Self-envelope of oscillatory sinc(x) function

Question

Show that equation of the curve tangentially contacting $ y= \dfrac{\sin x}{x} $ is given by $ x\,y= \pm 1 $ and find coordinates of these contact points.

EDIT1:

Apologies. I did not properly convey my question.. in haste. The above can be treated deleted or kept that way for relevance. Freshly asking the following question:

Just as for example we have precessing elliptical orbits as a result of a single differential equation that carry a self-envelope ( a circle touching all perigees) , I imagined that the differential equation of sinc(x) should have a procedure of capturing its own envelope/envelopes. So actually I wanted to be able to ask how to find self-envelopes given its derivative/DE:

$$ \dfrac{dy}{dx}= \dfrac{\cos x}{x}-\dfrac{\sin x }{x^2} $$

But how now to take off.. we have no single parameter to eliminate and so on.

Regards

Narasimham

Answer 1

HINT

According to the suggestions given in the comments by Jack and Arthur, we need to check that

• $\frac{\sin x}x=\pm \frac1x$ determine intersection points

• at that point the two curves are tangent (by derivative)

Answer 2

As said by others, the "tangentially contacting" curve is not unique, and we will just show that $ xy=\pm1$ is a possible solution.

We express the tangency condition by equating the function values and first derivative values:

$$\begin{cases}\dfrac{\sin x}{x}=\pm\dfrac1x, \\\dfrac{x\cos x-\sin x}{x^2}=\mp\dfrac1{x^2}.\end{cases}$$

We can reject the point $x=0$. Then the first equation gives $\color{green}{\sin x=\pm1}$, which implies $\cos x=0$. With this, the second equation is fulfilled.

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Variant:

The equations of the two curves can be written in the forms

$$xy=\sin x,\\xy=\pm1$$ and the equality of the derivatives of $y$ is implied by the equality of the derivatives of $xy$ (as $(xy)'=y+xy'$). The system is just

$$\begin{cases}\sin x=\pm1, \\\cos x=0.\end{cases}$$

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Update:

Turn the equation to that of a family by introducing a phase shift, and eliminate this parameter:

$$\begin{cases}xy=\sin(x+t),\\0=\cos(x+t)\end{cases}$$ gives you an envelope:

$$xy=\pm1.$$

A rescaling

$$xy=\sin(tx)$$ would also work.

But other parameterizations can give other envelopes.