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If you plot these two relations:

$$y=cos(x)$$ $$x^2 + y^2 = 1$$

The cosine wave seems to "hug" the unit circle shown below. enter image description here

I'm wondering the calculus explanation for this, as I believe it must have to do with the relationship between the implicit derivative of the unit circle, where the slope of the tangent line at any point is found using the negative cotangent function. But this doesn't seem to explain why the unit circle is wrapped around the cosine function like this as the derivative is asymptotic at $\frac{\pi}{2}$. And even then, is there a geometric intuition for why they behave so similarly around 0?

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  • $\begingroup$ When $x=0$ then $y=1$ which is true for both $\endgroup$ – Fawad Mar 22 '17 at 5:16
  • $\begingroup$ @Fawad I totally understand that, I'm saying there has to be a close knit tie to the two derivatives around 0. It's almost like a taylor approximation. $\endgroup$ – rb612 Mar 22 '17 at 5:17
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Two curves $y=f(x)$ and $y=g(x)$ will meet at a value $x=a$ if $f(a)=g(a)$. They will have the same tangent (which you could think of as a "very light hug" if also $f'(a)=g'(a)$. They will hug each other better if also $f''(a)=g''(a)$, which means that the curvatures are equal. This is also sometimes known as "second order contact". They will hug even more closely if also $f'''(a)=g'''(a)$, and so on. In this case you have $a=0$ and $$\displaylines{ f(x)=\cos x\ ,\quad g(x)=(1-x^2)^{1/2}\ ,\quad f(0)=1=g(0)\cr f'(x)=-\sin x\ ,\quad g'(x)=-x(1-x^2)^{-1/2}\ ,\quad f'(0)=0=g'(0)\cr f''(x)=-\cos x\ ,\quad g''(x)=-(1-x^2)^{-3/2}\ ,\quad f''(0)=-1=g''(0)\cr f'''(x)=\sin x\ ,\quad g'''(x)=-3x(1-x^2)^{-5/2}\ ,\quad f'''(0)=0=g'''(0)\cr f^{(4)}(x)=\cos x\ ,\quad g^{(4)}(x)=-3(1+4x^2)(1-x^2)^{-7/2}\ ,\quad f^{(4)}(0)=1\ne g^{(4)}(0)\ .\cr}$$ So the curves have third order contact.

As you commented, it is related to Taylor series. From the above calculations you can write down the Taylor series for $f$ and $g$ and see that they are equal up to third order terms.

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What you have here is what is called an osculating circle: a circle tangent to a curve at a point such that the curve has the same curvature as the circle.

To cut a long story short, for a curve of the form $y=f(x)$, the (unsigned) curvature at the point $(x,f(x))$ is given by $$ \frac{\lvert f''(x) \rvert}{(1+f'(x)^2)^{3/2}}, $$ and the curvature of a circle (i.e. $y=b \pm \sqrt{R^2-(x-a)^2}$) is then easily computed to be $1/R$ (or perhaps more clearly, the curvature is defined as the curvature of osculating circle and the definition that a circle of radius $R$ has curvature $1/R$).

In this case, $$ \cos'0 = \sin{0} = 0, \quad \cos''0 = -\cos{0} = -1, $$ so the curvature of the cosine is $1$, the same as the circle of radius $1$ you have drawn.

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