Je bent je ingevulde velden bij deze pagina aan het verwijderen. Ben je zeker dat je dit wilt doen?
You are erasing your filled-in fields on this page. Are you sure that is what you want?
Nieuwe Versie BeschikbaarNew Version Available
Er is een update van deze pagina. Als je update naar de meest recente versie, verlies je mogelijk je huidige antwoorden voor deze pagina. Hoe wil je verdergaan ?
There is an updated version of this page. If you update to the most recent version, then your current progress on this page will be erased. Regardless, your record of completion will remain. How would you like to proceed?
There is a notion of derivatives of forms that generalizes the notions of classical
derivative, gradient, curl, and divergence. The derivative of a smooth \(k\)-form is a
smooth \((k+1)\)-form.
In \(\mathbb {R}\), if \(\omega = x^3 + \sin x\), then
\[ d \omega = ( 3x^2 + \cos x )\, dx \]
In \(\mathbb {R}^2\), if \(\omega = x^2 \cos y - y e^x\), then
\[ d \omega = ( 2x \cos y - y e^x )\, dx + (- x^2 \sin y - e^x) \, dy \]
In \(\mathbb {R}^3\), if \(\omega = z^3 e^x + 2 z \cos y \), then
\[ d \omega = e^x \, dx - 2 z \sin y \, dy + (3 z^2 e^x + 2 \cos y) \, dz \]
In \(\mathbb {R}^2\), if \(\omega = x y^2 \, dx + y \cos x \, dy \), then
\begin{align*} d \omega & = y^2 \, dx \wedge dx + 2 xy \, dy \wedge dx - y \sin x \, dx \wedge dy + \cos x \, dy \wedge dy \\ & = (- y \sin x - 2 xy ) \, dx \wedge dy \end{align*}
because the terms \(dx \wedge dx \) and \(dy \wedge dy\) are zero. Note that if we think of \(\omega \) as the vector field
\[ F = \langle x y^2 , y \cos x \rangle ,\]
then
\[ \text {curl} (F) = - y \sin x - 2 xy \]
In \(\mathbb {R}^3\), if \(\omega = yz \, dx + y e^z \, dy + x^2 \, dz \), then
\begin{align*} d \omega & = z \, dy \wedge dx + y \, dz \wedge dx + e^z \, dy \wedge dy + y e^z \, dz \wedge dy + 2x \, dx \wedge dz \\ & = - y e^z \, dy \wedge dz + ( y - 2x ) \, dz \wedge dx - z \, dx \wedge dy \end{align*}
Note that if we think of \(\omega \) as the vector field
\[ F = \langle yz , y e^z , x^2 \rangle ,\]
then
\[ \text {Curl} (F) = \langle - y e^z , y - 2x , -z \rangle \]
In \(\mathbb {R}^3\), if \(\omega = x^2 y \, dy \wedge dz + e^y \cos z \, dz \wedge dx + x ^2 \sin z \, dx \wedge dy \), then
\begin{align*} d \omega & = 2xy \, dx \wedge dy \wedge dz + e^y \cos z \, dy \wedge dz \wedge dx + x^2 \cos z \, dz \wedge dx \wedge dy \\& = ( 2xy + e^y \cos z + x^2 \cos z )\, dx \wedge dy \wedge dz \end{align*}
Note that if we think of \(\omega \) as the vector field
\[ F = \langle x^2 y , e^y \cos z , x ^2 \sin z \rangle ,\]
then
\[ \text {div} (F) = 2xy + e^y \cos z + x^2 \cos z \]
In \(\mathbb {R}\), if \(\omega = f(x)\), then
\[ d \omega = f' (x)\, dx \]
In \(\mathbb {R}^2\), if \(\omega = f(x,y)\), then
\[ d \omega = \frac {\partial f}{\partial x} \, dx + \frac {\partial f}{\partial y} \, dy \]
In \(\mathbb {R}^3\), if \(\omega = f(x,y,z) \), then
Fundamental Theorem of Calculus Let \(\omega \) be a smooth \(k\)-form in \(\mathbb {R}^n\), and \(M\subset \mathbb {R}^n\) an oriented
\((k+1)\)-dimensional manifold with boundary. Then
MTH 255H