Mixing
Calculating the work done of a force field
$begingroup$
A force field in 3-space is given by the formula $$bar{F} (x, y, z) = (x − yz)bar{i} + (y − xz)bar{j} +(x(1−y) +z^2)bar{k}$$ Calculate the work done by $bar{F}$ in moving a particle once around the triangle with vertices $(0, 0, 0), (1, 1, 1), (1, 1, 0)$ in that order.
I'm really not sure how to go about doing this one. Any help would be much appreciated. Thank you.
integration multivariable-calculus line-integrals
asked Jan 28 '17 at 15:50
TheMaster47xTheMaster47x
696
$endgroup$
add a comment |
$begingroup$
A force field in 3-space is given by the formula $$bar{F} (x, y, z) = (x − yz)bar{i} + (y − xz)bar{j} +(x(1−y) +z^2)bar{k}$$ Calculate the work done by $bar{F}$ in moving a particle once around the triangle with vertices $(0, 0, 0), (1, 1, 1), (1, 1, 0)$ in that order.
I'm really not sure how to go about doing this one. Any help would be much appreciated. Thank you.
integration multivariable-calculus line-integrals
asked Jan 28 '17 at 15:50
TheMaster47xTheMaster47x
696
$endgroup$
add a comment |
$begingroup$
A force field in 3-space is given by the formula $$bar{F} (x, y, z) = (x − yz)bar{i} + (y − xz)bar{j} +(x(1−y) +z^2)bar{k}$$ Calculate the work done by $bar{F}$ in moving a particle once around the triangle with vertices $(0, 0, 0), (1, 1, 1), (1, 1, 0)$ in that order.
I'm really not sure how to go about doing this one. Any help would be much appreciated. Thank you.
integration multivariable-calculus line-integrals
asked Jan 28 '17 at 15:50
TheMaster47xTheMaster47x
696
$endgroup$
A force field in 3-space is given by the formula $$bar{F} (x, y, z) = (x − yz)bar{i} + (y − xz)bar{j} +(x(1−y) +z^2)bar{k}$$ Calculate the work done by $bar{F}$ in moving a particle once around the triangle with vertices $(0, 0, 0), (1, 1, 1), (1, 1, 0)$ in that order.
I'm really not sure how to go about doing this one. Any help would be much appreciated. Thank you.
integration multivariable-calculus line-integrals
integration multivariable-calculus line-integrals
asked Jan 28 '17 at 15:50
TheMaster47xTheMaster47x
696
asked Jan 28 '17 at 15:50
TheMaster47xTheMaster47x
696
asked Jan 28 '17 at 15:50
TheMaster47xTheMaster47x
696
asked Jan 28 '17 at 15:50
TheMaster47xTheMaster47x
696
asked Jan 28 '17 at 15:50
TheMaster47xTheMaster47x
696
696
add a comment |
add a comment |
1 Answer
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You can split your triangle into three line segments, calculate the work of $F$ along each segment and add the results. I'll demonstrate the calculation for the first segment.
The line going from $(0,0,0)$ to $(1,1,1)$ can be parametrized as
$$gamma(t) = t(1,1,1) = t vec{i} + t vec{j} + tvec{k}$$
for $0 leq t leq 1$. Hence
$$ int_{gamma} vec{F} ,dvec{r} = int_0^1 vec{F}(gamma(t)) cdot dot{vec{gamma}}(t) , dt = int_0^1 ((t - t^2) vec{i} + (t - t^2) vec{j} + (t(1-t) + t^2) vec{k}) cdot (vec{i} + vec{j} + vec{k}) , dt
\
= int_0^1 (t - t^2) + (t - t^2) + t(1-t) + t^2 , dt = int_0^1 (3t - 2t^2) , dt = left[ frac{3}{2}t^2 - frac{2}{3}t^3right]^{t=1}_{t =0} = frac{5}{6}.$$
answered Jan 28 '17 at 16:05
levaplevap
47.4k33274
$endgroup$
$begingroup$
Going from $(1,1,1)$ to $(1,1,0)$ would $γ(t)=-tvec{k}$ and would the limits still remain the same?
$endgroup$
– TheMaster47x
Jan 28 '17 at 16:10
$begingroup$
No. You want $gamma(0) = (1,1,1)$ and $gamma(1) = (1,1,0)$ so the correct parametrization is $gamma(t) = (1 - t)(1,1,1) + t(1,1,0) = (1,1,1-t) = (1,1,1) - t(0,0,1) = vec{i} + vec{j} + (1 - t)vec{k}$.
$endgroup$
– levap
Jan 28 '17 at 16:12
$begingroup$
I'm going to try to get my head around that, the limits are all $0≤t≤1$?
$endgroup$
– TheMaster47x
Jan 28 '17 at 16:26
$begingroup$
Yep. In general, a parametrization of a line which starts at $vec{p}$ at time $t = 0$ and ends at $vec{q}$ at time $t = 1$ is $gamma(t) = (1 - t) vec{p} + t vec{q}$.
$endgroup$
– levap
Jan 28 '17 at 16:31
$begingroup$
Just to verify does $-frac{1}{3}$ sound right for the second line and $-1$ for the third with a final answer of $-frac{1}{2}$?
$endgroup$
– TheMaster47x
Jan 28 '17 at 17:04
|
show 3 more comments
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$begingroup$
You can split your triangle into three line segments, calculate the work of $F$ along each segment and add the results. I'll demonstrate the calculation for the first segment.
The line going from $(0,0,0)$ to $(1,1,1)$ can be parametrized as
$$gamma(t) = t(1,1,1) = t vec{i} + t vec{j} + tvec{k}$$
for $0 leq t leq 1$. Hence
$$ int_{gamma} vec{F} ,dvec{r} = int_0^1 vec{F}(gamma(t)) cdot dot{vec{gamma}}(t) , dt = int_0^1 ((t - t^2) vec{i} + (t - t^2) vec{j} + (t(1-t) + t^2) vec{k}) cdot (vec{i} + vec{j} + vec{k}) , dt
\
= int_0^1 (t - t^2) + (t - t^2) + t(1-t) + t^2 , dt = int_0^1 (3t - 2t^2) , dt = left[ frac{3}{2}t^2 - frac{2}{3}t^3right]^{t=1}_{t =0} = frac{5}{6}.$$
answered Jan 28 '17 at 16:05
levaplevap
47.4k33274
$endgroup$
$begingroup$
Going from $(1,1,1)$ to $(1,1,0)$ would $γ(t)=-tvec{k}$ and would the limits still remain the same?
$endgroup$
– TheMaster47x
Jan 28 '17 at 16:10
$begingroup$
No. You want $gamma(0) = (1,1,1)$ and $gamma(1) = (1,1,0)$ so the correct parametrization is $gamma(t) = (1 - t)(1,1,1) + t(1,1,0) = (1,1,1-t) = (1,1,1) - t(0,0,1) = vec{i} + vec{j} + (1 - t)vec{k}$.
$endgroup$
– levap
Jan 28 '17 at 16:12
$begingroup$
I'm going to try to get my head around that, the limits are all $0≤t≤1$?
$endgroup$
– TheMaster47x
Jan 28 '17 at 16:26
$begingroup$
Yep. In general, a parametrization of a line which starts at $vec{p}$ at time $t = 0$ and ends at $vec{q}$ at time $t = 1$ is $gamma(t) = (1 - t) vec{p} + t vec{q}$.
$endgroup$
– levap
Jan 28 '17 at 16:31
$begingroup$
Just to verify does $-frac{1}{3}$ sound right for the second line and $-1$ for the third with a final answer of $-frac{1}{2}$?
$endgroup$
– TheMaster47x
Jan 28 '17 at 17:04
|
show 3 more comments
$begingroup$
You can split your triangle into three line segments, calculate the work of $F$ along each segment and add the results. I'll demonstrate the calculation for the first segment.
The line going from $(0,0,0)$ to $(1,1,1)$ can be parametrized as
$$gamma(t) = t(1,1,1) = t vec{i} + t vec{j} + tvec{k}$$
for $0 leq t leq 1$. Hence
$$ int_{gamma} vec{F} ,dvec{r} = int_0^1 vec{F}(gamma(t)) cdot dot{vec{gamma}}(t) , dt = int_0^1 ((t - t^2) vec{i} + (t - t^2) vec{j} + (t(1-t) + t^2) vec{k}) cdot (vec{i} + vec{j} + vec{k}) , dt
\
= int_0^1 (t - t^2) + (t - t^2) + t(1-t) + t^2 , dt = int_0^1 (3t - 2t^2) , dt = left[ frac{3}{2}t^2 - frac{2}{3}t^3right]^{t=1}_{t =0} = frac{5}{6}.$$
answered Jan 28 '17 at 16:05
levaplevap
47.4k33274
$endgroup$
$begingroup$
Going from $(1,1,1)$ to $(1,1,0)$ would $γ(t)=-tvec{k}$ and would the limits still remain the same?
$endgroup$
– TheMaster47x
Jan 28 '17 at 16:10
$begingroup$
No. You want $gamma(0) = (1,1,1)$ and $gamma(1) = (1,1,0)$ so the correct parametrization is $gamma(t) = (1 - t)(1,1,1) + t(1,1,0) = (1,1,1-t) = (1,1,1) - t(0,0,1) = vec{i} + vec{j} + (1 - t)vec{k}$.
$endgroup$
– levap
Jan 28 '17 at 16:12
$begingroup$
I'm going to try to get my head around that, the limits are all $0≤t≤1$?
$endgroup$
– TheMaster47x
Jan 28 '17 at 16:26
$begingroup$
Yep. In general, a parametrization of a line which starts at $vec{p}$ at time $t = 0$ and ends at $vec{q}$ at time $t = 1$ is $gamma(t) = (1 - t) vec{p} + t vec{q}$.
$endgroup$
– levap
Jan 28 '17 at 16:31
$begingroup$
Just to verify does $-frac{1}{3}$ sound right for the second line and $-1$ for the third with a final answer of $-frac{1}{2}$?
$endgroup$
– TheMaster47x
Jan 28 '17 at 17:04
|
show 3 more comments
$begingroup$
You can split your triangle into three line segments, calculate the work of $F$ along each segment and add the results. I'll demonstrate the calculation for the first segment.
The line going from $(0,0,0)$ to $(1,1,1)$ can be parametrized as
$$gamma(t) = t(1,1,1) = t vec{i} + t vec{j} + tvec{k}$$
for $0 leq t leq 1$. Hence
$$ int_{gamma} vec{F} ,dvec{r} = int_0^1 vec{F}(gamma(t)) cdot dot{vec{gamma}}(t) , dt = int_0^1 ((t - t^2) vec{i} + (t - t^2) vec{j} + (t(1-t) + t^2) vec{k}) cdot (vec{i} + vec{j} + vec{k}) , dt
\
= int_0^1 (t - t^2) + (t - t^2) + t(1-t) + t^2 , dt = int_0^1 (3t - 2t^2) , dt = left[ frac{3}{2}t^2 - frac{2}{3}t^3right]^{t=1}_{t =0} = frac{5}{6}.$$
answered Jan 28 '17 at 16:05
levaplevap
47.4k33274
$endgroup$
You can split your triangle into three line segments, calculate the work of $F$ along each segment and add the results. I'll demonstrate the calculation for the first segment.
The line going from $(0,0,0)$ to $(1,1,1)$ can be parametrized as
$$gamma(t) = t(1,1,1) = t vec{i} + t vec{j} + tvec{k}$$
for $0 leq t leq 1$. Hence
$$ int_{gamma} vec{F} ,dvec{r} = int_0^1 vec{F}(gamma(t)) cdot dot{vec{gamma}}(t) , dt = int_0^1 ((t - t^2) vec{i} + (t - t^2) vec{j} + (t(1-t) + t^2) vec{k}) cdot (vec{i} + vec{j} + vec{k}) , dt
\
= int_0^1 (t - t^2) + (t - t^2) + t(1-t) + t^2 , dt = int_0^1 (3t - 2t^2) , dt = left[ frac{3}{2}t^2 - frac{2}{3}t^3right]^{t=1}_{t =0} = frac{5}{6}.$$
answered Jan 28 '17 at 16:05
levaplevap
47.4k33274
answered Jan 28 '17 at 16:05
levaplevap
47.4k33274
answered Jan 28 '17 at 16:05
levaplevap
47.4k33274
answered Jan 28 '17 at 16:05
levaplevap
47.4k33274
47.4k33274
$begingroup$
Going from $(1,1,1)$ to $(1,1,0)$ would $γ(t)=-tvec{k}$ and would the limits still remain the same?
$endgroup$
– TheMaster47x
Jan 28 '17 at 16:10
$begingroup$
No. You want $gamma(0) = (1,1,1)$ and $gamma(1) = (1,1,0)$ so the correct parametrization is $gamma(t) = (1 - t)(1,1,1) + t(1,1,0) = (1,1,1-t) = (1,1,1) - t(0,0,1) = vec{i} + vec{j} + (1 - t)vec{k}$.
$endgroup$
– levap
Jan 28 '17 at 16:12
$begingroup$
I'm going to try to get my head around that, the limits are all $0≤t≤1$?
$endgroup$
– TheMaster47x
Jan 28 '17 at 16:26
$begingroup$
Yep. In general, a parametrization of a line which starts at $vec{p}$ at time $t = 0$ and ends at $vec{q}$ at time $t = 1$ is $gamma(t) = (1 - t) vec{p} + t vec{q}$.
$endgroup$
– levap
Jan 28 '17 at 16:31
$begingroup$
Just to verify does $-frac{1}{3}$ sound right for the second line and $-1$ for the third with a final answer of $-frac{1}{2}$?
$endgroup$
– TheMaster47x
Jan 28 '17 at 17:04
|
show 3 more comments
$begingroup$
Going from $(1,1,1)$ to $(1,1,0)$ would $γ(t)=-tvec{k}$ and would the limits still remain the same?
$endgroup$
– TheMaster47x
Jan 28 '17 at 16:10
$begingroup$
No. You want $gamma(0) = (1,1,1)$ and $gamma(1) = (1,1,0)$ so the correct parametrization is $gamma(t) = (1 - t)(1,1,1) + t(1,1,0) = (1,1,1-t) = (1,1,1) - t(0,0,1) = vec{i} + vec{j} + (1 - t)vec{k}$.
$endgroup$
– levap
Jan 28 '17 at 16:12
$begingroup$
I'm going to try to get my head around that, the limits are all $0≤t≤1$?
$endgroup$
– TheMaster47x
Jan 28 '17 at 16:26
$begingroup$
Yep. In general, a parametrization of a line which starts at $vec{p}$ at time $t = 0$ and ends at $vec{q}$ at time $t = 1$ is $gamma(t) = (1 - t) vec{p} + t vec{q}$.
$endgroup$
– levap
Jan 28 '17 at 16:31
$begingroup$
Just to verify does $-frac{1}{3}$ sound right for the second line and $-1$ for the third with a final answer of $-frac{1}{2}$?
$endgroup$
– TheMaster47x
Jan 28 '17 at 17:04
$begingroup$
Going from $(1,1,1)$ to $(1,1,0)$ would $γ(t)=-tvec{k}$ and would the limits still remain the same?
$endgroup$
– TheMaster47x
Jan 28 '17 at 16:10
$begingroup$
Going from $(1,1,1)$ to $(1,1,0)$ would $γ(t)=-tvec{k}$ and would the limits still remain the same?
$endgroup$
– TheMaster47x
Jan 28 '17 at 16:10
$begingroup$
No. You want $gamma(0) = (1,1,1)$ and $gamma(1) = (1,1,0)$ so the correct parametrization is $gamma(t) = (1 - t)(1,1,1) + t(1,1,0) = (1,1,1-t) = (1,1,1) - t(0,0,1) = vec{i} + vec{j} + (1 - t)vec{k}$.
$endgroup$
– levap
Jan 28 '17 at 16:12
$begingroup$
No. You want $gamma(0) = (1,1,1)$ and $gamma(1) = (1,1,0)$ so the correct parametrization is $gamma(t) = (1 - t)(1,1,1) + t(1,1,0) = (1,1,1-t) = (1,1,1) - t(0,0,1) = vec{i} + vec{j} + (1 - t)vec{k}$.
$endgroup$
– levap
Jan 28 '17 at 16:12
$begingroup$
I'm going to try to get my head around that, the limits are all $0≤t≤1$?
$endgroup$
– TheMaster47x
Jan 28 '17 at 16:26
$begingroup$
I'm going to try to get my head around that, the limits are all $0≤t≤1$?
$endgroup$
– TheMaster47x
Jan 28 '17 at 16:26
$begingroup$
Yep. In general, a parametrization of a line which starts at $vec{p}$ at time $t = 0$ and ends at $vec{q}$ at time $t = 1$ is $gamma(t) = (1 - t) vec{p} + t vec{q}$.
$endgroup$
– levap
Jan 28 '17 at 16:31
$begingroup$
Yep. In general, a parametrization of a line which starts at $vec{p}$ at time $t = 0$ and ends at $vec{q}$ at time $t = 1$ is $gamma(t) = (1 - t) vec{p} + t vec{q}$.
$endgroup$
– levap
Jan 28 '17 at 16:31
$begingroup$
Just to verify does $-frac{1}{3}$ sound right for the second line and $-1$ for the third with a final answer of $-frac{1}{2}$?
$endgroup$
– TheMaster47x
Jan 28 '17 at 17:04
$begingroup$
Just to verify does $-frac{1}{3}$ sound right for the second line and $-1$ for the third with a final answer of $-frac{1}{2}$?
$endgroup$
– TheMaster47x
Jan 28 '17 at 17:04
|
show 3 more comments
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