Question

Consider the paraboloid f (x comma y )equals14 minus StartFraction x squared Over 4 EndFraction minus StartFraction y squared Over 16 EndFraction and the point Upper P (2 StartRoot 2 EndRoot comma 8 )on the level curve f (x comma y )equals8. Compute the slope of the line tangent to the level curve at P and verify that the tangent line is orthogonal to the gradient at that point.

Answer 1

Answer:

See steps below

Step-by-step explanation:

We have the paraboloid

$\bf f(x,y)=14-\frac{x^2}{4}-\frac{y^2}{16}$

The point

$\bf P(x,y)=(2\sqrt{2},8)$

on the level curve f(x,y) = 8

The slope of the line tangent to the level curve at P is given by

$\bf  -\frac{f_x(2\sqrt{2},8)}{f_y(2\sqrt{2},8)}$

Let's then compute the partial derivatives

$\bf f_x(x,y)=-\frac{x}{2}$

$\bf f_y(x,y)=-\frac{y}{8}$

so, the slope of the tangent at P would be

$\bf  -\frac{f_x(2\sqrt{2},8)}{f_y(2\sqrt{2},8)}=-\frac{-\sqrt{2}}{-1}=-\sqrt{2}$

Hence, the line

$\bf y=-\sqrt{2}x+b$

is tangent to the level curve. As we know the line passes through $\bf (2\sqrt{2},8)$, then

$\bf 8=-\sqrt{2}(2\sqrt{2})+b\rightarrow b=16$

and the line  

$\bf y=-\sqrt{2}x+16$ is the tangent to the level curve at point $\bf (2\sqrt{2},8)$

In order to find a vector that lies on this line, we take any two points on the line and subtract them.

For example, (0,16) and $\bf (\sqrt{2},14)$

The vector $\bf (\sqrt{2},-2)$ is parallel to line.

We have now to see that it is orthogonal to the gradient at

$\bf (2\sqrt{2},8)$.

To verify this, we have to show that their inner product is 0

the gradient at $\bf (2\sqrt{2},8)$ is

$\bf (f_x(2\sqrt{2},8),f_y(2\sqrt{2},8))=(-\sqrt{2},-1)$

so the inner product is

$\bf (\sqrt{2},-2)\circ(-\sqrt{2},-1)=-2+2=0$