Question

Does the function f(x,y)=x^{2}+y^{2} satisfy the
two dimensional Laplace's equation? Explain why/why not

And then calculate the gradient of g(x,y) at points g(x,y)=(0,1), (1,0), (0,-1) and (-1,0) and indicate by little arrows the directions in which these gradient vectors point.

Answer #1

find the work done by the force field f(x,y)= <
x2+y2, -x > on a particle that moves along
the curve c: x2+y2=1, counterclockwise from
(0,1) to (-1,0)

Minimize f (x, y) = x2 + y2 subject to the
constraint function g (x, y) = x2y−16 = 0

Given the function f(x, y, z) = (x2 + y2 +
z2 )−1/2
a) what is the gradient at the point (12,0,16)?
b) what is the directional derivative of f in the direction of
the vector u = (1,1,1) at the point (12,0,16)?

show that the function f(x,y,z) =
1/√(x2+y2+z2) provides the
equation fxx + fyy + fzz = 0, called the 3−D Laplace equation.

Calculate ∫ ∫S f(x,y,z)dS for the given surface and function.
x2+y2+z2=144, 6≤z≤12; f(x,y,z)=z2(x2+y2+z2)−1.

f(x, y) = x2 + y2 + 2xy + 6.
1. Find all the local extremas.
2. Does the function f has an absolute max or min on R2 ?
3. Draw E = {(x, y) ∈ R2; x >=0; y >=0; x + y<=1}.
4. Explain why f has an absolute max and min on E and find
them.

Is the function f(x,y)=x−yx+y continuous at the point (−1,−1)?
If not, why is the function not continuous?
Select the correct answer below:
A. Yes
B. No, because lim(x,y)→(−1,1)x−yx+y=−1 and f(0,0)=0.
C. No, because lim(x,y)→(−1,1)x−yx+y does not exist and f(0,0)
does not exist.
D. No, because lim(x,y)→(0,0)x2−y2x2+y2=1 and f(0,0)=0.

F(x,y) = (x2+y3+xy,
x3-y2)
a) Find the linearization of F at the point (-1,-1)
b) Explain that F has an inverse function G defined in an area
of (1, −2) such that
that G (1, −2) = (−1, −1), and write down the linearization to G in
(1, −2)

Find all positive integers x and y that satisfy x2 -
y2= 2018? Are there any positive integers x and y such
that x3 − y3 = 2018? Explain.

Consider a function f(x; y) =
2x2y
x4 + y2 .
(a) Find lim
(x;y)!(1;1)
f(x; y).
(b) Find an equation of the level curve to f(x; y) that passes
through the point (1; 1).
(c) Show that f(x; y) has no limits as (x; y) approaches (0;
0).

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 2 minutes ago

asked 12 minutes ago

asked 22 minutes ago

asked 33 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago

asked 3 hours ago

asked 4 hours ago