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The question in the textbook gives the graph of $y^2=\sin{\frac{\pi}{2}x}$ which looks like this. enter image description here

It then asks to prove that although the graph looks like it is made from circles, it is not. I'm not sure how to prove this exactly. I tried to show this by finding the derivative which is $\frac{dy}{dx}=\frac{\pi\cos{(\pi/2)x}}{4y}$. Then I thought since the numerator isn't a linear function it wouldn't be a circle but I'm not sure if that conclusion makes any sense. Could someone clarify how could I disprove that the graph is made of circles?

Pen and Paper
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  • your work is correct – YOu will not know Nov 26 '21 at 11:10
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    $y^2=\sin\dfrac{\pi}{2}x$. Consider closed part of graph for $x\in[0;2]$. Minimum $x$ is 0 maximum $x$ is 2. Minimum $y$ is -1, maximum $y$ is 1. If it was circle then its center must be $(1;0)$ and radius must be $1$, then its equation must be $y^2+(x-1)^2=1\Rightarrow y^2=2x-x^2$. – Ivan Kaznacheyeu Nov 26 '21 at 11:17
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    Other way to prove is to consider 4 points of closed part of graph, which form no inscribed quadrilateral. For example: $(0,0)$, $(1,1)$, $(2,0)$, $(1/2,1/2^{1/4})$. – Ivan Kaznacheyeu Nov 26 '21 at 11:24
  • Probably the simplest way is just to notice that if it was a circle then the coordinates would have to satisfy the equation of a circle. For example, for $0 \leq x \leq 2,$ we need every point $(x, y)$ on the curve to satisfy some equation like $(x-1)^2+y^2=1.$ But we have an expression for $y^2$ as a function of $\sin$, so this implies that $\sin$ is a polynomial, which is obviously false because $\sin$ has infinitely many roots. – Will R Nov 26 '21 at 11:50

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Here's a more geometric approach. The pieces are identical since $\sin\left(\frac{\pi}{2}x\right)$ is periodic so it's sufficient to show any one of them is not a circle. Consider $x\in[0,2]$ and suppose to the contrary that the graph were a circle with center $O$. Points $A=(0,0)$ and $B=(2,0)$ are on the graph so $O$ must lie on the perpendicular bisector of segment $AB$, which is the line $x=1$. Similarly, $C=(1,1)$ and $D=(1,-1)$ are on the graph so O must be on the line $y=0$, the perpendicular bisector of segment $CD$. Thus, the lines $x=1$ and $y=0$ intersect at $O$, i.e. $O=(1,0)$, and the radius is equal to $OA=1$. Point $E=\left(\frac 13,\frac{1}{\sqrt 2}\right)$ is on the graph as well. From the distance formula, we have $$OE=\sqrt{\left(1-\frac 13\right)^2+\left(\frac{1}{\sqrt 2}\right)^2 }=\sqrt{\frac{17}{18}}\ne 1,$$ which is a contradiction.

bjorn93
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