How to integrate the curve












2














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




















  • 1




    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 $(1,0)inBbb R^2$)
    – Masacroso
    Dec 5 '18 at 2:58








  • 2




    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
    – Masacroso
    Dec 5 '18 at 3:04
















2














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




















  • 1




    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 $(1,0)inBbb R^2$)
    – Masacroso
    Dec 5 '18 at 2:58








  • 2




    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
    – Masacroso
    Dec 5 '18 at 3:04














2












2








2







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















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




    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 $(1,0)inBbb R^2$)
    – Masacroso
    Dec 5 '18 at 2:58








  • 2




    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
    – Masacroso
    Dec 5 '18 at 3:04














  • 1




    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 $(1,0)inBbb R^2$)
    – Masacroso
    Dec 5 '18 at 2:58








  • 2




    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
    – Masacroso
    Dec 5 '18 at 3:04








1




1




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 $(1,0)inBbb R^2$)
– Masacroso
Dec 5 '18 at 2:58






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 $(1,0)inBbb R^2$)
– Masacroso
Dec 5 '18 at 2:58






2




2




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
– Masacroso
Dec 5 '18 at 3:04




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
– Masacroso
Dec 5 '18 at 3:04










1 Answer
1






active

oldest

votes


















2














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























  • 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.
    – Naochi
    Dec 5 '18 at 3:29










  • 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?
    – Naochi
    Dec 5 '18 at 3:37










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










  • 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?
    – Naochi
    Dec 5 '18 at 4:01










  • @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 $x$
    – Masacroso
    Dec 5 '18 at 4:04













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1 Answer
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2














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























  • 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.
    – Naochi
    Dec 5 '18 at 3:29










  • 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?
    – Naochi
    Dec 5 '18 at 3:37










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










  • 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?
    – Naochi
    Dec 5 '18 at 4:01










  • @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 $x$
    – Masacroso
    Dec 5 '18 at 4:04


















2














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























  • 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.
    – Naochi
    Dec 5 '18 at 3:29










  • 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?
    – Naochi
    Dec 5 '18 at 3:37










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










  • 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?
    – Naochi
    Dec 5 '18 at 4:01










  • @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 $x$
    – Masacroso
    Dec 5 '18 at 4:04
















2












2








2






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














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












  • 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.
    – Naochi
    Dec 5 '18 at 3:29










  • 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?
    – Naochi
    Dec 5 '18 at 3:37










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










  • 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?
    – Naochi
    Dec 5 '18 at 4:01










  • @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 $x$
    – Masacroso
    Dec 5 '18 at 4:04




















  • 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.
    – Naochi
    Dec 5 '18 at 3:29










  • 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?
    – Naochi
    Dec 5 '18 at 3:37










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










  • 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?
    – Naochi
    Dec 5 '18 at 4:01










  • @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 $x$
    – Masacroso
    Dec 5 '18 at 4:04


















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.
– Naochi
Dec 5 '18 at 3:29




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.
– Naochi
Dec 5 '18 at 3:29












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?
– Naochi
Dec 5 '18 at 3:37




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?
– Naochi
Dec 5 '18 at 3:37












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




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












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?
– Naochi
Dec 5 '18 at 4:01




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?
– Naochi
Dec 5 '18 at 4:01












@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 $x$
– Masacroso
Dec 5 '18 at 4:04






@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 $x$
– Masacroso
Dec 5 '18 at 4:04




















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Berounka

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Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
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Sphinx de Gizeh

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Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
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Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

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3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
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