Find the distance from the point P(0,0,1) to the surface z=x^2+2y^2

Question

Jovanni Salinas · Accepted Answer

Find the distance from the point P(0,0,1) to the surface

z = x^{2} + 2 y^{2}

Let

d (x, y, z) = \sqrt{(x - 0)^{2} + (y - 0)^{2} + (z - 1)^{2}}

be the distance from point P.
We need to find the shortest distance wr.t constant

z = f (x, y) = x^{2} + 2 y^{2} i . e . z - x^{2} - 2 y^{2} = 0

d is always positive , so , for simplicity we can work on

d^{2} = x^{2} + y^{2} + (z - 1)^{2}

.
Subject to

z - x^{2} - 2 y^{2} = 0

Lagrangian function

L (x, u, z, λ) = d (x, y, z) + λ g (x, y, z) = x^{2} + y^{2} + (z - 1)^{2} + λ (z - x^{2} - 2 y^{2})

where

λ

is lagrange multiplier
By the lagrangian method we need to solve

\frac{\partial L}{\partial x} = 0, \frac{\partial L}{\partial y} = 0, \frac{\partial L}{\partial z} = 0, \frac{\partial L}{\partial λ} = 0 i . e 2 x - 2 λ x = 0 2 y - 4 λ y = 0 2 (z - 1) + λ = 0 z - x^{2} - 2 y^{2} = 0 2 x (1 - λ) = 0 2 y (1 - 2 λ) = 0

when

x = y = z = 0 \Rightarrow d (x, y, z) = 1 x = \pm \frac{1}{\sqrt{2}}, y = 0, z = \frac{1}{2} d (x, y, z) = \sqrt{\frac{1}{2} + 0 + (\frac{1}{2} - 1)^{2}} = \sqrt{\frac{1}{2} + \frac{1}{4}} = \frac{\sqrt{3}}{2} x = 0, y^{2} = \frac{3}{8}, z = \frac{3}{4} d (x, y, z) = \sqrt{0 + \frac{3}{8} + (\frac{3}{4} - 1)^{2}} = \sqrt{\frac{3}{8} + \frac{1}{16}} = \frac{\sqrt{7}}{4}

shortest distance is from (0,0,0)

Answered question