Extremum Problem

Question

How big is the circle?

My first steps:

$ x^2 + y^2 =r^2$
$ f(x)=y=e^{-x^2}$

Substitute $y^2$ in $x^2 + y^2 =r^2.$
So, $x^2 + e^{-2x^2} =r^2$

Is this way correct? Because after calculating the first derivative $2x -4xe^{-2x^2}$ and so on my solution at the end is $x^2 = \dfrac {\ln(2)}{2}, $ and this is not correct.

@LarsBaumann: It looks as though you may simply be mis-interpreting what you've got so far. Your result for $x^2$ isn't the radius of the circle. It's the square of the $x$ component of the intersection point. You still have to calculate $\sqrt{x^2+y^2}$ to get the radius, if that's what your question is asking. — Adrian Keister, Aug 30 '18 at 15:45
Based on your comment to Mostafa Ayaz’ answer, it appears that the problem is asking you to find the area of the circle, not its radius. — amd, Aug 30 '18 at 20:14

Mostafa Ayaz · Answer 1 · 2018-08-30T17:02:36.093

1

Your solution is right. Clearly the circle and the curve meet at $(x,y)$ in first quadrant where $$x^2+y^2=r^2\\y=e^{-x^2}\\-2xe^{-x^2}=\dfrac{-x}{\sqrt{r^2-x^2}}$$which means that $$2y=\dfrac{1}{\sqrt {r^2-x^2}}=\dfrac{1}{y}$$therefore $$y=\dfrac{\sqrt 2}{2}$$and $$x=\sqrt{\dfrac{\ln 2}{2}}$$which means that $$r=\sqrt{\dfrac{1+\ln 2}{2}}\approx 0.9200943377$$

edited Aug 30 '18 at 17:02

answered Aug 30 '18 at 15:21

Mostafa Ayaz

31,924

Should it not be $$y=\pm\frac{1}{2}$$ etc? – Dr. Sonnhard Graubner Aug 30 '18 at 15:25
@Dr.SonnhardGraubner Mostafa's answer specifies that the point being considered is in the first quadrant. – Billy Kalfus Aug 30 '18 at 15:40
The right answer for $r^2$ is $ \frac {\pi}{2} * (ln(2)+1)$ , but i can´t find a way to solve this. – Lars Baumann Aug 30 '18 at 16:05
@LarsBaumann That’s the area of the circle, not its radius squared. – amd Aug 30 '18 at 20:12

Extremum Problem

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