concepts review 201

line of intersection between 2 planes

n1 x n2 = v (direction for the line)

get a point on both planes for parametrization

how many normal vectors do planes have

infinitely many

tangent line to a curve r(t)

(r’(t)s + r(t) = L

*be careful with your parameters!

shadow (or projection onto a plane)

rearrange equations to eliminate the variable

(setting = 0 is only for cross sections)

sin/cos parameters

DO NOT FORGET BOUNDS!

speed function s(t) = |v(t)|

the speed is increasing, s’(t) > 0

decreasing speed is s’(t) > 0

when you see a common denominator, pull it out!

tan(pi/4)

1

velocity =

speed * direction

direction = v/|v|

speed = |v|

dont forget

remove extraneous solutions after squaring

level curve

C = f(x,y)

when completing the square

no coefficients in the squared terms!

the set of all possible constants C that could make a level set

all of the C values represent the range of a higher dimension function

level curves are subsets of the domain

implicit differentiation

dz/dx = -Fx/Fz

dz/dy = -Fy/Fz

range for arctan

(-pi/2, pi/2)

fyx = fxy

!!

D = fxxfyy - (fxy)²

if D < 0, saddle point

if D = 0, inconclusive

if D > 0 and fxx < 0, local max

if D > 0 and fxx > 0, local min

parallel planes

x + y + z = 1

x + y + z = C (for any constant C)

distance from a point to a plane

|plane(point)| / |n|

sums of squares

can they be 0??

“highest and lowest” points optimization

minimize/maximize the z coordinate with 2 constraints

gradf = k(gradg) + y (gradh)

optimization

-use lagrange to find possible critical points

-make sure they agree with the constraint

-plug into f(x,y) and test

the maximum directional derivative for a function is the same as

the maximum rate of change of the function at that point

at the max directional derivative

u points in the same direction as grad f

*think about dot product rules

***u is a direction vector, think velocity

the level curve of a function is ALWAYS

PERPENDICULAR to its gradient

limits DNE

when they cannot be approached by infinitely many curves

convert to polar to prove existence of a limit

gradients match function behavior

ex. perpendicular functions have perpendicular gradients

2ne derivative test (Hession)

only for local mins and local max

x² + y² =

r²

when given a boundary for a max/min, there can be critical points ON the boundary

plug the boundary into the OG function and solve for critical points if possible

linearization

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

*same as tangent plane

approximates the given surface at point (a,b)

1/0

+infinity

ln(0)

-infinity

ln(1)

0

e^0

1

cos0

1

sin0

0

plane eq

n dot (point on plane)

arctanu

1 / (1 + u²)

scalar component of a projection

(u*v) / |v|

proj u onto v

proj of big onto small

(u*v) / (v*v) * v

triple scalar product

( u x v) * w

magnitude is the volume of a parallelepiped

open

consists of all interior points

closed

contains its boundary points

bounded

the region lies in a disk of finite radius

unbounded

does not live in a disk of finite radius

to find the point where a line intersects a plane

*parametrize the line

*plug into plane eq

*solve for t

distance from a point to a plane

|PS * (n/|n|)|

*the distance from point S to the plane is the length of the vector projection proj n PS

the directional derivative is the rate of change of f

with respect to distance in the direction of u

f increases most rapidly when

Duf = +|gradF|

f decreases most rapidly when

Duf = -|gradF|

at the direction orthogonal to the gradient

the directional derivative is 0

extraneous solutions

be mindful of squaring!!!

sqrt domain restriction

>= 0

1/x domain restriction

x ≠ 0

ln(x)

x > 0

interior point

center of a disk of finite radius

boundary point

disk contains points in and out of small radius r

“closest to x”

distance formula needed

boundary critical points only on a

closed, unbounded function

to justify the existence/nonexistence of an absolute max or min

use limits, set one variable to 0

the gradient vector (gradf)

points in the direction of steepest increase

remember to include any restrictions

in your graphs, drawings, boundaries, max and min, etc.

e^(-infinity)

0

1/infinity

0

infinity/infinity

indeterminate form, keep going

level curves must use domain restrictions

exclude stuff not in the domain

explain limits in terms of

the infinite paths required to get there

extreme value theorem

justify max/min absolute

*if a real function is continuous on a closed and bounded subset, then f is guaranteed to attain both an absolute maximum and an absolute minimum