Are the bounds of t always [0, 1] for line integrals?
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I was given the task to find the line integral $int _C (x+y)ds$ where $C$ is the line segment from $(0,1,1)$ to $(3, 2, 2)$.
I parameterised $C$ as $3tvec{i}+(1+t)vec{j}+(1+t)vec{k}$, which means that $vec{r'}(t)=3vec{i}+vec{j}+vec{k}$ and that $||vec{r'}(t)||=sqrt11$.
This gave me the integral $sqrt11 int(4t+1)dt$, but then I wasn't sure what the bounds of $t$ are supposed to be. In class, we've been using $t space∈space [0,1]$, and if I use that, I get $3sqrt11$.
The only thing I'm uncertain of here is why the bounds of $t$ go from $0$ to $1$. And if they don't, how can I determine what the bounds are? (Also, if I made any other mistakes in my calculation, please correct me!) Thank you very much!
multivariable-calculus line-integrals bounds-of-integration
edited Dec 4 at 1:34
epimorphic
2,72331533
asked Dec 4 at 1:25
GreyZhang
61
add a comment |
up vote
1
down vote
favorite
I was given the task to find the line integral $int _C (x+y)ds$ where $C$ is the line segment from $(0,1,1)$ to $(3, 2, 2)$.
I parameterised $C$ as $3tvec{i}+(1+t)vec{j}+(1+t)vec{k}$, which means that $vec{r'}(t)=3vec{i}+vec{j}+vec{k}$ and that $||vec{r'}(t)||=sqrt11$.
This gave me the integral $sqrt11 int(4t+1)dt$, but then I wasn't sure what the bounds of $t$ are supposed to be. In class, we've been using $t space∈space [0,1]$, and if I use that, I get $3sqrt11$.
The only thing I'm uncertain of here is why the bounds of $t$ go from $0$ to $1$. And if they don't, how can I determine what the bounds are? (Also, if I made any other mistakes in my calculation, please correct me!) Thank you very much!
multivariable-calculus line-integrals bounds-of-integration
edited Dec 4 at 1:34
epimorphic
2,72331533
asked Dec 4 at 1:25
GreyZhang
61
2
It depends on your parameterization of the curve. Does $t = 0$ correspond to the initial point and $t = 1$ correspond to the terminal point? If yes, cool. If no, use other bounds.
– T. Bongers
Dec 4 at 1:26
@T.Bongers when you say 'correspond,' do you mean if I substitute t=0 and t=1 back into the parameterisation, I get the initial and terminal points respectively? Thank you!
– GreyZhang
Dec 4 at 1:28
Yes. ${}{}{}{}{}$
– T. Bongers
Dec 4 at 1:28
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I was given the task to find the line integral $int _C (x+y)ds$ where $C$ is the line segment from $(0,1,1)$ to $(3, 2, 2)$.
I parameterised $C$ as $3tvec{i}+(1+t)vec{j}+(1+t)vec{k}$, which means that $vec{r'}(t)=3vec{i}+vec{j}+vec{k}$ and that $||vec{r'}(t)||=sqrt11$.
This gave me the integral $sqrt11 int(4t+1)dt$, but then I wasn't sure what the bounds of $t$ are supposed to be. In class, we've been using $t space∈space [0,1]$, and if I use that, I get $3sqrt11$.
The only thing I'm uncertain of here is why the bounds of $t$ go from $0$ to $1$. And if they don't, how can I determine what the bounds are? (Also, if I made any other mistakes in my calculation, please correct me!) Thank you very much!
multivariable-calculus line-integrals bounds-of-integration
edited Dec 4 at 1:34
epimorphic
2,72331533
asked Dec 4 at 1:25
GreyZhang
61
I was given the task to find the line integral $int _C (x+y)ds$ where $C$ is the line segment from $(0,1,1)$ to $(3, 2, 2)$.
I parameterised $C$ as $3tvec{i}+(1+t)vec{j}+(1+t)vec{k}$, which means that $vec{r'}(t)=3vec{i}+vec{j}+vec{k}$ and that $||vec{r'}(t)||=sqrt11$.
This gave me the integral $sqrt11 int(4t+1)dt$, but then I wasn't sure what the bounds of $t$ are supposed to be. In class, we've been using $t space∈space [0,1]$, and if I use that, I get $3sqrt11$.
The only thing I'm uncertain of here is why the bounds of $t$ go from $0$ to $1$. And if they don't, how can I determine what the bounds are? (Also, if I made any other mistakes in my calculation, please correct me!) Thank you very much!
multivariable-calculus line-integrals bounds-of-integration
multivariable-calculus line-integrals bounds-of-integration
edited Dec 4 at 1:34
epimorphic
2,72331533
asked Dec 4 at 1:25
GreyZhang
61
edited Dec 4 at 1:34
epimorphic
2,72331533
asked Dec 4 at 1:25
GreyZhang
61
edited Dec 4 at 1:34
epimorphic
2,72331533
edited Dec 4 at 1:34
epimorphic
2,72331533
edited Dec 4 at 1:34
epimorphic
2,72331533
2,72331533
asked Dec 4 at 1:25
GreyZhang
61
asked Dec 4 at 1:25
GreyZhang
61
asked Dec 4 at 1:25
GreyZhang
61
61
2
It depends on your parameterization of the curve. Does $t = 0$ correspond to the initial point and $t = 1$ correspond to the terminal point? If yes, cool. If no, use other bounds.
– T. Bongers
Dec 4 at 1:26
@T.Bongers when you say 'correspond,' do you mean if I substitute t=0 and t=1 back into the parameterisation, I get the initial and terminal points respectively? Thank you!
– GreyZhang
Dec 4 at 1:28
Yes. ${}{}{}{}{}$
– T. Bongers
Dec 4 at 1:28
add a comment |
2
It depends on your parameterization of the curve. Does $t = 0$ correspond to the initial point and $t = 1$ correspond to the terminal point? If yes, cool. If no, use other bounds.
– T. Bongers
Dec 4 at 1:26
@T.Bongers when you say 'correspond,' do you mean if I substitute t=0 and t=1 back into the parameterisation, I get the initial and terminal points respectively? Thank you!
– GreyZhang
Dec 4 at 1:28
Yes. ${}{}{}{}{}$
– T. Bongers
Dec 4 at 1:28
2
2
It depends on your parameterization of the curve. Does $t = 0$ correspond to the initial point and $t = 1$ correspond to the terminal point? If yes, cool. If no, use other bounds.
– T. Bongers
Dec 4 at 1:26
It depends on your parameterization of the curve. Does $t = 0$ correspond to the initial point and $t = 1$ correspond to the terminal point? If yes, cool. If no, use other bounds.
– T. Bongers
Dec 4 at 1:26
@T.Bongers when you say 'correspond,' do you mean if I substitute t=0 and t=1 back into the parameterisation, I get the initial and terminal points respectively? Thank you!
– GreyZhang
Dec 4 at 1:28
@T.Bongers when you say 'correspond,' do you mean if I substitute t=0 and t=1 back into the parameterisation, I get the initial and terminal points respectively? Thank you!
– GreyZhang
Dec 4 at 1:28
Yes. ${}{}{}{}{}$
– T. Bongers
Dec 4 at 1:28
Yes. ${}{}{}{}{}$
– T. Bongers
Dec 4 at 1:28
add a comment |
1 Answer
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In short this is dependent on the choice of parametrization. As a simple example, let's set up the line integral of $x+y$ along the line segment $C$ connecting $(0,0)$ and $(1,1)$ in $mathbb{R}^2$. We can parametrize $C$ as $C(t)=(t,t)$ for $0le tle 1$ or we can parametrize $C$ "more quickly" using $C(t)=(2t,2t)$ for $0le tle frac{1}{2}$. Using the first parametrization:
$$ int_C (x+y)ds=int_0^1x(t)frac{dx}{dt}cdot dt+int_0^1 y(t)frac{dy}{dt}cdot dt=int_0^1tcdot dt+int_0^1tcdot dt=int_0^12tcdot dt=t^2bigg|_0^1=1.$$
Using the second parametrization:
$$ int_C (x+y)ds=int_0^{frac{1}{2}}x(t)frac{dx}{dt}cdot dt+int_0^frac{1}{2} y(t)frac{dy}{dt}cdot dt=int_0^{frac{1}{2}}4tcdot dt+int_0^{frac{1}{2}}4tcdot dt=int_0^{frac{1}{2}}8tcdot dt=4t^2bigg|_0^{frac{1}{2}}=1.$$
So, the integral does not care about the parametrization. More particularly, we can see that the $frac{dx}{dt}$ and $frac{dy}{dt}$ terms "correct" the error introduced by reparametrization.
This is actually a special case of the Change of Variables formula, which in this case tells us that given a reparametrization $varphi:[a,b]to [c,d]$ then
$$ int_c^df(x)dx=int_a^b f(varphi(t))cdot frac{dvarphi}{dt}dt.$$
answered Dec 4 at 1:45
Antonios-Alexandros Robotis
9,10541640
add a comment |
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1 Answer
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In short this is dependent on the choice of parametrization. As a simple example, let's set up the line integral of $x+y$ along the line segment $C$ connecting $(0,0)$ and $(1,1)$ in $mathbb{R}^2$. We can parametrize $C$ as $C(t)=(t,t)$ for $0le tle 1$ or we can parametrize $C$ "more quickly" using $C(t)=(2t,2t)$ for $0le tle frac{1}{2}$. Using the first parametrization:
$$ int_C (x+y)ds=int_0^1x(t)frac{dx}{dt}cdot dt+int_0^1 y(t)frac{dy}{dt}cdot dt=int_0^1tcdot dt+int_0^1tcdot dt=int_0^12tcdot dt=t^2bigg|_0^1=1.$$
Using the second parametrization:
$$ int_C (x+y)ds=int_0^{frac{1}{2}}x(t)frac{dx}{dt}cdot dt+int_0^frac{1}{2} y(t)frac{dy}{dt}cdot dt=int_0^{frac{1}{2}}4tcdot dt+int_0^{frac{1}{2}}4tcdot dt=int_0^{frac{1}{2}}8tcdot dt=4t^2bigg|_0^{frac{1}{2}}=1.$$
So, the integral does not care about the parametrization. More particularly, we can see that the $frac{dx}{dt}$ and $frac{dy}{dt}$ terms "correct" the error introduced by reparametrization.
This is actually a special case of the Change of Variables formula, which in this case tells us that given a reparametrization $varphi:[a,b]to [c,d]$ then
$$ int_c^df(x)dx=int_a^b f(varphi(t))cdot frac{dvarphi}{dt}dt.$$
answered Dec 4 at 1:45
Antonios-Alexandros Robotis
9,10541640
add a comment |
up vote
0
down vote
In short this is dependent on the choice of parametrization. As a simple example, let's set up the line integral of $x+y$ along the line segment $C$ connecting $(0,0)$ and $(1,1)$ in $mathbb{R}^2$. We can parametrize $C$ as $C(t)=(t,t)$ for $0le tle 1$ or we can parametrize $C$ "more quickly" using $C(t)=(2t,2t)$ for $0le tle frac{1}{2}$. Using the first parametrization:
$$ int_C (x+y)ds=int_0^1x(t)frac{dx}{dt}cdot dt+int_0^1 y(t)frac{dy}{dt}cdot dt=int_0^1tcdot dt+int_0^1tcdot dt=int_0^12tcdot dt=t^2bigg|_0^1=1.$$
Using the second parametrization:
$$ int_C (x+y)ds=int_0^{frac{1}{2}}x(t)frac{dx}{dt}cdot dt+int_0^frac{1}{2} y(t)frac{dy}{dt}cdot dt=int_0^{frac{1}{2}}4tcdot dt+int_0^{frac{1}{2}}4tcdot dt=int_0^{frac{1}{2}}8tcdot dt=4t^2bigg|_0^{frac{1}{2}}=1.$$
So, the integral does not care about the parametrization. More particularly, we can see that the $frac{dx}{dt}$ and $frac{dy}{dt}$ terms "correct" the error introduced by reparametrization.
This is actually a special case of the Change of Variables formula, which in this case tells us that given a reparametrization $varphi:[a,b]to [c,d]$ then
$$ int_c^df(x)dx=int_a^b f(varphi(t))cdot frac{dvarphi}{dt}dt.$$
answered Dec 4 at 1:45
Antonios-Alexandros Robotis
9,10541640
add a comment |
up vote
0
down vote
up vote
0
down vote
In short this is dependent on the choice of parametrization. As a simple example, let's set up the line integral of $x+y$ along the line segment $C$ connecting $(0,0)$ and $(1,1)$ in $mathbb{R}^2$. We can parametrize $C$ as $C(t)=(t,t)$ for $0le tle 1$ or we can parametrize $C$ "more quickly" using $C(t)=(2t,2t)$ for $0le tle frac{1}{2}$. Using the first parametrization:
$$ int_C (x+y)ds=int_0^1x(t)frac{dx}{dt}cdot dt+int_0^1 y(t)frac{dy}{dt}cdot dt=int_0^1tcdot dt+int_0^1tcdot dt=int_0^12tcdot dt=t^2bigg|_0^1=1.$$
Using the second parametrization:
$$ int_C (x+y)ds=int_0^{frac{1}{2}}x(t)frac{dx}{dt}cdot dt+int_0^frac{1}{2} y(t)frac{dy}{dt}cdot dt=int_0^{frac{1}{2}}4tcdot dt+int_0^{frac{1}{2}}4tcdot dt=int_0^{frac{1}{2}}8tcdot dt=4t^2bigg|_0^{frac{1}{2}}=1.$$
So, the integral does not care about the parametrization. More particularly, we can see that the $frac{dx}{dt}$ and $frac{dy}{dt}$ terms "correct" the error introduced by reparametrization.
This is actually a special case of the Change of Variables formula, which in this case tells us that given a reparametrization $varphi:[a,b]to [c,d]$ then
$$ int_c^df(x)dx=int_a^b f(varphi(t))cdot frac{dvarphi}{dt}dt.$$
answered Dec 4 at 1:45
Antonios-Alexandros Robotis
9,10541640
In short this is dependent on the choice of parametrization. As a simple example, let's set up the line integral of $x+y$ along the line segment $C$ connecting $(0,0)$ and $(1,1)$ in $mathbb{R}^2$. We can parametrize $C$ as $C(t)=(t,t)$ for $0le tle 1$ or we can parametrize $C$ "more quickly" using $C(t)=(2t,2t)$ for $0le tle frac{1}{2}$. Using the first parametrization:
$$ int_C (x+y)ds=int_0^1x(t)frac{dx}{dt}cdot dt+int_0^1 y(t)frac{dy}{dt}cdot dt=int_0^1tcdot dt+int_0^1tcdot dt=int_0^12tcdot dt=t^2bigg|_0^1=1.$$
Using the second parametrization:
$$ int_C (x+y)ds=int_0^{frac{1}{2}}x(t)frac{dx}{dt}cdot dt+int_0^frac{1}{2} y(t)frac{dy}{dt}cdot dt=int_0^{frac{1}{2}}4tcdot dt+int_0^{frac{1}{2}}4tcdot dt=int_0^{frac{1}{2}}8tcdot dt=4t^2bigg|_0^{frac{1}{2}}=1.$$
So, the integral does not care about the parametrization. More particularly, we can see that the $frac{dx}{dt}$ and $frac{dy}{dt}$ terms "correct" the error introduced by reparametrization.
This is actually a special case of the Change of Variables formula, which in this case tells us that given a reparametrization $varphi:[a,b]to [c,d]$ then
$$ int_c^df(x)dx=int_a^b f(varphi(t))cdot frac{dvarphi}{dt}dt.$$
answered Dec 4 at 1:45
Antonios-Alexandros Robotis
9,10541640
answered Dec 4 at 1:45
Antonios-Alexandros Robotis
9,10541640
answered Dec 4 at 1:45
Antonios-Alexandros Robotis
9,10541640
answered Dec 4 at 1:45
Antonios-Alexandros Robotis
9,10541640
9,10541640
add a comment |
add a comment |
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It depends on your parameterization of the curve. Does $t = 0$ correspond to the initial point and $t = 1$ correspond to the terminal point? If yes, cool. If no, use other bounds.
– T. Bongers
Dec 4 at 1:26
@T.Bongers when you say 'correspond,' do you mean if I substitute t=0 and t=1 back into the parameterisation, I get the initial and terminal points respectively? Thank you!
– GreyZhang
Dec 4 at 1:28
Yes. ${}{}{}{}{}$
– T. Bongers
Dec 4 at 1:28