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I have tried to find envelope for $$x \sin \theta - y \cos \theta + z = a \theta$$ First I find derivative w.r.t. $\theta$ $$F(\theta)=x \sin \theta - y \cos \theta + z - a \theta = 0$$ $$\frac{\partial F(\theta)}{\partial \theta}=y \sin \theta + x \cos \theta - a = 0$$ Then by solving these two above equation in order to eliminate parameter $\theta$, I find values of $\sin \theta$ and $\cos \theta$ $$\sin \theta = \frac{ax\theta + ay - xz}{x^2 + y^2}$$ and $$\cos \theta = \frac{ax - ay\theta + yz}{x^2 + y^2}$$ Finally to find value of $\theta$ I squares and add above two equations then I get a quadratic equation in $\theta$ as $$a^2\theta^2 - 2az\theta + a^2 + z^2 - x^2 - y^2=0$$ Then solving this for $\theta$ by applying quadratic formula, I get condition for real values of $\theta$ i.e.

$\theta$ is real only if $$x^2+y^2 \ge a^2$$

Now here is my question. Either $x^2+y^2 \ge a^2$ is required envelope or something more to do? Because $x^2+y^2 \ge a^2$ is an equality in which we have eliminated parameter $\theta$.

Syed Muhammad Asad
  • 485

1 Answers1

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Hint:

You get rid of the annoying trigonometric functions by

$$(z-a\theta)^2+a^2=x^2+y^2.$$

From this you draw $\theta$, which you plug in one of the initial equations. Not a really nice expression.

Alternatively, in cylindrical coordinates $(\phi,\rho)$ we rewrite

$$\begin{cases}\rho\sin(\phi-\theta)=z-a\theta,\\\rho\cos(\phi-\theta)=a.\end{cases}$$

Then

$$\phi=\theta+\arccos\frac a\rho=\frac{z\pm\sqrt{\rho^2-a^2}}a+\arccos\frac a\rho.$$