Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the surface $S$ (in $\mathbb R^3$) given by the equation $z=f(x,y)=\frac32(x^2+y^2)$. How can I find the shortest distance from a point $p=(a,b,c)$ on $S$ to the point $(0,0,1)$.

This is what I have done: Define $d(a,b,c)=a^2+b^2+(c-1)^2$, for all points $p=(a,b,c)\in S$. Then $\sqrt d$ is the distance from $S$ to $(0,0,1)$. I think that the method of Lagrange multipliers is the easiest way to solve my question, but how can I find the Lagrangian function? Or is there an easier way to find the shortest distance?

share|improve this question
    
What do you mean by "shortest distance"? –  Mercy Jul 26 '12 at 21:39

5 Answers 5

Let $q=(a,b,c)$ be the one of the closest to $p$ point of $S$. Since $q\in S$ we have $$ c=\frac{3}{2}(a^2+b^2)\tag{1} $$ On the other hand the vector $pq=(a,b,c-1)$ is orthogonal to $S$ (because $q$ is the closest to $p$ point of $S$), hence $pq$ is collinear to the normal $n$ to the surface $S$ at the point $q$. This normal is easily computable $$ n=(-3a,-3b,1) $$ Since $pq$ and $n$ are collinear vectors $$ \frac{-3a}{a}=\frac{-3b}{b}=\frac{1}{c-1}\tag{2} $$ The rest is clear.

share|improve this answer

I think that the method of Lagrange multipliers is the easiest way to solve my question, but how can I find the Lagrangian function?

As shown by other answers and in note 1 there are easier ways to find the shortest distance, but here is a detailed solution using the method of Lagrange multipliers. You need to find the minimum of the distance function

$$\begin{equation} \sqrt{d(x,y,z)}=\sqrt{x^{2}+y^{2}+(z-1)^{2}} \tag{1a} \end{equation}$$

subject to the constraint given by the surface equation $z=f(x,y)=\frac32(x^2+y^2)$ $$\begin{equation} g(x,y,z)=z-\frac{3}{2}\left( x^{2}+y^{2}\right) =0. \tag{2} \end{equation}$$ Since $\sqrt{d(x,y,z)}$ increases with $d(x,y,z)$ you can simplify the computations if you find the minimum of $$\begin{equation} d(x,y,z)=x^{2}+y^{2}+(z-1)^{2} \tag{1b} \end{equation}$$ subject to the same constraint $(2)$. The Lagrangian function is then defined by $$\begin{eqnarray} L\left( x,y,z,\lambda \right) &=&d(x,y,z)+\lambda g(x,y,z) \\ L\left( x,y,z,\lambda \right) &=&x^{2}+y^{2}+(z-1)^{2}+\lambda \left( z- \frac{3}{2}\left( x^{2}+y^{2}\right) \right), \tag{3} \end{eqnarray}$$ where $\lambda $ is the Lagrange multiplier. By this method you need to solve the following system $$\begin{equation} \left\{ \frac{\partial L}{\partial x}=0,\frac{\partial L}{\partial y}=0, \frac{\partial L}{\partial z}=0,\frac{\partial L}{\partial \lambda } =0,\right. \tag{4} \end{equation}$$ which results in

$$\begin{eqnarray} \left\{ \begin{array}{c} 2x+3\lambda x=0 \\ 2y+3\lambda y=0 \\ 2z-2-\lambda =0 \\ -z+\frac{3}{2}\left( x^{2}+y^{2}\right) =0 \end{array} \right. &\Leftrightarrow &\left\{ \begin{array}{c} x=0\vee 2+3\lambda =0 \\ y=0\vee 2+3\lambda =0 \\ 2z-2-\lambda =0 \\ -z+\frac{3}{2}\left( x^{2}+y^{2}\right) =0 \end{array} \right. \\ &\Leftrightarrow &\left\{ \begin{array}{c} x=0 \\ y=0 \\ \lambda =2 \\ z=0 \end{array} \right. \vee \left\{ \begin{array}{c} \lambda =-2/3 \\ z=2/3 \\ x^{2}+y^{2}=4/9 \end{array} \right. \tag{5} \end{eqnarray}$$

For $x=y=x=0$ we get $\sqrt{d(0,0,0)}=1$. And for $x^2+y^2=4/9,z=2/3$ we get the minimum distance subject to the given conditions $$\begin{equation} \underset{g(x,y,z)=0}\min \sqrt{d(x,y,z)}=\sqrt{\frac{4}{9}+(\frac{2}{3}-1)^{2}}=\frac{1}{3}\sqrt{5}. \tag{6} \end{equation}$$

It is attained on the intersection of the surface $z=\frac{3}{2}\left( x^{2}+y^{2}\right) $ with the vertical cylinder $x^{2}+y^{2}=\frac{4}{9}$ or equivalently with the horizontal plane $z=\frac{2}{3}$.

enter image description here $$\text{Plane }z=\frac{2}{3} \text{(blue) and surface }z=\frac{3}{2}\left( x^{2}+y^{2}\right) $$

Notes.

  1. As the solution depends only on the sum $r^{2}=x^{2}+y^{2}$ we could just find $$\begin{equation} \min d(r)=r^{2}+(\frac{3}{2}r^{2}-1)^{2}. \tag{7} \end{equation}$$
  2. The surface $z=\frac{3}{2}\left( x^{2}+y^{2}\right) $ is a surface of revolution around the $z$ axis.
share|improve this answer

Here's another approach. The distance between the surface and the point can be expressed solely in terms of $r=\sqrt{x^2+y^2}$, $$d(r) = \sqrt{1-2 r^2+\frac{9}{4} r^4}.$$ Extremizing $d(r)$ with respect to $r$ we find $$r\left(r^2-\frac{4}{9}\right) = 0,$$ so $r=0$ or $r=2/3$. But $d(0) = 1$ and $d(2/3) = \sqrt{5}/3$. Thus, the shortest distance between the point and the surface is $\sqrt{5}/3$. This is the distance between the point and the circle $(x,y,\frac{3}{2}r^2) = (x,y,2/3)$, where $x^2+y^2 = 4/9$.

share|improve this answer

The Lagrangian is $L(x,y,z,\lambda)=d(x,y,z)+\lambda(z-\frac32(x^2+y^2))$. Compute the partial derivatives of $L$ with respect to $x,y,z$ and $\lambda$. Setting all the partial derivatives equal to $0$ gives $x=y=0, z=\frac32(x^2+y^2)$ and $\lambda=\frac23$. This gives critical points $(0,0,0)$ and $(0,0,\frac23)$. Plugging these points into $d$ gives the minimal distance from a point on $S$ to $(0,0,1)$.

share|improve this answer
1  
According to my computations the second critical point is not $(0,0,2/3)$. Instead of $x=y=0$, it should be the circle $x^2+y^2=4/9$ located at $z=2/3$. –  Américo Tavares Jul 27 '12 at 19:52

Let $d = a^2 + b^2$ and,

$$f(d) = a^2 + b^2 + (c-1)^2 = d + \left ( \frac{3}{2}d -1 \right )^2.$$

Thus, setting $f'(d) = (9/2)d-2 = 0$ gives us the critical point $d^* = 4/9$ and so, $$f(d^*) = \frac{5}{9}.$$

Hence, the distance you're looking for is $$\sqrt{f(d^*)} = \frac{\sqrt{5}}{3}.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.