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:See steps belowStep-by-step explanation:We have the paraboloid [tex]\bf f(x,y)=14-\frac{x^2}{4}-\frac{y^2}{16}[/tex] The point [tex]\bf P(x,y)=(2\sqrt{2},8)[/tex] on the level curve f(x,y) = 8 The slope of the line tangent to the level curve at P is given by [tex]\bf  -\frac{f_x(2\sqrt{2},8)}{f_y(2\sqrt{2},8)}[/tex] Let's then compute the partial derivatives [tex]\bf f_x(x,y)=-\frac{x}{2}[/tex] [tex]\bf f_y(x,y)=-\frac{y}{8}[/tex] so, the slope of the tangent at P would be [tex]\bf  -\frac{f_x(2\sqrt{2},8)}{f_y(2\sqrt{2},8)}=-\frac{-\sqrt{2}}{-1}=-\sqrt{2}[/tex] Hence, the line [tex]\bf y=-\sqrt{2}x+b[/tex] is tangent to the level curve. As we know the line passes through [tex]\bf (2\sqrt{2},8)[/tex], then [tex]\bf 8=-\sqrt{2}(2\sqrt{2})+b\rightarrow b=16[/tex] and the line  [tex]\bf y=-\sqrt{2}x+16[/tex] is the tangent to the level curve at point [tex]\bf (2\sqrt{2},8)[/tex] 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 [tex]\bf (\sqrt{2},14)[/tex] The vector [tex]\bf (\sqrt{2},-2)[/tex] is parallel to line. We have now to see that it is orthogonal to the gradient at [tex]\bf (2\sqrt{2},8)[/tex]. To verify this, we have to show that their inner product is 0 the gradient at [tex]\bf (2\sqrt{2},8)[/tex] is [tex]\bf (f_x(2\sqrt{2},8),f_y(2\sqrt{2},8))=(-\sqrt{2},-1)[/tex] so the inner product is [tex]\bf (\sqrt{2},-2)\circ(-\sqrt{2},-1)=-2+2=0[/tex]