

Is this what you were thinking about?:


This would give me a graph in quadrant II.


This would graph quadrant IV.

Date: 12/28/96 at 20:09:59
From: Doctor Pete
Subject: Re: Functions

Hmm....  while I might agree that the equations you have given are
technically correct, I think I should point out a few things.  

First, note that Sqrt[-x]^2 = Sqrt[x]^2 = |x|, the absolute value 
of x.  Then Sqrt[-x]^4 = Sqrt[x]^4 = |x|^2 = x^2, because the square 
of a number is always non-negative.  So there is no sense in 
differentiating between the two; that is, for all values of x:

Sqrt[1-Sqrt[-x]^4] = Sqrt[1-Sqrt[x]^4] = Sqrt[1-x^2] 

Second, the purpose of introducing the function h(x) = |x|/x is to try 
to do what you have done by splitting up g(x) into two cases.  (I will 
call g(x) the function that we wish to obtain, and f(x) = Sqrt[1-x^2] 
as I have defined in the previous message.)  In order to get the 
desired function, we could have simply written:

             /   f(x)  =   Sqrt[1-x^2]  ,  x < 0
     g(x) = -|
             \  -f(x)  =  -Sqrt[1-x^2]  ,  x > 0

This definition is not always desirable. You can use h(x) to do what 
the minus sign is doing in the second case, because h(x) = 1 when 
x > 0, and -1 when x < 0.  So -h(x) = -1 when x > 0, and -h(x) = 1 
when x < 0.  This gives you the correct sign, so in order to merge the 
two cases into one, we take the product -f(x)h(x), which gives:

     g(x) = -f(x)h(x) = -Sqrt[1-x^2] |x|/x

This is the function we're looking for.  If x < 0, then |x|/x = -1, 
and g(x) = Sqrt[1-x^2].  If x > 0, then |x|/x = 1, and 
g(x) = -Sqrt[1-x^2].

Also note that g(x), the function we were looking for, is 
discontinuous at 0, and is defined on the domain [-1,0) U (0,1].  
f(x) = Sqrt[1-x^2], however, is defined, and continuous, on the entire 
interval [-1,1].  So obtaining g(x) from f(x) requires the 
introduction of a function h(x) that has a discontinuity at 0.  
Indeed, h(x) = |x|/x is not continuous at 0, however you may wish to 
define h(0).

-Doctor Pete,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Basic Algebra
High School Functions

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