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Forums
Mathematics
Calculus
Minima/maxima of a circle
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[QUOTE="notgoodatphysics, post: 6862138, member: 733439"] We have the equation for a circle, and its derivative: $$(y-a)^2 + (x-a)^2 = r^2$$ $$\frac{dy}{dx} = \frac{a-x}{(r^{2}-(x-a)^2)^{1/2}} = 0$$ 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. [/QUOTE]
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Forums
Mathematics
Calculus
Minima/maxima of a circle
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