MATH 251 Lecture 13

« previous | Wednesday, February 15, 2012 | next »

Gradient (Cont'd)

Vector quantity of a function:

• Points in the direction of steepest ascent of $f$ (perpendicular to level curves).
  • This direction is orthogonal to the surface... Why? (see below)
• Magnitude is the rate of change.

Example

Find the gradient of $f(x,y,z)=x^2+y^2+z^2$ at $(2,1,3)$:

$\begin{array}{rl}\mathrm{\nabla }f& =⟨2x,2y,2z⟩\\ \text{at }(2,1,3)& =⟨4,2,-6⟩\end{array}$

(does WebAssign want the unit vector in that direction?)

$\widehat{\mathrm{\nabla }f}=\frac{⟨2,1,-3⟩}{\sqrt{14}}$

Find the magitude of the rate of change in that direction.

$\Vert \mathrm{\nabla }f\Vert =2\sqrt{14}$

Tangent Plane to a surface

$\mathrm{\nabla }f$ acts as the normal vector to the plane:

Plane tangent to the surface $f(x,y,z)=x^2+y^2-z^2$ at the point (4,0,1).

$\mathrm{\nabla }f(4,0,1)=⟨8,0,-2⟩$

Therefore, the tangent plane at (4,0,1) is $0=8(x-4)+0(y-0)-2(z-1)$