Question

Find the surface whose tangent plane cut off an intercept of constant length k from the axis z

Answer 1

Let $f(x,y,z)=0$ be the equation of the surface, $P(x_P, y_P, z_P)$ - arbitrary point on it.

The equation of the tangent plane at $P$ is

$\frac{\partial f}{\partial x}(P)(x-x_P) + \frac{\partial f}{\partial y}(P)(y-y_P)+\frac{\partial f}{\partial z}(P)(z-z_P)=0$

This plane intersects the axis z at the point $Q=(0,0,z_Q)$ , and by the condition $z_Q=k$ is independent on $P$ .

Since $Q$ lies on the tangent plane, then

$\frac{\partial f}{\partial x}(P)(-x_P) + \frac{\partial f}{\partial y}(P)(-y_P)+\frac{\partial f}{\partial z}(P)(k-z_P)=0$

Consider cylindrical coordinates $(r, \theta, z)$ . Then $x=r\cos\theta, y=r\sin\theta$ ,

$\frac{\partial}{\partial r}(x,y,z)=\frac{\partial}{\partial r}(r\cos\theta, r\sin\theta, z)=(\cos\theta, \sin\theta, 0) = (\frac{x}{r}, \frac{y}{r}, 0)$ .

Hence

$\frac{\partial}{\partial r}=\frac{x}{r}\frac{\partial}{\partial x} + \frac{y}{r}\frac{\partial}{\partial y}$

$x \frac{\partial}{\partial x} + y \frac{\partial}{\partial y}=r \frac{\partial}{\partial r}$

Therefore, we can rewrite the obtained equation as follows:

$\frac{\partial f}{\partial z}(P)(k-z_P)=\frac{\partial f}{\partial x}(P)x_P + \frac{\partial f}{\partial y}(P)y_P =r \frac{\partial f}{\partial r}(P)$

The first partial solution to this equation is, evidently, the coordinate function $\theta$.

However, it is a multi-valued function on $\mathbb{R}^3 \setminus Oz$ . So, we should take $f_1 = e^{i\theta}$ instead of it.

The second partial solution can be found of the form $f_2 = z + g(r)$, where $g(r)$ is unknown function.

$r \frac{\partial f_2}{\partial r} = rg'(r)$

$k \frac{\partial f_2}{\partial r} = k$

Hence $rg'(r)=k$ and $g(r)= k\log r$ (partial solution).

The functions $f_1$ and $f_2$ are functionally independent, therefore the general solution can be written as

$f(x,y,z) = \Phi(f_1, f_2)= \Phi(e^{i\theta}, z+k\log r)$ , where $\Phi$ is arbitrary smooth function on $\mathbb{R}^3 \setminus Oz$ .

Answer. $f(r, \theta, z) = \Phi(f_1, f_2)= \Phi(e^{i\theta}, z+k\log r)$ (in cylindrical coordinates)