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I have $f(x,y)=x^2-y^2$ and need to find absolute values in the region defined by $x^2+y^2 \le 1$

I have a solution that my teacher gave me, where $r(t)= (\cos t, \sin t)$ with $0 \le t \le 2\pi$

$$f (r (t))= g (t)=\cos^2 (t)-\sin^2 (t)$$

And its derivative: $$g'(t)=-4 \cos (t) \sin (t)$$

Trying to find critical points:

$$-4 \cos(t) \sin (t) = 0$$ Then we get that cosine will be $0$ or sine will be $0$.

If $\cos (t)=0 $ then $t=\pi/2$ or $t=3\pi/2$ . Then points will be $(0;1)$ and $(0;-1)$

If $\sin (t)=0$ then $t=0$ or $t=2\pi$ but those are not within the region (why??), or $t=\pi$ and point is $(-1;0)$

I don't understand how exactly he got those $3$ points. Also, why he says $t=0$ and $t=\pi$ are not in the region. Thanks.

Floella
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2 Answers2

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Use polar coordinates we have $x = r\cos \theta, y = r\sin \theta$ with $0 \le r \le 1, 0 \le \theta \le 2\pi$. Thus $f(x,y) = x^2 - y^2 $ becomes $f(r,\theta) = r^2\cos(2\theta)$. We see that $f_{\text{min}} = -1, f_{\text{max}} = 1$ occur at $r = 1, \theta = \dfrac{\pi}{2}, \dfrac{3\pi}{2}$, and $r = 1, \theta = 0,\pi$ respectively.

DeepSea
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  • But he's not using polar coordinates in the solution he gave us. How did he get to those points without them? – Floella Dec 04 '16 at 00:09
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Hint: $$f(0,y)\leq f(x,y)\leq f(x,0).$$

So $f(0,\pm 1)=\min_{-1\leq y\leq 1}f(0,y)=\min_{x^2+y^2\leq1}f(x,y),\text{ and}$ $\max_{x^2+y^2\leq1}f(x,y)=\max_{-1\leq x\leq 1}f(x,0)=f(\pm1,0).$