Formula of curvature not defined in arc length
As a title I have some trouble to understand the formula of curvature that professor gave to us during the lesson. I have already tried to read the book he suggested and searched on the internet but I did not find what I was looking for...
Firstable he gave us the definition of curvature using the arc lenght.
Given the curve $gamma(t): [a,b] to mathbb{R}$, the curvature vector $vec k(s)$ is defined as $vec k(s)=frac{d}{ds}tau(s)=frac{d^2}{ds^2}gamma(s)$
Later, since usually arc lenght parametrization is less commonly used, he wrote down this:
Thank to the chain rule and the definition of function $s(t):[a,b] to [0,L(gamma)]$ in the arc lenght parametrization we obtain:
$$partial_s = frac{1}{|gamma'(t)|}partial_t$$
Now we using the definition of curvature vector $vec k(s)$:
$$partial_s^2 = vec k = frac{1}{|gamma'(t)|}partial_t gamma_s = frac{1}{|gamma'(t)|} left( frac{gamma'(t)}{|gamma'(t)|} right)' = frac{gamma''(t)- langle gamma''(t),tau(t) rangle cdot tau(t)}{|gamma'(t)|^2} $$
I do not understand the last equality, esentially for two reasons.
- The math process that the professor has followed to write down this result (I do not know how to do a derivate of a quotient in more than one variable)
- I know that the derivate of a vector is another vector orthogonal to the first. For this reason I think that the scalar product should be zero.
In the end he said that
$$P_{gamma(x)}^{perp}left( frac{gamma''(t)}{|gamma'(t)|^2} right) = frac{gamma''(t)- langle gamma''(t),tau(t) rangle cdot tau(t)}{|gamma'(t)|^2}$$
By definition:
$$P_{gamma(x)}^{perp}(eta) := eta - langle eta, tau rangle space tau $$
where $eta$ is a casual vector in $mathbb{R^n}$ that has its application point on the curve; $tau$ is the tangent vector at the point curve $x$.
As the professor said $P_{gamma(x)}^{perp}(eta)$ is the projection of vector $eta$ to the normal vector of the curve point $x$, but I can not see why.
However I'm not pretty sure about the last definition and I can't find it nowhere (because I do not even know its name). Can somebody teach me what $P_{gamma(x)}^{perp}(eta)$ is, please?
multivariable-calculus curvature
edited Nov 29 '18 at 22:41
Bernard
118k639112
asked Nov 29 '18 at 22:39
user515933user515933
897
add a comment |
As a title I have some trouble to understand the formula of curvature that professor gave to us during the lesson. I have already tried to read the book he suggested and searched on the internet but I did not find what I was looking for...
Firstable he gave us the definition of curvature using the arc lenght.
Given the curve $gamma(t): [a,b] to mathbb{R}$, the curvature vector $vec k(s)$ is defined as $vec k(s)=frac{d}{ds}tau(s)=frac{d^2}{ds^2}gamma(s)$
Later, since usually arc lenght parametrization is less commonly used, he wrote down this:
Thank to the chain rule and the definition of function $s(t):[a,b] to [0,L(gamma)]$ in the arc lenght parametrization we obtain:
$$partial_s = frac{1}{|gamma'(t)|}partial_t$$
Now we using the definition of curvature vector $vec k(s)$:
$$partial_s^2 = vec k = frac{1}{|gamma'(t)|}partial_t gamma_s = frac{1}{|gamma'(t)|} left( frac{gamma'(t)}{|gamma'(t)|} right)' = frac{gamma''(t)- langle gamma''(t),tau(t) rangle cdot tau(t)}{|gamma'(t)|^2} $$
I do not understand the last equality, esentially for two reasons.
- The math process that the professor has followed to write down this result (I do not know how to do a derivate of a quotient in more than one variable)
- I know that the derivate of a vector is another vector orthogonal to the first. For this reason I think that the scalar product should be zero.
In the end he said that
$$P_{gamma(x)}^{perp}left( frac{gamma''(t)}{|gamma'(t)|^2} right) = frac{gamma''(t)- langle gamma''(t),tau(t) rangle cdot tau(t)}{|gamma'(t)|^2}$$
By definition:
$$P_{gamma(x)}^{perp}(eta) := eta - langle eta, tau rangle space tau $$
where $eta$ is a casual vector in $mathbb{R^n}$ that has its application point on the curve; $tau$ is the tangent vector at the point curve $x$.
As the professor said $P_{gamma(x)}^{perp}(eta)$ is the projection of vector $eta$ to the normal vector of the curve point $x$, but I can not see why.
However I'm not pretty sure about the last definition and I can't find it nowhere (because I do not even know its name). Can somebody teach me what $P_{gamma(x)}^{perp}(eta)$ is, please?
multivariable-calculus curvature
edited Nov 29 '18 at 22:41
Bernard
118k639112
asked Nov 29 '18 at 22:39
user515933user515933
897
add a comment |
As a title I have some trouble to understand the formula of curvature that professor gave to us during the lesson. I have already tried to read the book he suggested and searched on the internet but I did not find what I was looking for...
Firstable he gave us the definition of curvature using the arc lenght.
Given the curve $gamma(t): [a,b] to mathbb{R}$, the curvature vector $vec k(s)$ is defined as $vec k(s)=frac{d}{ds}tau(s)=frac{d^2}{ds^2}gamma(s)$
Later, since usually arc lenght parametrization is less commonly used, he wrote down this:
Thank to the chain rule and the definition of function $s(t):[a,b] to [0,L(gamma)]$ in the arc lenght parametrization we obtain:
$$partial_s = frac{1}{|gamma'(t)|}partial_t$$
Now we using the definition of curvature vector $vec k(s)$:
$$partial_s^2 = vec k = frac{1}{|gamma'(t)|}partial_t gamma_s = frac{1}{|gamma'(t)|} left( frac{gamma'(t)}{|gamma'(t)|} right)' = frac{gamma''(t)- langle gamma''(t),tau(t) rangle cdot tau(t)}{|gamma'(t)|^2} $$
I do not understand the last equality, esentially for two reasons.
- The math process that the professor has followed to write down this result (I do not know how to do a derivate of a quotient in more than one variable)
- I know that the derivate of a vector is another vector orthogonal to the first. For this reason I think that the scalar product should be zero.
In the end he said that
$$P_{gamma(x)}^{perp}left( frac{gamma''(t)}{|gamma'(t)|^2} right) = frac{gamma''(t)- langle gamma''(t),tau(t) rangle cdot tau(t)}{|gamma'(t)|^2}$$
By definition:
$$P_{gamma(x)}^{perp}(eta) := eta - langle eta, tau rangle space tau $$
where $eta$ is a casual vector in $mathbb{R^n}$ that has its application point on the curve; $tau$ is the tangent vector at the point curve $x$.
As the professor said $P_{gamma(x)}^{perp}(eta)$ is the projection of vector $eta$ to the normal vector of the curve point $x$, but I can not see why.
However I'm not pretty sure about the last definition and I can't find it nowhere (because I do not even know its name). Can somebody teach me what $P_{gamma(x)}^{perp}(eta)$ is, please?
multivariable-calculus curvature
edited Nov 29 '18 at 22:41
Bernard
118k639112
asked Nov 29 '18 at 22:39
user515933user515933
897
As a title I have some trouble to understand the formula of curvature that professor gave to us during the lesson. I have already tried to read the book he suggested and searched on the internet but I did not find what I was looking for...
Firstable he gave us the definition of curvature using the arc lenght.
Given the curve $gamma(t): [a,b] to mathbb{R}$, the curvature vector $vec k(s)$ is defined as $vec k(s)=frac{d}{ds}tau(s)=frac{d^2}{ds^2}gamma(s)$
Later, since usually arc lenght parametrization is less commonly used, he wrote down this:
Thank to the chain rule and the definition of function $s(t):[a,b] to [0,L(gamma)]$ in the arc lenght parametrization we obtain:
$$partial_s = frac{1}{|gamma'(t)|}partial_t$$
Now we using the definition of curvature vector $vec k(s)$:
$$partial_s^2 = vec k = frac{1}{|gamma'(t)|}partial_t gamma_s = frac{1}{|gamma'(t)|} left( frac{gamma'(t)}{|gamma'(t)|} right)' = frac{gamma''(t)- langle gamma''(t),tau(t) rangle cdot tau(t)}{|gamma'(t)|^2} $$
I do not understand the last equality, esentially for two reasons.
- The math process that the professor has followed to write down this result (I do not know how to do a derivate of a quotient in more than one variable)
- I know that the derivate of a vector is another vector orthogonal to the first. For this reason I think that the scalar product should be zero.
In the end he said that
$$P_{gamma(x)}^{perp}left( frac{gamma''(t)}{|gamma'(t)|^2} right) = frac{gamma''(t)- langle gamma''(t),tau(t) rangle cdot tau(t)}{|gamma'(t)|^2}$$
By definition:
$$P_{gamma(x)}^{perp}(eta) := eta - langle eta, tau rangle space tau $$
where $eta$ is a casual vector in $mathbb{R^n}$ that has its application point on the curve; $tau$ is the tangent vector at the point curve $x$.
As the professor said $P_{gamma(x)}^{perp}(eta)$ is the projection of vector $eta$ to the normal vector of the curve point $x$, but I can not see why.
However I'm not pretty sure about the last definition and I can't find it nowhere (because I do not even know its name). Can somebody teach me what $P_{gamma(x)}^{perp}(eta)$ is, please?
multivariable-calculus curvature
multivariable-calculus curvature
edited Nov 29 '18 at 22:41
Bernard
118k639112
asked Nov 29 '18 at 22:39
user515933user515933
897
edited Nov 29 '18 at 22:41
Bernard
118k639112
asked Nov 29 '18 at 22:39
user515933user515933
897
edited Nov 29 '18 at 22:41
Bernard
118k639112
edited Nov 29 '18 at 22:41
Bernard
118k639112
edited Nov 29 '18 at 22:41
Bernard
118k639112
118k639112
asked Nov 29 '18 at 22:39
user515933user515933
897
asked Nov 29 '18 at 22:39
user515933user515933
897
asked Nov 29 '18 at 22:39
user515933user515933
897
897
add a comment |
add a comment |
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