1
$\begingroup$

Consider a smooth surface given by the function $z = f(x,y)$, such that the partial derivatives of $f(x,y)$ exist. Suppose $Q$ is a point that does not lie on the surface, and $P$ is the nearest point on the surface to $Q$. Show that the line through $P$ and $Q$ is perpendicular to the surface at $P$.

2 Answers 2

0

Let $Q=(\alpha,\beta,\gamma)$. For any point $S=(x,y,f(x,y))$ on the surface define the function

$$g(x,y)=(x-\alpha)^2+(y-\beta)^2+(f(x,y)-\gamma)^2$$

The function $g(x,y)$ represents the square of the distance between $S$ and $Q$.

Let $P=(x_0,y_0,f(x_0,y_0)$ denote the nearest point the surface $Q$.

Then $g(x_0,y_0)$ is the minimum value of $g(x,y)$.

Now $$g_x(x,y)=2(x-\alpha)+2(f(x,y)-\gamma)f_x(x,y)$$ $$g_y(x,y)=2(y-\beta)+2(f(x,y)-\gamma)f_y(x,y)$$

so $$0=2(x_0-\alpha)+2(f(x_0,y_0)-\gamma)f_x(x_0,y_0) \tag{1}$$ $$0=2(y_0-\beta)+2(f(x_0,y_0)-\gamma)f_y(x_0,y_0) \tag{2}$$

The equation of the tangent plane at $(x_0,y_0)$ is given by $$z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$$

which is standard scalar form $ax+by+cz=d$, is expressed as $$f_x(x_0,y_0)x+f_y(x_0,y_0)y-z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$$

From the above equation a normal vector for the plane is read off as being $$n=(f_x(x_0,y_0), f_y(x_0,y_0), -1)$$

Now, $$\overrightarrow{QP}=(x_0-\alpha,y_0-\beta,f(x_0,y_0)-\gamma)$$ $$\overrightarrow{QP}=(\gamma-f(x_0,y_0))(f_x(x_0,y_0), f_y(x_0,y_0), -1)$$ $$\overrightarrow{QP}=(\gamma-f(x_0,y_0))n$$

Where $equation (1)(2)$ have been used in going from the first line to the second. Sice $Q$ does not lie on the surface, $(\gamma-f(x_0,y_0))\neq0$. It follows that $n$ and $\overrightarrow{QP}$ are parallel, so the line through $P$ and $Q$ is perpendicular to the surface at $P$.

0

The question is incorrect I am afraid. $Q$ also must be at a minimum distance to the surface.

The required normal direction comes from cross-product $ \partial z / \partial x$ X $ \partial z / \partial y $

  • 0
    This question is from a exercise of the text book, So I assume its correct maybe. – 2017-01-03
  • 0
    Textbooks are not always correct. $PQ$ should be common normal. – 2017-01-03
  • 0
    Ok maybe its a wrong question, anyways I was wondering if we could find a tangent line using partial derivative then we can find a normal line, which would be perpendicular to the tangent line. If that makes sense? – 2017-01-03
  • 0
    Sorry, but how is the question incorrect? “... $P$ is the nearest point on the surface to $Q$” says to me that the distance is minimal. – 2017-01-03
  • 0
    @amd do you know how to tackle this problem? – 2017-01-04
  • 0
    @Reboot That depends on what you’re allowed to use. The first, and likely easiest, thing that comes to my mind is to set this up as a constrained minimization problem to be solved via a Lagrange multiplier. – 2017-01-04
  • 0
    @amd here is the solution from my professor. – 2017-02-04