- #1

notgoodatphysics

- 8

- 0

$$(y-a)^2 + (x-a)^2 = r^2$$

$$\frac{dy}{dx} = \frac{a-x}{(r^{2}-(x-a)^2)^{1/2}} = 0$$

So ##x = a## then he subs it into the original equation to get the max/min.

Why does ##x = a## give the points of minima/ maxima if we didn’t know it was a circle in this case? Is there a specific rule? It’s not really explained well in the book.

It’s from calculus made easy and he says “Since no value whatever of x will make the denominator infinite, the only condition to give zero is x = a”

I’m pretty confused what is meant by that statement.