# Is there a function that

1. Apr 13, 2004

### the1024b

Is there a function that represents a curved line like a circle but that is only represented on the (+,+) zone of the graph?

2. Apr 13, 2004

### Nexus[Free-DC]

Sure. Try this:

$$let x^* = (\sqrt{x})^2$$. This is equal to x for x>0 and undefined for x<0.

Now, your quarter circle is given by:

$$f(x) = \sqrt{1-(x^*)^2}$$

3. Apr 13, 2004

### the1024b

and whats the derivative of that function?

Maybe i should put this on calculus?

4. Apr 14, 2004

### Chen

The derivative is the same as that of a regular circle, except that you use $(\sqrt{x})^2$ instead of just x:

$$f'(x) = -\frac{(\sqrt{x})^2}{\sqrt{1-(\sqrt{x})^4}}$$

5. Apr 14, 2004

### uart

There's no problem in putting restriction on the domain of the function as part of it's definition. Just use,

$$f(x) = +\sqrt{r^2 - x^2} : 0 \le x \le r$$

6. Apr 14, 2004

### matt grime

You could just declare that the function "defining" the circle is undefined for negative x and takes the positive square root to avoid all this competely unnecessary behaviour.

7. Apr 14, 2004

### Integral

Staff Emeritus
Perhaps I am missing something, what is wrong with
$$f(x) = +\sqrt{r^2 - (x-h)^2} +k : h-r \le x \le h+r :k ,h\ge r$$

Last edited: Apr 14, 2004