Associated Topics || Dr. Math Home || Search Dr. Math

### 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
```

```
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
g(x).

-Doctor Pete,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

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

Is this what you were thinking about?:

f(x)=sqrt[1-(sqrt(-x))^4]

This would give me a graph in quadrant II.

f(x)=-sqrt[1-(sqrt(x))^4]

```

```
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!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Functions

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search