Are the bounds of t always [0, 1] for line integrals?












1












$begingroup$


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







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  • 2




    $begingroup$
    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.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 1:26










  • $begingroup$
    @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!
    $endgroup$
    – GreyZhang
    Dec 4 '18 at 1:28










  • $begingroup$
    Yes. ${}{}{}{}{}$
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 1:28
















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    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.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 1:26










  • $begingroup$
    @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!
    $endgroup$
    – GreyZhang
    Dec 4 '18 at 1:28










  • $begingroup$
    Yes. ${}{}{}{}{}$
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 1:28














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













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edited Dec 4 '18 at 1:34









epimorphic

2,74231533




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asked Dec 4 '18 at 1:25









GreyZhangGreyZhang

61




61








  • 2




    $begingroup$
    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.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 1:26










  • $begingroup$
    @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!
    $endgroup$
    – GreyZhang
    Dec 4 '18 at 1:28










  • $begingroup$
    Yes. ${}{}{}{}{}$
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 1:28














  • 2




    $begingroup$
    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.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 1:26










  • $begingroup$
    @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!
    $endgroup$
    – GreyZhang
    Dec 4 '18 at 1:28










  • $begingroup$
    Yes. ${}{}{}{}{}$
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 1:28








2




2




$begingroup$
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.
$endgroup$
– T. Bongers
Dec 4 '18 at 1:26




$begingroup$
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.
$endgroup$
– T. Bongers
Dec 4 '18 at 1:26












$begingroup$
@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!
$endgroup$
– GreyZhang
Dec 4 '18 at 1:28




$begingroup$
@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!
$endgroup$
– GreyZhang
Dec 4 '18 at 1:28












$begingroup$
Yes. ${}{}{}{}{}$
$endgroup$
– T. Bongers
Dec 4 '18 at 1:28




$begingroup$
Yes. ${}{}{}{}{}$
$endgroup$
– T. Bongers
Dec 4 '18 at 1:28










1 Answer
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$begingroup$

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
$$ int_c^df(x)dx=int_a^b f(varphi(t))cdot frac{dvarphi}{dt}dt.$$






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    $begingroup$

    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
    $$ int_c^df(x)dx=int_a^b f(varphi(t))cdot frac{dvarphi}{dt}dt.$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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
      $$ int_c^df(x)dx=int_a^b f(varphi(t))cdot frac{dvarphi}{dt}dt.$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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
        $$ int_c^df(x)dx=int_a^b f(varphi(t))cdot frac{dvarphi}{dt}dt.$$






        share|cite|improve this answer









        $endgroup$



        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
        $$ int_c^df(x)dx=int_a^b f(varphi(t))cdot frac{dvarphi}{dt}dt.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 1:45









        Antonios-Alexandros RobotisAntonios-Alexandros Robotis

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