For each vector field F~ (x, y) = hP(x, y), Q(x, y)i, find a
function f(x,...
For each vector field F~ (x, y) = hP(x, y), Q(x, y)i, find a
function f(x, y) such that F~ (x, y) = ∇f(x, y) = h ∂f ∂x , ∂f ∂y i
by integrating P and Q with respect to the appropriate variables
and combining answers. Then use that potential function to directly
calculate the given line integral (via the Fundamental Theorem of
Line Integrals):
a) F~ 1(x, y) = hx 2 , y2 i Z C F~ 1...
1. a) Let F(x,y) = hcosy,−xsiny + 2yi. Show that F is
conservative, and find a...
1. a) Let F(x,y) = hcosy,−xsiny + 2yi. Show that F is
conservative, and find a function
φ such that ∇φ(x,y) = F(x,y).
b) Use the result from part a) to find
R
C F · Tds, where C is given by r(t) = ht,πti,0 ≤
t ≤ 1.
(a) Is the vector field F = <e^(−x) cos y, e^(−x) sin y>
conservative?
(b) If...
(a) Is the vector field F = <e^(−x) cos y, e^(−x) sin y>
conservative?
(b) If so, find the associated potential function φ.
(c) Evaluate Integral C F*dr, where C is the straight line path
from (0, 0) to (2π, 2π).
(d) Write the expression for the line integral as a single
integral without using the fundamental theorem of calculus.
Is the function f(x,y)=x−yx+y continuous at the point (−1,−1)?
If not, why is the function not...
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.
Given the force field F(x, y) = (x − y, 4x + y^2 ), find the...
Given the force field F(x, y) = (x − y, 4x + y^2 ), find the
work done to move along a line segment from (0, 0) to (2,0), along
a line segment from (2,0) to (0,1), and then along another line to
the point (−2, 0). Show your work.
You are given that the function f(x,y)=8x2+y2+2x2y+3 has first
partials fx(x,y)=16x+4xy and fy(x,y)=2y+2x2, and has second...
You are given that the function f(x,y)=8x2+y2+2x2y+3 has first
partials fx(x,y)=16x+4xy and fy(x,y)=2y+2x2, and has second
partials fxx(x,y)=16+4y, fxy(x,y)=4x and fyy(x,y)=2. Consider the
point (0,0). Which one of the following statements is true?
A. (0,0) is not a critical point of f(x,y).
B. f(x,y) has a saddle point at (0,0).
C. f(x,y) has a local maximum at (0,0).
D. f(x,y) has a local minimum at (0,0).
E. The second derivative test provides no information about the
behaviour of f(x,y) at...
Let F(x,y,z) = yzi + xzj + (xy+2z)k
show that vector field F is conservative by...
Let F(x,y,z) = yzi + xzj + (xy+2z)k
show that vector field F is conservative by finding a function f
such that
and use that to evaluate
where C is any path from (1,0,-2) to (4,6,3)
Consider the vector force field given by F⃗ = 〈2x + y, 3y +
x〉
(a)...
Consider the vector force field given by F⃗ = 〈2x + y, 3y +
x〉
(a) Let C1 be the straight line segment from (2, 0) to (−2,
0).
Directly compute ∫ C1 F⃗ · d⃗r (Do not use Green’s Theorem or
the Fundamental Theorem of Line Integration)
(b) Is the vector field F⃗ conservative? If it is not
conservative, explain why. If it is conservative, find its
potential function f(x, y)
Let C2 be the arc of the half-circle...
Evaluate H C F · dr, if F(x, y, z) = yi + 2xj + yzk,...
Evaluate H C F · dr, if F(x, y, z) = yi + 2xj + yzk, and C is
the curve of intersection of the part of the paraboliod z = 1 − x 2
− y 2 in the first octant (x ≥ 0, y ≥ 0, z ≥ 0) with the coordinate
planes x = 0, y = 0 and z = 0, oriented counterclockwise when
viewed from above. The answer is pi/4+4/15
1. (a) Determine whether or not F is a conservative vector
field. If it is, find...
1. (a) Determine whether or not F is a conservative vector
field. If it is, find the potential function for F.
(b) Evaluate R C1 F · dr and R C2 F · dr where C1 is the
straight line path from (0, −1) to (3, 0), while C2 is the union of
two straight line paths: first piece from (0, −1) to (0, 0) and
then second piece from (0, 0) to (3, 0). (When applicable, use the
Fundamental...