Calculus 3 Exam #2

Front 1

Linearization

Back 1

L(x,y) = f(a,b) + fx(a,b) (x - a) + fy(a,b) (y - b)

Front 2

Equation of Tangent Plane

Back 2

z - z0 = fx(x0,y0) (x - x0) + fy(x0,y0) (y - y0)

Front 3

Gradient Vector

Back 3

< fx, fy, fz >

Front 4

Directional Derivative

Back 4

Du f(x,y,z) = ▽f (x,y,z) * u

where u = vector v / magnitude of v

Front 5

Second Derivative Test

Back 5

D = fxx(a,b) * fyy(a,b) - [ fxy(a,b) ]^2

Front 6

Give the conditions for max and min.

Back 6

If D > 0, and fxx(a,b) > 0 , then it is a minimum.
If D > 0, and fxx(a,b) < 0, then it is a maximum.
If D < 0, it is a saddle point.

Front 7

How do you find extrema?

Back 7

1. Find fx and fy.
2. Set equal to 0 to find critical points.
3. Find equations of lines for the boundaries.
4, Check the endpoints. If slope is negative then highest number is minimum and lowest number is maximum.
5. These are all candidates for absolute maximum and absolute minimum.
6. Compare to find which one it is.