Line integral path independence proof check The Next CEO of Stack OverflowFinding work via...
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Line integral path independence proof check
The Next CEO of Stack OverflowFinding work via Line Integralsfinding the parametric path for line integralparameterise line segment or functionCompute line integral for vector function along the curveFinding work done by force with line integralsBasic Line Integral for Workderiving the formula of the torsion of a curveCalculation of the line integral of a vector field along a pathTried a line integral and then did it using Stokes theorem, but I'm getting different results…Computing the line integral of the oriented curve
$begingroup$
Find the work done by the force $F(x, y, z) = (x^4y^5, x^3)$ along the curve
C given by the part of the graph of $y$ = $(x^3)$ from $(0, 0)$ to $(-1, -1)$.
I first checked for independence, which did not work.
Next I parameterized the curve by $r(t) = [x(t),y(t)]$,
begin{align*}
x(t) &= t, \
y(t)&=t^3
end{align*}
which has
$$dr= (1, 3t^2). $$
Computing the work is then
$$int_0^{-1} (t^{19},t^3)cdot (1,3t^2),dt = 11/20.$$
Is this correct?
calculus proof-verification line-integrals
edited Mar 16 at 10:45
Benjamin
631419
asked Mar 16 at 5:12
MasterYoshiMasterYoshi
807
$endgroup$
add a comment |
$begingroup$
Find the work done by the force $F(x, y, z) = (x^4y^5, x^3)$ along the curve
C given by the part of the graph of $y$ = $(x^3)$ from $(0, 0)$ to $(-1, -1)$.
I first checked for independence, which did not work.
Next I parameterized the curve by $r(t) = [x(t),y(t)]$,
begin{align*}
x(t) &= t, \
y(t)&=t^3
end{align*}
which has
$$dr= (1, 3t^2). $$
Computing the work is then
$$int_0^{-1} (t^{19},t^3)cdot (1,3t^2),dt = 11/20.$$
Is this correct?
calculus proof-verification line-integrals
edited Mar 16 at 10:45
Benjamin
631419
asked Mar 16 at 5:12
MasterYoshiMasterYoshi
807
$endgroup$
add a comment |
$begingroup$
Find the work done by the force $F(x, y, z) = (x^4y^5, x^3)$ along the curve
C given by the part of the graph of $y$ = $(x^3)$ from $(0, 0)$ to $(-1, -1)$.
I first checked for independence, which did not work.
Next I parameterized the curve by $r(t) = [x(t),y(t)]$,
begin{align*}
x(t) &= t, \
y(t)&=t^3
end{align*}
which has
$$dr= (1, 3t^2). $$
Computing the work is then
$$int_0^{-1} (t^{19},t^3)cdot (1,3t^2),dt = 11/20.$$
Is this correct?
calculus proof-verification line-integrals
edited Mar 16 at 10:45
Benjamin
631419
asked Mar 16 at 5:12
MasterYoshiMasterYoshi
807
$endgroup$
Find the work done by the force $F(x, y, z) = (x^4y^5, x^3)$ along the curve
C given by the part of the graph of $y$ = $(x^3)$ from $(0, 0)$ to $(-1, -1)$.
I first checked for independence, which did not work.
Next I parameterized the curve by $r(t) = [x(t),y(t)]$,
begin{align*}
x(t) &= t, \
y(t)&=t^3
end{align*}
which has
$$dr= (1, 3t^2). $$
Computing the work is then
$$int_0^{-1} (t^{19},t^3)cdot (1,3t^2),dt = 11/20.$$
Is this correct?
calculus proof-verification line-integrals
calculus proof-verification line-integrals
edited Mar 16 at 10:45
Benjamin
631419
asked Mar 16 at 5:12
MasterYoshiMasterYoshi
807
edited Mar 16 at 10:45
Benjamin
631419
asked Mar 16 at 5:12
MasterYoshiMasterYoshi
807
edited Mar 16 at 10:45
Benjamin
631419
edited Mar 16 at 10:45
Benjamin
631419
edited Mar 16 at 10:45
Benjamin
631419
631419
asked Mar 16 at 5:12
MasterYoshiMasterYoshi
807
asked Mar 16 at 5:12
MasterYoshiMasterYoshi
807
asked Mar 16 at 5:12
MasterYoshiMasterYoshi
807
807
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Looks fine to me. By "check independence" I assume you mean that you computed that $nabla times F neq 0$ so that $F$ has no chance of being integrable.
As a sanity check, when $x<0,y<0$ then $F$ points to the bottom-left, so that a positive work along the given curve segment is expected.
answered Mar 16 at 7:34
user7530user7530
35.1k761114
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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oldest
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votes
$begingroup$
Looks fine to me. By "check independence" I assume you mean that you computed that $nabla times F neq 0$ so that $F$ has no chance of being integrable.
As a sanity check, when $x<0,y<0$ then $F$ points to the bottom-left, so that a positive work along the given curve segment is expected.
answered Mar 16 at 7:34
user7530user7530
35.1k761114
$endgroup$
add a comment |
$begingroup$
Looks fine to me. By "check independence" I assume you mean that you computed that $nabla times F neq 0$ so that $F$ has no chance of being integrable.
As a sanity check, when $x<0,y<0$ then $F$ points to the bottom-left, so that a positive work along the given curve segment is expected.
answered Mar 16 at 7:34
user7530user7530
35.1k761114
$endgroup$
add a comment |
$begingroup$
Looks fine to me. By "check independence" I assume you mean that you computed that $nabla times F neq 0$ so that $F$ has no chance of being integrable.
As a sanity check, when $x<0,y<0$ then $F$ points to the bottom-left, so that a positive work along the given curve segment is expected.
answered Mar 16 at 7:34
user7530user7530
35.1k761114
$endgroup$
Looks fine to me. By "check independence" I assume you mean that you computed that $nabla times F neq 0$ so that $F$ has no chance of being integrable.
As a sanity check, when $x<0,y<0$ then $F$ points to the bottom-left, so that a positive work along the given curve segment is expected.
answered Mar 16 at 7:34
user7530user7530
35.1k761114
answered Mar 16 at 7:34
user7530user7530
35.1k761114
answered Mar 16 at 7:34
user7530user7530
35.1k761114
answered Mar 16 at 7:34
user7530user7530
35.1k761114
35.1k761114
add a comment |
add a comment |
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