12.6-13.1

Question 1

how to find the directional derivative at (a,b) in the direction of (c,d)

Answer 1

gradient evaluated at (a,b) dotted into the unit vector from (a,b) to (c,d)

Question 2

what must be true of the vector you are taking the directional derivative in the direction of?

Answer 2

It must be a unit vector

Question 3

when does the funciton have the maximum rate of increase on its directional derivative

Answer 3

when the unit vector u is in the direction of the gradient vector. its max increase value is the magnitude of the gradient

Question 4

when does the function have the maximum rate of decrease on its directional derivative

Answer 4

when the unit vector u is in the NEGATIVE direction of the gradient vector. its max increase value is the NEGATIVE magnitude of the gradient

Question 5

how to verify that the gradient vector at a point is orthogonal to the level curve at that point.

Answer 5

1. find gradient at point,
2. use slope to find a vector proportional to the curve at that point
3. the cross product must be 0

Question 6

how to find the equation of a line tangent to level curve at a point.

Answer 6

fx(x-`a)+fy(y-b)

Question 7

2 ways a surface can be defined

Answer 7

explicitly z=f(x,y)
implicitly f(z,y,z) = 0

Question 8

how to find the equation of a tangent plane at a given point implicitly

Answer 8

1. define it implicitly

2. fx(pt)(x-a)+fy(pt)(y-b)+fz(pt)(y-c)

Question 9

find all the points on a plane that are horizontal(or fit the bill)

Answer 9

1. gradient of the equation

2. set the gradient equal to the qualification vector, In this case <0,0,c>

Question 10

how to find the equation of a tangent plane at a point when it is defined explicitly

Answer 10

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

Question 11

find the linear approximation at a

Answer 11

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

Question 12

differential

Answer 12

dz = fx(pt)dx+fy(pt)dy

Question 13

what can the differential do?

Answer 13

approximate the change of things

make sure to study

Question 14

how to find critical points on an equation

Answer 14

gradient set equal to 0

Question 15

equation for second derivative test

Answer 15

D(pt) = fxx(pt)fyy(pt) - fxy^2

Question 16

if D>0 and Fxx<0

Answer 16

local max at a,b

Question 17

local max at a,b

Answer 17

if D>0 and Fxx<0

Question 18

D(pt) =0

Answer 18

inconclusive

Question 19

inconclusive

Answer 19

D(pt) =0

Question 20

if D>0 and Fxx>0

Answer 20

local min

Question 21

local min

Answer 21

if D<0 and Fxx>0

Question 22

D<0

Answer 22

saddle point

Question 23

saddle point

Answer 23

D<0

Question 24

how to find the absolute maximums and minimums

Answer 24

1. find all the critical points
2. find the max and min values on the boundaries
3. max value of f at the rest is max. min is relative to same

Question 25

how to find the critical points on the boundaries

Answer 25

derive and set equal to 0;

Question 26

legrange multipliers

Answer 26

grad f = lambda grad g

Question 27

when setting up the legrange multiplieers what cant i forget

Answer 27

The constraint equation is the final equation,

and grad of constraint is also multiplied by lambda

Question 28

how to do a double integral

Answer 28

integrate the inner part then the outer part. pay atention to the bounds and crap.

Question 29

what is the average value of a function over a region

Answer 29

(1/area of r )double int(f(x,y) dA

Question 30

how to find the normal vector of a surface at a point

Answer 30

gradient evaluated at the point

Question 31

Consider the line perpendicular to the surface z=x2+y2 at the point where x=3 and y=−4. Find a vector parametric equation for this line in terms of the parameter t.

Answer 31

1. find z
2. find the gradient of the equation.
3. parameterize by multiplying t by the grad at that point and then adding the initial value

Question 32

how to find a linearization at a point

Answer 32

1. gradient at the point

2. multiply each gradient by the (variable - initial term.

Question 33

(1 pt) A car is driving northwest at v mph across a sloping plain whose height, in feet above sea level, at a point N miles north and E miles east of a city is given by
h(N,E)=2000+175N+25E.
(a) At what rate is the height above sea level changing with respect to distance in the direction the car is driving?
rate =
(150/sqrt(2))
feet/mile
(b) Express the rate of change of the height of the car with respect to time in terms of v.
rate =
150v/sqrt(2)
feet/hour

Answer 33

a. ‘
1. directional derivative to find the rate of the height change in that direction
b. multiply a by b

Question 34

logically speaking how does the gradient relate to the level curves

Answer 34

the gradient points perpendicular to the level curves in the direction of the greatest increase