Page:Calculus Made Easy.pdf/123

knowledge; let us set about finding what value of $x$ will make $y$ a maximum or a minimum, by the process of differentiating and equating to zero.

${\frac {dy}{dx}}={\frac {1}{2}}{\frac {1}{\sqrt {r^{2}-(x-a)^{2}}}}\times {(2a-2x)}$,

which reduces to

${\frac {dy}{dx}}={\frac {a-x}{\sqrt {r^{2}-(x-a)^{2}}}}$.

Then the condition for $y$ being maximum or minimum is:

${\frac {a-x}{\sqrt {r^{2}-(x-a)^{2}}}}=0$.

Since no value whatever of $x$ will make the denominator infinite, the only condition to give zero is

$x=a$.

Inserting this value in the original equation for the circle, we find

$y={\sqrt {r^{2}}}+b$;

and as the root of $r^{2}$ is either $+r$ or $-r$, we have two resulting values of $y$,

${\begin{cases}y=b+r\\y=b-r.\end{cases}}$

The first of these is the maximum, at the top; the second the minimum, at the bottom.