How to integrate the curve












2












$begingroup$


We have a Curve $C:vec{x}left(tright)=begin{pmatrix}1-2t^2\ tend{pmatrix}$ Now you have to calculate $int _Cvec{F}left(vec{x}right)dvec{x}$ for




  • $vec{F}left(vec{x}right)=begin{pmatrix}1\ 0end{pmatrix}$


  • $vec{F}left(vec{x}right)=begin{pmatrix}x_2\ x_1end{pmatrix}$



There are many more practice problems just like this and I'd like to try and solve them but I don't know how to start.



Do I make $1-2t^2 =1$ or $t=0$?



What does $vec{F}left(vec{x}right)=begin{pmatrix}x_2\ x_1end{pmatrix}$ even mean?



Question Pic






calculus integration definite-integrals vectors vector-fields







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    there $x:=(x_1,x_2,ldots,x_n)$, assuming that $F:Bbb R^ntoBbb R^m$, so $F(x)=F(x_1,x_2,ldots,x_n)=(x_2,x_1)$ is a function from $Bbb R^ntoBbb R^2$. By the other hand $F(x)=(1,0)$ also goes from $Bbb R^n$ to $Bbb R^2$ (it is a constant function, it maps every $xinBbb R^n$ to the vector $(1,0)inBbb R^2$)
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:58








  • 2




    $begingroup$
    seeing the definition of $x(t)$ I guess that $n=2$ for this exercise, otherwise the composition $(Fcirc x)(t)=F(x(t))$ would not be defined
    $endgroup$
    – Masacroso
    Dec 5 '18 at 3:04
















2












$begingroup$


We have a Curve $C:vec{x}left(tright)=begin{pmatrix}1-2t^2\ tend{pmatrix}$ Now you have to calculate $int _Cvec{F}left(vec{x}right)dvec{x}$ for




  • $vec{F}left(vec{x}right)=begin{pmatrix}1\ 0end{pmatrix}$


  • $vec{F}left(vec{x}right)=begin{pmatrix}x_2\ x_1end{pmatrix}$



There are many more practice problems just like this and I'd like to try and solve them but I don't know how to start.



Do I make $1-2t^2 =1$ or $t=0$?



What does $vec{F}left(vec{x}right)=begin{pmatrix}x_2\ x_1end{pmatrix}$ even mean?



Question Pic






calculus integration definite-integrals vectors vector-fields







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    there $x:=(x_1,x_2,ldots,x_n)$, assuming that $F:Bbb R^ntoBbb R^m$, so $F(x)=F(x_1,x_2,ldots,x_n)=(x_2,x_1)$ is a function from $Bbb R^ntoBbb R^2$. By the other hand $F(x)=(1,0)$ also goes from $Bbb R^n$ to $Bbb R^2$ (it is a constant function, it maps every $xinBbb R^n$ to the vector $(1,0)inBbb R^2$)
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:58








  • 2




    $begingroup$
    seeing the definition of $x(t)$ I guess that $n=2$ for this exercise, otherwise the composition $(Fcirc x)(t)=F(x(t))$ would not be defined
    $endgroup$
    – Masacroso
    Dec 5 '18 at 3:04














2












2








2





$begingroup$


We have a Curve $C:vec{x}left(tright)=begin{pmatrix}1-2t^2\ tend{pmatrix}$ Now you have to calculate $int _Cvec{F}left(vec{x}right)dvec{x}$ for




  • $vec{F}left(vec{x}right)=begin{pmatrix}1\ 0end{pmatrix}$


  • $vec{F}left(vec{x}right)=begin{pmatrix}x_2\ x_1end{pmatrix}$



There are many more practice problems just like this and I'd like to try and solve them but I don't know how to start.



Do I make $1-2t^2 =1$ or $t=0$?



What does $vec{F}left(vec{x}right)=begin{pmatrix}x_2\ x_1end{pmatrix}$ even mean?



Question Pic






calculus integration definite-integrals vectors vector-fields







share|cite|improve this question











$endgroup$




We have a Curve $C:vec{x}left(tright)=begin{pmatrix}1-2t^2\ tend{pmatrix}$ Now you have to calculate $int _Cvec{F}left(vec{x}right)dvec{x}$ for




  • $vec{F}left(vec{x}right)=begin{pmatrix}1\ 0end{pmatrix}$


  • $vec{F}left(vec{x}right)=begin{pmatrix}x_2\ x_1end{pmatrix}$



There are many more practice problems just like this and I'd like to try and solve them but I don't know how to start.



Do I make $1-2t^2 =1$ or $t=0$?



What does $vec{F}left(vec{x}right)=begin{pmatrix}x_2\ x_1end{pmatrix}$ even mean?



Question Pic






calculus integration definite-integrals vectors vector-fields




calculus integration definite-integrals vectors vector-fields






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 4:33







Naochi

















asked Dec 5 '18 at 2:50









NaochiNaochi

626




626








  • 1




    $begingroup$
    there $x:=(x_1,x_2,ldots,x_n)$, assuming that $F:Bbb R^ntoBbb R^m$, so $F(x)=F(x_1,x_2,ldots,x_n)=(x_2,x_1)$ is a function from $Bbb R^ntoBbb R^2$. By the other hand $F(x)=(1,0)$ also goes from $Bbb R^n$ to $Bbb R^2$ (it is a constant function, it maps every $xinBbb R^n$ to the vector $(1,0)inBbb R^2$)
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:58








  • 2




    $begingroup$
    seeing the definition of $x(t)$ I guess that $n=2$ for this exercise, otherwise the composition $(Fcirc x)(t)=F(x(t))$ would not be defined
    $endgroup$
    – Masacroso
    Dec 5 '18 at 3:04














  • 1




    $begingroup$
    there $x:=(x_1,x_2,ldots,x_n)$, assuming that $F:Bbb R^ntoBbb R^m$, so $F(x)=F(x_1,x_2,ldots,x_n)=(x_2,x_1)$ is a function from $Bbb R^ntoBbb R^2$. By the other hand $F(x)=(1,0)$ also goes from $Bbb R^n$ to $Bbb R^2$ (it is a constant function, it maps every $xinBbb R^n$ to the vector $(1,0)inBbb R^2$)
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:58








  • 2




    $begingroup$
    seeing the definition of $x(t)$ I guess that $n=2$ for this exercise, otherwise the composition $(Fcirc x)(t)=F(x(t))$ would not be defined
    $endgroup$
    – Masacroso
    Dec 5 '18 at 3:04








1




1




$begingroup$
there $x:=(x_1,x_2,ldots,x_n)$, assuming that $F:Bbb R^ntoBbb R^m$, so $F(x)=F(x_1,x_2,ldots,x_n)=(x_2,x_1)$ is a function from $Bbb R^ntoBbb R^2$. By the other hand $F(x)=(1,0)$ also goes from $Bbb R^n$ to $Bbb R^2$ (it is a constant function, it maps every $xinBbb R^n$ to the vector $(1,0)inBbb R^2$)
$endgroup$
– Masacroso
Dec 5 '18 at 2:58






$begingroup$
there $x:=(x_1,x_2,ldots,x_n)$, assuming that $F:Bbb R^ntoBbb R^m$, so $F(x)=F(x_1,x_2,ldots,x_n)=(x_2,x_1)$ is a function from $Bbb R^ntoBbb R^2$. By the other hand $F(x)=(1,0)$ also goes from $Bbb R^n$ to $Bbb R^2$ (it is a constant function, it maps every $xinBbb R^n$ to the vector $(1,0)inBbb R^2$)
$endgroup$
– Masacroso
Dec 5 '18 at 2:58






2




2




$begingroup$
seeing the definition of $x(t)$ I guess that $n=2$ for this exercise, otherwise the composition $(Fcirc x)(t)=F(x(t))$ would not be defined
$endgroup$
– Masacroso
Dec 5 '18 at 3:04




$begingroup$
seeing the definition of $x(t)$ I guess that $n=2$ for this exercise, otherwise the composition $(Fcirc x)(t)=F(x(t))$ would not be defined
$endgroup$
– Masacroso
Dec 5 '18 at 3:04










1 Answer
1






active

oldest

votes


















2












$begingroup$

Note: if you dont know the notation $fcirc g$ just mean the composition of functions $f$ and $g$, that is, $(fcirc g)(s)=f(g(s))$.



There is something that could confuse you: the symbol $x$ is used for two different purposes in the exercise




  1. to denote a vector on the domain of $F$, and


  2. to denote a function from $Bbb RtoBbb R^2$.



Just substituting we have that



$$(Fcirc x)(t)=F(1-2t^2,1)$$



where we assumed here that the domain of $F$ is some subset of $Bbb R^2$ (probably $Bbb R^2$ itself) because the image of the function $x$ is in $Bbb R^2$.



Now, for the first case we have that $(Fcirc x)(t)=(1,0)$, that is, every $tinBbb R$ defines some $x(t)inBbb R^2$, thus $F(x(t))=(1,0)$ because $F$ is already constant.



For the second case we have that $(Fcirc x)(t)=(t,1-2t^2)$, assuming that the notation $x_k$ refer to the coordinates of the vector $x$.



Can you evaluate $int_C F(x)cdot dx$ in each case now?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok I will try to do it now. I can evaluate it in the first case. But I am still confused about the second one.
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:29










  • $begingroup$
    Ok I still do not understand where am I supposed to use the (1, 0). Before the integration? Will it be used to find t?
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:37










  • $begingroup$
    $int _0^1-4tdt=-2$ This is what I understood from your explanation, I multiply F(x) with the derivative of x(t). Is that correct? @Masacroso
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:43










  • $begingroup$
    Ok nvm I could do it :) Thanks for the explanation. So for the second case it would be $int _0^1begin{pmatrix}t\ 1-2t^2end{pmatrix}cdot begin{pmatrix}-4t\ 1end{pmatrix}$ which would then be $int _0^1-4t^2+1-2t^2dt=-1$ No?
    $endgroup$
    – Naochi
    Dec 5 '18 at 4:01










  • $begingroup$
    @Naochi we have that $$int_C F(x)cdot dx=int_a^b (Fcirc x)(t)cdot x'(t), dt$$ assuming that $[a,b]$ is the domain of the function $x$ that parametrizes the curve $C$. Yes, what you did is right if $[0,1]$ is the domain of $x$
    $endgroup$
    – Masacroso
    Dec 5 '18 at 4:04













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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Note: if you dont know the notation $fcirc g$ just mean the composition of functions $f$ and $g$, that is, $(fcirc g)(s)=f(g(s))$.



There is something that could confuse you: the symbol $x$ is used for two different purposes in the exercise




  1. to denote a vector on the domain of $F$, and


  2. to denote a function from $Bbb RtoBbb R^2$.



Just substituting we have that



$$(Fcirc x)(t)=F(1-2t^2,1)$$



where we assumed here that the domain of $F$ is some subset of $Bbb R^2$ (probably $Bbb R^2$ itself) because the image of the function $x$ is in $Bbb R^2$.



Now, for the first case we have that $(Fcirc x)(t)=(1,0)$, that is, every $tinBbb R$ defines some $x(t)inBbb R^2$, thus $F(x(t))=(1,0)$ because $F$ is already constant.



For the second case we have that $(Fcirc x)(t)=(t,1-2t^2)$, assuming that the notation $x_k$ refer to the coordinates of the vector $x$.



Can you evaluate $int_C F(x)cdot dx$ in each case now?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok I will try to do it now. I can evaluate it in the first case. But I am still confused about the second one.
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:29










  • $begingroup$
    Ok I still do not understand where am I supposed to use the (1, 0). Before the integration? Will it be used to find t?
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:37










  • $begingroup$
    $int _0^1-4tdt=-2$ This is what I understood from your explanation, I multiply F(x) with the derivative of x(t). Is that correct? @Masacroso
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:43










  • $begingroup$
    Ok nvm I could do it :) Thanks for the explanation. So for the second case it would be $int _0^1begin{pmatrix}t\ 1-2t^2end{pmatrix}cdot begin{pmatrix}-4t\ 1end{pmatrix}$ which would then be $int _0^1-4t^2+1-2t^2dt=-1$ No?
    $endgroup$
    – Naochi
    Dec 5 '18 at 4:01










  • $begingroup$
    @Naochi we have that $$int_C F(x)cdot dx=int_a^b (Fcirc x)(t)cdot x'(t), dt$$ assuming that $[a,b]$ is the domain of the function $x$ that parametrizes the curve $C$. Yes, what you did is right if $[0,1]$ is the domain of $x$
    $endgroup$
    – Masacroso
    Dec 5 '18 at 4:04


















2












$begingroup$

Note: if you dont know the notation $fcirc g$ just mean the composition of functions $f$ and $g$, that is, $(fcirc g)(s)=f(g(s))$.



There is something that could confuse you: the symbol $x$ is used for two different purposes in the exercise




  1. to denote a vector on the domain of $F$, and


  2. to denote a function from $Bbb RtoBbb R^2$.



Just substituting we have that



$$(Fcirc x)(t)=F(1-2t^2,1)$$



where we assumed here that the domain of $F$ is some subset of $Bbb R^2$ (probably $Bbb R^2$ itself) because the image of the function $x$ is in $Bbb R^2$.



Now, for the first case we have that $(Fcirc x)(t)=(1,0)$, that is, every $tinBbb R$ defines some $x(t)inBbb R^2$, thus $F(x(t))=(1,0)$ because $F$ is already constant.



For the second case we have that $(Fcirc x)(t)=(t,1-2t^2)$, assuming that the notation $x_k$ refer to the coordinates of the vector $x$.



Can you evaluate $int_C F(x)cdot dx$ in each case now?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok I will try to do it now. I can evaluate it in the first case. But I am still confused about the second one.
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:29










  • $begingroup$
    Ok I still do not understand where am I supposed to use the (1, 0). Before the integration? Will it be used to find t?
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:37










  • $begingroup$
    $int _0^1-4tdt=-2$ This is what I understood from your explanation, I multiply F(x) with the derivative of x(t). Is that correct? @Masacroso
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:43










  • $begingroup$
    Ok nvm I could do it :) Thanks for the explanation. So for the second case it would be $int _0^1begin{pmatrix}t\ 1-2t^2end{pmatrix}cdot begin{pmatrix}-4t\ 1end{pmatrix}$ which would then be $int _0^1-4t^2+1-2t^2dt=-1$ No?
    $endgroup$
    – Naochi
    Dec 5 '18 at 4:01










  • $begingroup$
    @Naochi we have that $$int_C F(x)cdot dx=int_a^b (Fcirc x)(t)cdot x'(t), dt$$ assuming that $[a,b]$ is the domain of the function $x$ that parametrizes the curve $C$. Yes, what you did is right if $[0,1]$ is the domain of $x$
    $endgroup$
    – Masacroso
    Dec 5 '18 at 4:04
















2












2








2





$begingroup$

Note: if you dont know the notation $fcirc g$ just mean the composition of functions $f$ and $g$, that is, $(fcirc g)(s)=f(g(s))$.



There is something that could confuse you: the symbol $x$ is used for two different purposes in the exercise




  1. to denote a vector on the domain of $F$, and


  2. to denote a function from $Bbb RtoBbb R^2$.



Just substituting we have that



$$(Fcirc x)(t)=F(1-2t^2,1)$$



where we assumed here that the domain of $F$ is some subset of $Bbb R^2$ (probably $Bbb R^2$ itself) because the image of the function $x$ is in $Bbb R^2$.



Now, for the first case we have that $(Fcirc x)(t)=(1,0)$, that is, every $tinBbb R$ defines some $x(t)inBbb R^2$, thus $F(x(t))=(1,0)$ because $F$ is already constant.



For the second case we have that $(Fcirc x)(t)=(t,1-2t^2)$, assuming that the notation $x_k$ refer to the coordinates of the vector $x$.



Can you evaluate $int_C F(x)cdot dx$ in each case now?






share|cite|improve this answer











$endgroup$



Note: if you dont know the notation $fcirc g$ just mean the composition of functions $f$ and $g$, that is, $(fcirc g)(s)=f(g(s))$.



There is something that could confuse you: the symbol $x$ is used for two different purposes in the exercise




  1. to denote a vector on the domain of $F$, and


  2. to denote a function from $Bbb RtoBbb R^2$.



Just substituting we have that



$$(Fcirc x)(t)=F(1-2t^2,1)$$



where we assumed here that the domain of $F$ is some subset of $Bbb R^2$ (probably $Bbb R^2$ itself) because the image of the function $x$ is in $Bbb R^2$.



Now, for the first case we have that $(Fcirc x)(t)=(1,0)$, that is, every $tinBbb R$ defines some $x(t)inBbb R^2$, thus $F(x(t))=(1,0)$ because $F$ is already constant.



For the second case we have that $(Fcirc x)(t)=(t,1-2t^2)$, assuming that the notation $x_k$ refer to the coordinates of the vector $x$.



Can you evaluate $int_C F(x)cdot dx$ in each case now?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 3:29

























answered Dec 5 '18 at 3:26









MasacrosoMasacroso

13k41746




13k41746












  • $begingroup$
    Ok I will try to do it now. I can evaluate it in the first case. But I am still confused about the second one.
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:29










  • $begingroup$
    Ok I still do not understand where am I supposed to use the (1, 0). Before the integration? Will it be used to find t?
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:37










  • $begingroup$
    $int _0^1-4tdt=-2$ This is what I understood from your explanation, I multiply F(x) with the derivative of x(t). Is that correct? @Masacroso
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:43










  • $begingroup$
    Ok nvm I could do it :) Thanks for the explanation. So for the second case it would be $int _0^1begin{pmatrix}t\ 1-2t^2end{pmatrix}cdot begin{pmatrix}-4t\ 1end{pmatrix}$ which would then be $int _0^1-4t^2+1-2t^2dt=-1$ No?
    $endgroup$
    – Naochi
    Dec 5 '18 at 4:01










  • $begingroup$
    @Naochi we have that $$int_C F(x)cdot dx=int_a^b (Fcirc x)(t)cdot x'(t), dt$$ assuming that $[a,b]$ is the domain of the function $x$ that parametrizes the curve $C$. Yes, what you did is right if $[0,1]$ is the domain of $x$
    $endgroup$
    – Masacroso
    Dec 5 '18 at 4:04




















  • $begingroup$
    Ok I will try to do it now. I can evaluate it in the first case. But I am still confused about the second one.
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:29










  • $begingroup$
    Ok I still do not understand where am I supposed to use the (1, 0). Before the integration? Will it be used to find t?
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:37










  • $begingroup$
    $int _0^1-4tdt=-2$ This is what I understood from your explanation, I multiply F(x) with the derivative of x(t). Is that correct? @Masacroso
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:43










  • $begingroup$
    Ok nvm I could do it :) Thanks for the explanation. So for the second case it would be $int _0^1begin{pmatrix}t\ 1-2t^2end{pmatrix}cdot begin{pmatrix}-4t\ 1end{pmatrix}$ which would then be $int _0^1-4t^2+1-2t^2dt=-1$ No?
    $endgroup$
    – Naochi
    Dec 5 '18 at 4:01










  • $begingroup$
    @Naochi we have that $$int_C F(x)cdot dx=int_a^b (Fcirc x)(t)cdot x'(t), dt$$ assuming that $[a,b]$ is the domain of the function $x$ that parametrizes the curve $C$. Yes, what you did is right if $[0,1]$ is the domain of $x$
    $endgroup$
    – Masacroso
    Dec 5 '18 at 4:04


















$begingroup$
Ok I will try to do it now. I can evaluate it in the first case. But I am still confused about the second one.
$endgroup$
– Naochi
Dec 5 '18 at 3:29




$begingroup$
Ok I will try to do it now. I can evaluate it in the first case. But I am still confused about the second one.
$endgroup$
– Naochi
Dec 5 '18 at 3:29












$begingroup$
Ok I still do not understand where am I supposed to use the (1, 0). Before the integration? Will it be used to find t?
$endgroup$
– Naochi
Dec 5 '18 at 3:37




$begingroup$
Ok I still do not understand where am I supposed to use the (1, 0). Before the integration? Will it be used to find t?
$endgroup$
– Naochi
Dec 5 '18 at 3:37












$begingroup$
$int _0^1-4tdt=-2$ This is what I understood from your explanation, I multiply F(x) with the derivative of x(t). Is that correct? @Masacroso
$endgroup$
– Naochi
Dec 5 '18 at 3:43




$begingroup$
$int _0^1-4tdt=-2$ This is what I understood from your explanation, I multiply F(x) with the derivative of x(t). Is that correct? @Masacroso
$endgroup$
– Naochi
Dec 5 '18 at 3:43












$begingroup$
Ok nvm I could do it :) Thanks for the explanation. So for the second case it would be $int _0^1begin{pmatrix}t\ 1-2t^2end{pmatrix}cdot begin{pmatrix}-4t\ 1end{pmatrix}$ which would then be $int _0^1-4t^2+1-2t^2dt=-1$ No?
$endgroup$
– Naochi
Dec 5 '18 at 4:01




$begingroup$
Ok nvm I could do it :) Thanks for the explanation. So for the second case it would be $int _0^1begin{pmatrix}t\ 1-2t^2end{pmatrix}cdot begin{pmatrix}-4t\ 1end{pmatrix}$ which would then be $int _0^1-4t^2+1-2t^2dt=-1$ No?
$endgroup$
– Naochi
Dec 5 '18 at 4:01












$begingroup$
@Naochi we have that $$int_C F(x)cdot dx=int_a^b (Fcirc x)(t)cdot x'(t), dt$$ assuming that $[a,b]$ is the domain of the function $x$ that parametrizes the curve $C$. Yes, what you did is right if $[0,1]$ is the domain of $x$
$endgroup$
– Masacroso
Dec 5 '18 at 4:04






$begingroup$
@Naochi we have that $$int_C F(x)cdot dx=int_a^b (Fcirc x)(t)cdot x'(t), dt$$ assuming that $[a,b]$ is the domain of the function $x$ that parametrizes the curve $C$. Yes, what you did is right if $[0,1]$ is the domain of $x$
$endgroup$
– Masacroso
Dec 5 '18 at 4:04




















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