Math 233 Midterm #2

f(x,y)

input an a and y value

get a z value (height)

Domain

What points can be plugged in?

x and y values

steps to finding domain

1. look for restrictions:

no 0 in denominator, no negative square roots, ln and log must be greater than 0

2. solve for restriction

make equation with values to create a line on the xy plane, ex: x+y+1 >0 → y >-x-1

3. understand the equation

circle, line, parabola?

4. find what part of the region to shade by plugging in a practice point

typically try (0,0)

5. understand if boundary line is apart of shaded region

Range

all the possible outputs of the function (heights)

steps to find range

1. does the function have a min or max value?

ex: √9-x²-y² cant be more than 9 because you cannot have a neg sqrt

2. find the lowest and highest outputs

ex: √9-x²-y², max when x=y=0, min when x²+y²=9

3. make range interval

for example, [0,3]

level curves

set z to a constant number to understand what the surface looks like at that height

***like 3D printing

1. set equal to constant

2. solve for x and y

3. identify shape

4. graph multiple curves

5. convert to 3d graph

understand limits in multi

in 3D, there are infinitely many ways to go through a point (a,b)

the limit only exists if ALL paths go through that same point and are equal

how to approach limits

1. Try direct substitution: Plug in the point. If you get a defined number, that's the limit.

2. If you get an indeterminate form (like (0/0)), simplify the expression, then plug in

3. Try two paths test: For example, (y=0), (x=0), or (y=mx).

4. If two paths give different limits, the limit does not exist (DNE).

5. If the paths agree, the limit might exist, but this alone doesn't prove it.

6. To prove the limit exists, use a valid method such as the squeeze theorem, polar coordinates, or bounding the expression.

finding limit with option 1: direct subsitution

just plug in x and y values of the point your’e approaching and solve

finding limit with option 2: factoring

if plugging in gives you an indeterminate form, break down the equation into smaller parts and try to cancel things out

finding limit with option 3: rationalize the denominator

if sqrt is causing indeterminate form, multiply by the conjugate to simply the equation

the conjugate of √A - B is √A + B

remember difference of squares identity: a²-b² = (a-b)(a+b)

rlly just multiply out the sqrt in the denominator

finding limit with option 4: squeeze theorem

If you can trap a function between two functions that approach the same limit, the trapped function must approach that limit too

if both upper and lower bounds are approaching the same limit, so will the one in between

finding limit with option 5: convert to polar coords

rewrite x and y as polar coords to make understanding them easier

x = rcosθ, y=rsinθ, x² +y² =r²

allows you to work with one variable so you can use L'hopital's rule!

Two Paths Test

The limit does NOT exist if the limit along one path is different to the limit along another

want a path that goes through the origin → use a vector

example: y=x → r(t) = <t,t>, or along the x-axis r(t)= <t, 0>

Potentials paths for TPT

x axis: (t,0)

y axis: (0,t)

Diagonally through origin: (t,t), (t,-t), (-t,t) (-t,-t)

parabolas: <t²,t>, <t³,t>

continuity rules

a function with two variables is continuous at a point if:

f(a,b) exists

lim as (x,y) approaches (a,b) exists

f(a,b) is equal to the lim as (x,y) approaches (a,b)

Partial derivative

focus on one variable and take its derivative leaving all else constant

remember quotient rule: h(x) = f(x)/g(x) → h’(x) = f’(x)g(x) - f(x)g’(x) / (g(x))²

implicit differentiation

lets you find partial derivatives when a variable like z is mixed into an equation and isn't already isolated

take derivative of chosen variable

when deriving z, dont forget to multiply by ∂z/∂x

move z/∂x to one side

clairaut’s theorem

fxy = fyx in our class

this is because theyre always goig to be continous for us?

limti def of partial derivatives

shows instantaneous rate of change in x or y for h

differentiability

f(xy) is differentiable if fx exists and is continuous and fy exists and is continous

chain rule and making tree

use when you have several variables

multiply down the tree and add across

making the gradient

just each partial derivative in vector form

∇f= <fx, fy, fz>

directional derivative

make sure you have UNIT vector —> divide by magnitude

get dot product: fx(x,y)*a + fy(x,y)*b

tells you how quickly a function output is changing when you move in a specific direction (vector) from a specific point (x,y,z)

why can we interpret directional vector to mean things about the function

if Duf = ∇f * u —> dot product = |∇f||u|cos

vector is unit vector, |u|=1 —> |∇f|cos

angle can tell us alot about the fuction

different angle meaning w directional derivatives

θ = 0 —> cos0 = 1 —> Duf = ∇f —> ∇f is where the func is increasing as quickly as possible

θ = 180 —> cos 180 = -1 —> Duf = -∇f —> where func is decreasing the quickest

θ = 90 —> cos 90= 0 —> Duf = 0 —> func has no instantaneous change