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Determine the tangent vector at any point $p=(x^1, x^2)$ on the circle $(x^1)^2 +(x^2)^2 = a^2$
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Determine the tangent vector at any point $p=(x^1, x^2)$ on the circle $(x^1)^2 +(x^2)^2 = a^2$
The following problem above and its solution below was given in my class.
Solution: Parametric equations for the circle are $x^1 = acos t$ and $x^2 = a sin t$. The tangent vector $v$ is given by $$v_p = v^1e_1 +v^2e_2$$ where $e_1 = fracpartialpartial x^1$ and $e_1 = fracpartialpartial x^2$. Note that $$v^1 = fracdx^1dt = -x^2$$ and $$v^2 = fracdx^2dt = x^1$$ and so we arrive at $$v_p = -x^2 fracpartialpartial x^1 +x^1 fracpartialpartial x^2$$
Now I want to make the above solution as rigorous as possible.
Now the way I understand it (using the book Introduction to Smooth Manifolds by John Lee as a reference), by tangent vector my lecturer means geometric tangent vector which is the element $v_p :=(p, v) in mathbbR^n_p = p times mathbbR^n$.
Now $$T_p(mathbbR^n) = left w text is linear and w(f cdot g)= f(a)cdot w(g) + g(a)cdot w(f)right$$ so in words $T_p(mathbbR^n)$ is the set of all smooth linear maps from $mathbbR^n$ to $mathbbR$ satsifying the product rule above.
Some quick facts. For any $v_p in mathbbR^n_p$, the map $D_v|_p : C^infty(mathbbR^n) to mathbbR$ defined by $$D_v|_p (f) = sum_i=1^n v^i fracpartial fpartial x^i (p)$$ is a derivation. So basically for any point in $mathbbR^n_p$, the map which takes the directional derivative of a smooth function on $mathbbR^n$ in the direction $v$ at $p$ is a derivation.
Furthermore $T_p(mathbbR^n)$ as basis $$mathcalB = left_pright := left_pright$$ where $e_i = (0, dots, underbrace1_i^textth text position , dots, 0)$
Moreover the map $psi : mathbbR^n_p to T_p(mathbbR^n)$ defined by $$psi(v_p) = D_v|_p$$ is a vector space isomorphism. So $mathbbR^n_p$ is isomorphic to $T_p(mathbbR^n)$ as vector spaces.
Now with all of the above established, I'm trying to rephrase the above solution more formally.
I understand (please correct me if I'm wring) that the 'tangent vector' $v_p =(p, v)$ where $v = (v^1, v^2)$ is being identified with its image $psi(v_p) = v^1fracpartialpartial x^1 + v^2 fracpartialpartial x^2$
Now what I don't understand is how we determine $v^1$ and $v^2$ and why we need parametric equations for the circle. I don't see why $v^1$ and $v^2$ are calculated in the way they are.
calculus multivariable-calculus differential-geometry
asked Jul 25 at 10:03
Perturbative
3,48911039
 |Â
show 4 more comments
up vote
1
down vote
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Determine the tangent vector at any point $p=(x^1, x^2)$ on the circle $(x^1)^2 +(x^2)^2 = a^2$
The following problem above and its solution below was given in my class.
Solution: Parametric equations for the circle are $x^1 = acos t$ and $x^2 = a sin t$. The tangent vector $v$ is given by $$v_p = v^1e_1 +v^2e_2$$ where $e_1 = fracpartialpartial x^1$ and $e_1 = fracpartialpartial x^2$. Note that $$v^1 = fracdx^1dt = -x^2$$ and $$v^2 = fracdx^2dt = x^1$$ and so we arrive at $$v_p = -x^2 fracpartialpartial x^1 +x^1 fracpartialpartial x^2$$
Now I want to make the above solution as rigorous as possible.
Now the way I understand it (using the book Introduction to Smooth Manifolds by John Lee as a reference), by tangent vector my lecturer means geometric tangent vector which is the element $v_p :=(p, v) in mathbbR^n_p = p times mathbbR^n$.
Now $$T_p(mathbbR^n) = left w text is linear and w(f cdot g)= f(a)cdot w(g) + g(a)cdot w(f)right$$ so in words $T_p(mathbbR^n)$ is the set of all smooth linear maps from $mathbbR^n$ to $mathbbR$ satsifying the product rule above.
Some quick facts. For any $v_p in mathbbR^n_p$, the map $D_v|_p : C^infty(mathbbR^n) to mathbbR$ defined by $$D_v|_p (f) = sum_i=1^n v^i fracpartial fpartial x^i (p)$$ is a derivation. So basically for any point in $mathbbR^n_p$, the map which takes the directional derivative of a smooth function on $mathbbR^n$ in the direction $v$ at $p$ is a derivation.
Furthermore $T_p(mathbbR^n)$ as basis $$mathcalB = left_pright := left_pright$$ where $e_i = (0, dots, underbrace1_i^textth text position , dots, 0)$
Moreover the map $psi : mathbbR^n_p to T_p(mathbbR^n)$ defined by $$psi(v_p) = D_v|_p$$ is a vector space isomorphism. So $mathbbR^n_p$ is isomorphic to $T_p(mathbbR^n)$ as vector spaces.
Now with all of the above established, I'm trying to rephrase the above solution more formally.
I understand (please correct me if I'm wring) that the 'tangent vector' $v_p =(p, v)$ where $v = (v^1, v^2)$ is being identified with its image $psi(v_p) = v^1fracpartialpartial x^1 + v^2 fracpartialpartial x^2$
Now what I don't understand is how we determine $v^1$ and $v^2$ and why we need parametric equations for the circle. I don't see why $v^1$ and $v^2$ are calculated in the way they are.
calculus multivariable-calculus differential-geometry
asked Jul 25 at 10:03
Perturbative
3,48911039
I think the question is rather vague. What do you mean by "$textitthe tangent vector$" ? There are infinitely many vectors tangent to $(a,b)$. From what i understand, the problem probably ask about $textitthe tangent vector of a curve on the circle at a point (a,b)$. Have you guys learned this in class ?
– Sou
Jul 25 at 10:30
@Sou You are correct, that is what the problem is asking. We have learned this in class but very informally, the class I'm attending is basically an applied maths course (so few definitions no proofs etc) and I'm trying to learn all of the material from a pure maths perspective
– Perturbative
Jul 25 at 10:35
@Perturbative : could you use notations such as there would be no confusion between powers, indice for different functions and degrees of differentiation.
– JJacquelin
Jul 25 at 10:35
@JJacquelin I thought I was using the standard notation used in Differential Geometry. I apologize if there was such a confusion.
– Perturbative
Jul 25 at 10:38
Look section 8.6 in Tu's Introduction to Manifolds and p.68 of Lee's smooth manifold for the explanation (both in their 2nd ed). There is an example in Tu's which pretty much similar to your problem, and the rigorous treatment in Lee's. However, you should know what "push-forward" is.
– Sou
Jul 25 at 10:58
 |Â
show 4 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Determine the tangent vector at any point $p=(x^1, x^2)$ on the circle $(x^1)^2 +(x^2)^2 = a^2$
The following problem above and its solution below was given in my class.
Solution: Parametric equations for the circle are $x^1 = acos t$ and $x^2 = a sin t$. The tangent vector $v$ is given by $$v_p = v^1e_1 +v^2e_2$$ where $e_1 = fracpartialpartial x^1$ and $e_1 = fracpartialpartial x^2$. Note that $$v^1 = fracdx^1dt = -x^2$$ and $$v^2 = fracdx^2dt = x^1$$ and so we arrive at $$v_p = -x^2 fracpartialpartial x^1 +x^1 fracpartialpartial x^2$$
Now I want to make the above solution as rigorous as possible.
Now the way I understand it (using the book Introduction to Smooth Manifolds by John Lee as a reference), by tangent vector my lecturer means geometric tangent vector which is the element $v_p :=(p, v) in mathbbR^n_p = p times mathbbR^n$.
Now $$T_p(mathbbR^n) = left w text is linear and w(f cdot g)= f(a)cdot w(g) + g(a)cdot w(f)right$$ so in words $T_p(mathbbR^n)$ is the set of all smooth linear maps from $mathbbR^n$ to $mathbbR$ satsifying the product rule above.
Some quick facts. For any $v_p in mathbbR^n_p$, the map $D_v|_p : C^infty(mathbbR^n) to mathbbR$ defined by $$D_v|_p (f) = sum_i=1^n v^i fracpartial fpartial x^i (p)$$ is a derivation. So basically for any point in $mathbbR^n_p$, the map which takes the directional derivative of a smooth function on $mathbbR^n$ in the direction $v$ at $p$ is a derivation.
Furthermore $T_p(mathbbR^n)$ as basis $$mathcalB = left_pright := left_pright$$ where $e_i = (0, dots, underbrace1_i^textth text position , dots, 0)$
Moreover the map $psi : mathbbR^n_p to T_p(mathbbR^n)$ defined by $$psi(v_p) = D_v|_p$$ is a vector space isomorphism. So $mathbbR^n_p$ is isomorphic to $T_p(mathbbR^n)$ as vector spaces.
Now with all of the above established, I'm trying to rephrase the above solution more formally.
I understand (please correct me if I'm wring) that the 'tangent vector' $v_p =(p, v)$ where $v = (v^1, v^2)$ is being identified with its image $psi(v_p) = v^1fracpartialpartial x^1 + v^2 fracpartialpartial x^2$
Now what I don't understand is how we determine $v^1$ and $v^2$ and why we need parametric equations for the circle. I don't see why $v^1$ and $v^2$ are calculated in the way they are.
calculus multivariable-calculus differential-geometry
asked Jul 25 at 10:03
Perturbative
3,48911039
Determine the tangent vector at any point $p=(x^1, x^2)$ on the circle $(x^1)^2 +(x^2)^2 = a^2$
The following problem above and its solution below was given in my class.
Solution: Parametric equations for the circle are $x^1 = acos t$ and $x^2 = a sin t$. The tangent vector $v$ is given by $$v_p = v^1e_1 +v^2e_2$$ where $e_1 = fracpartialpartial x^1$ and $e_1 = fracpartialpartial x^2$. Note that $$v^1 = fracdx^1dt = -x^2$$ and $$v^2 = fracdx^2dt = x^1$$ and so we arrive at $$v_p = -x^2 fracpartialpartial x^1 +x^1 fracpartialpartial x^2$$
Now I want to make the above solution as rigorous as possible.
Now the way I understand it (using the book Introduction to Smooth Manifolds by John Lee as a reference), by tangent vector my lecturer means geometric tangent vector which is the element $v_p :=(p, v) in mathbbR^n_p = p times mathbbR^n$.
Now $$T_p(mathbbR^n) = left w text is linear and w(f cdot g)= f(a)cdot w(g) + g(a)cdot w(f)right$$ so in words $T_p(mathbbR^n)$ is the set of all smooth linear maps from $mathbbR^n$ to $mathbbR$ satsifying the product rule above.
Some quick facts. For any $v_p in mathbbR^n_p$, the map $D_v|_p : C^infty(mathbbR^n) to mathbbR$ defined by $$D_v|_p (f) = sum_i=1^n v^i fracpartial fpartial x^i (p)$$ is a derivation. So basically for any point in $mathbbR^n_p$, the map which takes the directional derivative of a smooth function on $mathbbR^n$ in the direction $v$ at $p$ is a derivation.
Furthermore $T_p(mathbbR^n)$ as basis $$mathcalB = left_pright := left_pright$$ where $e_i = (0, dots, underbrace1_i^textth text position , dots, 0)$
Moreover the map $psi : mathbbR^n_p to T_p(mathbbR^n)$ defined by $$psi(v_p) = D_v|_p$$ is a vector space isomorphism. So $mathbbR^n_p$ is isomorphic to $T_p(mathbbR^n)$ as vector spaces.
Now with all of the above established, I'm trying to rephrase the above solution more formally.
I understand (please correct me if I'm wring) that the 'tangent vector' $v_p =(p, v)$ where $v = (v^1, v^2)$ is being identified with its image $psi(v_p) = v^1fracpartialpartial x^1 + v^2 fracpartialpartial x^2$
Now what I don't understand is how we determine $v^1$ and $v^2$ and why we need parametric equations for the circle. I don't see why $v^1$ and $v^2$ are calculated in the way they are.
calculus multivariable-calculus differential-geometry
asked Jul 25 at 10:03
Perturbative
3,48911039
asked Jul 25 at 10:03
Perturbative
3,48911039
asked Jul 25 at 10:03
Perturbative
3,48911039
asked Jul 25 at 10:03
Perturbative
3,48911039
3,48911039
I think the question is rather vague. What do you mean by "$textitthe tangent vector$" ? There are infinitely many vectors tangent to $(a,b)$. From what i understand, the problem probably ask about $textitthe tangent vector of a curve on the circle at a point (a,b)$. Have you guys learned this in class ?
– Sou
Jul 25 at 10:30
@Sou You are correct, that is what the problem is asking. We have learned this in class but very informally, the class I'm attending is basically an applied maths course (so few definitions no proofs etc) and I'm trying to learn all of the material from a pure maths perspective
– Perturbative
Jul 25 at 10:35
@Perturbative : could you use notations such as there would be no confusion between powers, indice for different functions and degrees of differentiation.
– JJacquelin
Jul 25 at 10:35
@JJacquelin I thought I was using the standard notation used in Differential Geometry. I apologize if there was such a confusion.
– Perturbative
Jul 25 at 10:38
Look section 8.6 in Tu's Introduction to Manifolds and p.68 of Lee's smooth manifold for the explanation (both in their 2nd ed). There is an example in Tu's which pretty much similar to your problem, and the rigorous treatment in Lee's. However, you should know what "push-forward" is.
– Sou
Jul 25 at 10:58
 |Â
show 4 more comments
I think the question is rather vague. What do you mean by "$textitthe tangent vector$" ? There are infinitely many vectors tangent to $(a,b)$. From what i understand, the problem probably ask about $textitthe tangent vector of a curve on the circle at a point (a,b)$. Have you guys learned this in class ?
– Sou
Jul 25 at 10:30
@Sou You are correct, that is what the problem is asking. We have learned this in class but very informally, the class I'm attending is basically an applied maths course (so few definitions no proofs etc) and I'm trying to learn all of the material from a pure maths perspective
– Perturbative
Jul 25 at 10:35
@Perturbative : could you use notations such as there would be no confusion between powers, indice for different functions and degrees of differentiation.
– JJacquelin
Jul 25 at 10:35
@JJacquelin I thought I was using the standard notation used in Differential Geometry. I apologize if there was such a confusion.
– Perturbative
Jul 25 at 10:38
Look section 8.6 in Tu's Introduction to Manifolds and p.68 of Lee's smooth manifold for the explanation (both in their 2nd ed). There is an example in Tu's which pretty much similar to your problem, and the rigorous treatment in Lee's. However, you should know what "push-forward" is.
– Sou
Jul 25 at 10:58
I think the question is rather vague. What do you mean by "$textitthe tangent vector$" ? There are infinitely many vectors tangent to $(a,b)$. From what i understand, the problem probably ask about $textitthe tangent vector of a curve on the circle at a point (a,b)$. Have you guys learned this in class ?
– Sou
Jul 25 at 10:30
I think the question is rather vague. What do you mean by "$textitthe tangent vector$" ? There are infinitely many vectors tangent to $(a,b)$. From what i understand, the problem probably ask about $textitthe tangent vector of a curve on the circle at a point (a,b)$. Have you guys learned this in class ?
– Sou
Jul 25 at 10:30
@Sou You are correct, that is what the problem is asking. We have learned this in class but very informally, the class I'm attending is basically an applied maths course (so few definitions no proofs etc) and I'm trying to learn all of the material from a pure maths perspective
– Perturbative
Jul 25 at 10:35
@Sou You are correct, that is what the problem is asking. We have learned this in class but very informally, the class I'm attending is basically an applied maths course (so few definitions no proofs etc) and I'm trying to learn all of the material from a pure maths perspective
– Perturbative
Jul 25 at 10:35
@Perturbative : could you use notations such as there would be no confusion between powers, indice for different functions and degrees of differentiation.
– JJacquelin
Jul 25 at 10:35
@Perturbative : could you use notations such as there would be no confusion between powers, indice for different functions and degrees of differentiation.
– JJacquelin
Jul 25 at 10:35
@JJacquelin I thought I was using the standard notation used in Differential Geometry. I apologize if there was such a confusion.
– Perturbative
Jul 25 at 10:38
@JJacquelin I thought I was using the standard notation used in Differential Geometry. I apologize if there was such a confusion.
– Perturbative
Jul 25 at 10:38
Look section 8.6 in Tu's Introduction to Manifolds and p.68 of Lee's smooth manifold for the explanation (both in their 2nd ed). There is an example in Tu's which pretty much similar to your problem, and the rigorous treatment in Lee's. However, you should know what "push-forward" is.
– Sou
Jul 25 at 10:58
Look section 8.6 in Tu's Introduction to Manifolds and p.68 of Lee's smooth manifold for the explanation (both in their 2nd ed). There is an example in Tu's which pretty much similar to your problem, and the rigorous treatment in Lee's. However, you should know what "push-forward" is.
– Sou
Jul 25 at 10:58
 |Â
show 4 more comments
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I think the question is rather vague. What do you mean by "$textitthe tangent vector$" ? There are infinitely many vectors tangent to $(a,b)$. From what i understand, the problem probably ask about $textitthe tangent vector of a curve on the circle at a point (a,b)$. Have you guys learned this in class ?
– Sou
Jul 25 at 10:30
@Sou You are correct, that is what the problem is asking. We have learned this in class but very informally, the class I'm attending is basically an applied maths course (so few definitions no proofs etc) and I'm trying to learn all of the material from a pure maths perspective
– Perturbative
Jul 25 at 10:35
@Perturbative : could you use notations such as there would be no confusion between powers, indice for different functions and degrees of differentiation.
– JJacquelin
Jul 25 at 10:35
@JJacquelin I thought I was using the standard notation used in Differential Geometry. I apologize if there was such a confusion.
– Perturbative
Jul 25 at 10:38
Look section 8.6 in Tu's Introduction to Manifolds and p.68 of Lee's smooth manifold for the explanation (both in their 2nd ed). There is an example in Tu's which pretty much similar to your problem, and the rigorous treatment in Lee's. However, you should know what "push-forward" is.
– Sou
Jul 25 at 10:58