"Work integral" for vector field along a curve

Contents

(1a) Consider the vector field F = (y,-x,-z).

Find G = curl F. Is the vector field conservative?

Answer: We have curl F = (0,0,-2). Hence the vector field is not conservative.

syms x y z
F = [y,-x,-z];                     % given vector field
G = curl(F)                        % note: Matlab gives curl as column
vectorfield3(F,-1:.5:1,-1:.5:1,-1:1)
view(60,20)

G =
  0
  0
 -2

(1b) Let C1 be the curve from the point (1,0,1) to the point (0,1,-1) along the circle z=1, x^2+y^2=1. Find the work W for the curve C1.

Answer: The work is W = -pi/2 (we have to spend energy to move along this curve)

syms t real                        % parameter t
R = [cos(t),sin(t),1];             % parametrization R(t)
a = 0; b = pi/2;                   % interval for t

FR = subs(F,{x,y,z},{R(1),R(2),R(3)}) % in F substitute parametrization R for x,y,z
Rp = diff(R,t)                     % R'(t)
W = int(dot(FR,Rp),t,a,b)          % find work integral

vectorfield3(F,-1:.25:1,-1:.25:1,[0 1])    % plot vector field for z=0,1
hold on; alpha(0.5)
h = Ezplot3(R(1),R(2),R(3),[a,b]); % (need to use Ezplot3, fixes bugs in ezplot3)
set(h,'Linewidth',10,'Color','yellow')     % make curve thick, yellow
hold off; view(60,50)

FR =
[ sin(t), -cos(t), -1]
Rp =
[ -sin(t), cos(t), 0]
W =
-pi/2

(1c) Let C2 be the straight line from the point (1,0,1) to the point (0,1,1). Find the work W for the curve C2.

Answer: Now the work is W = -1, which is different from (1b) (not surprising as the vector field is not conservative)

A = [1,0,1]; B = [0,1,1];          % endpoints
R = A + t*(B-A)                    % parametrization of line from A to B
a = 0; b = 1;                      % interval for t

FR = subs(F,{x,y,z},{R(1),R(2),R(3)}) % in F substitute parametrization R for x,y,z
Rp = diff(R,t)                     % R'(t)
W = int(dot(FR,Rp),t,a,b)          % find work integral

vectorfield3(F,-1:.25:1,-1:.25:1,[0 1])    % plot vector field for z=0,1
hold on; alpha(0.5)
h = Ezplot3(R(1),R(2),R(3),double([a,b])); % need to use Ezplot3 since ezplot3 has bug
set(h,'Linewidth',10,'Color','yellow')     % make curve thick, yellow
hold off; view(60,50)

R =
[ 1 - t, t, 1]
FR =
[ t, t - 1, -1]
Rp =
[ -1, 1, 0]
W =
-1

(2a) Consider the vector field F = (x+y,x-y,-z).

Find G = curl F. Is the vector field conservative?

Answer: We have curl F = (0,0,0). Hence the vector field is conservative.

syms x y z
F = [x+y,x-y,-z];                     % given vector field
G = curl(F)                           % note: Matlab gives curl as column
vectorfield3(F,-1:.5:1,-1:.5:1,-1:1)
view(60,20)

G =
 0
 0
 0

(2b) Let C1 be the curve from the point (1,0,1) to the point (0,1,1) along the circle z=1, x^2+y^2=1. Find the work W for the curve C1.

Answer: The work is W = -1 (we have to spend energy to move along this curve)

syms t real                        % parameter t
Pi = sym('pi');
R = [cos(t),sin(t),1];             % parametrization R(t)
a = 0; b = Pi/2;                   % interval for t

FR = subs(F,{x,y,z},{R(1),R(2),R(3)}) % in F substitute parametrization R for x,y,z
Rp = diff(R,t)                     % R'(t)
W = int(dot(FR,Rp),t,a,b)          % find work integral

vectorfield3(F,-1:.25:1,-1:.25:1,[0 1])    % plot vector field for z=0,1
hold on; alpha(0.5)
h = Ezplot3(R(1),R(2),R(3),double([a,b])); % (need to use Ezplot3, fixes bugs in ezplot3)
set(h,'Linewidth',10,'Color','yellow')     % make curve thick, yellow
hold off; view(60,50)

FR =
[ cos(t) + sin(t), cos(t) - sin(t), -1]
Rp =
[ -sin(t), cos(t), 0]
W =
-1

(2c) Let C2 be the straight line from the point (1,0,1) to the point (0,1,1). Find the work W for the curve C2.

Answer: Now the work is W = -1, which is the same as (2b) (since the vector field is conservatitve)

A = [1,0,1]; B = [0,1,1];          % endpoints
R = A + t*(B-A)                    % parametrization of line from A to B
a = 0; b = 1;                      % interval for t

FR = subs(F,{x,y,z},{R(1),R(2),R(3)}) % in F substitute parametrization R for x,y,z
Rp = diff(R,t)                     % R'(t)
W = int(dot(FR,Rp),t,a,b)          % find work integral

vectorfield3(F,-1:.25:1,-1:.25:1,[0 1])    % plot vector field for z=0,1
hold on; alpha(0.5)
h = Ezplot3(R(1),R(2),R(3),double([a,b])); % need to use Ezplot3 since ezplot3 has bug
set(h,'Linewidth',10,'Color','yellow')     % make curve thick, yellow
hold off; view(60,50)

R =
[ 1 - t, t, 1]
FR =
[ 1, 1 - 2*t, -1]
Rp =
[ -1, 1, 0]
W =
-1

(2d) Find the work for any curve from (0,1,1) to (1,0,1) using the potential!

f = potential(F,[x y z])           % find the potential
W = subs(f,{x,y,z},{0,1,1}) - subs(f,{x,y,z},{1,0,1}) % f at end point minus f at initial point

f =
(x*(x + 2*y))/2 - y^2/2 - z^2/2
W =
-1

Published with MATLAB® R2013b