MATH 251 Lecture 13

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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 $(2,1,3)$:

${\begin{aligned}\nabla f&=\left\langle 2x,2y,2z\right\rangle \\{\mbox{at }}(2,1,3)&=\left\langle 4,2,-6\right\rangle \end{aligned}}$

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

${\hat {\nabla f}}={\frac {\left\langle 2,1,-3\right\rangle }{\sqrt {14}}}$

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

$\left\|\nabla f\right\|=2{\sqrt {14}}$

Tangent Plane to a surface

$\nabla f$ 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).

$\nabla f(4,0,1)=\left\langle 8,0,-2\right\rangle$

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