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Conservative Vector Fields -- Is this right?
G = <(1 + x)e^{x+y}, xe^{x+y}+2z, -2y>
Evaluate \int_{C}G.dR
where C is the path given by:
x = (1 - t)e^{t}, y = t, z = 2t, 1=>t>=0
First, i noticed that there is a scalar potential associated with several terms (not all) in
the above vector field:
\varphi = xe^{x + y}
\nabla\varphi = <(1 + x)e^{x + y}, xe^{x + y},0>
So, i then separated the initial vector field G into two parts:
G=<0, 2z, -2y> + <(1 + x)e^{x + y}, xe^{x + y},0>
First, I compute some derivatives and then evaluate the first line integral:
\frac{dc}{dt} =
{dx = -te^{t}dt
dy = 1dt
dz = 2dt}
so,
\int_{C}G_{1} . \frac{dc}{dt}dt =
\int^{1}_{0}<0, 2(2t), -2(t)> . <-te^{t}, 1, 2> =
\int0dt = 0
I then computed the second part of G, keeping in mind that this portion of the vector field is conservative. Therefore, i just made a path from point t = 0 to t = 1 with a straight line r, parametrized like so:
r = { x = 1-t, y = t, z = 2t}
\frac{dr}{dt} = {dx = -1, dy = 1, dz = 2}
my second integral then becomes;
\int^{1}_{0}<(1 + 1 - t)e^{1 - t + t}, (1-t)e^{1-t + t},0> . <-1, 1, 2> dt = \int^{1}_{0}<(2-t)e, (1-t)e,0> . <-1, 1, 2> dt =
So, is my final answer simply this integral evaluated, or am I doing this completely wrong? Sorry if this is a bad question, my book just has no example problems at all, and integrating the theory can be confusing. Any feedback would be appreciated greatly. Thank you
Homework Statement
G = <(1 + x)e^{x+y}, xe^{x+y}+2z, -2y>
Evaluate \int_{C}G.dR
where C is the path given by:
x = (1 - t)e^{t}, y = t, z = 2t, 1=>t>=0
Homework Equations
The Attempt at a Solution
First, i noticed that there is a scalar potential associated with several terms (not all) in
the above vector field:
\varphi = xe^{x + y}
\nabla\varphi = <(1 + x)e^{x + y}, xe^{x + y},0>
So, i then separated the initial vector field G into two parts:
G=<0, 2z, -2y> + <(1 + x)e^{x + y}, xe^{x + y},0>
First, I compute some derivatives and then evaluate the first line integral:
\frac{dc}{dt} =
{dx = -te^{t}dt
dy = 1dt
dz = 2dt}
so,
\int_{C}G_{1} . \frac{dc}{dt}dt =
\int^{1}_{0}<0, 2(2t), -2(t)> . <-te^{t}, 1, 2> =
\int0dt = 0
I then computed the second part of G, keeping in mind that this portion of the vector field is conservative. Therefore, i just made a path from point t = 0 to t = 1 with a straight line r, parametrized like so:
r = { x = 1-t, y = t, z = 2t}
\frac{dr}{dt} = {dx = -1, dy = 1, dz = 2}
my second integral then becomes;
\int^{1}_{0}<(1 + 1 - t)e^{1 - t + t}, (1-t)e^{1-t + t},0> . <-1, 1, 2> dt = \int^{1}_{0}<(2-t)e, (1-t)e,0> . <-1, 1, 2> dt =
So, is my final answer simply this integral evaluated, or am I doing this completely wrong? Sorry if this is a bad question, my book just has no example problems at all, and integrating the theory can be confusing. Any feedback would be appreciated greatly. Thank you