- #1
Cyrus
- 3,238
- 17
I can't figure this out, help me NOW! Just kidding.
So anyways, here's the question:
Part of stokes theorem has the following in it:
[tex] \frac {\partial } {\partial x} ( Q + R \frac{\partial z}{\partial y} ) [/tex]
Which is written as:
[tex] \frac { \partial Q }{\partial x} + \frac { \partial Q }{\partial z} \frac { \partial z }{\partial x} + \frac { \partial R }{\partial x} \frac { \partial z }{\partial y}+ \frac { \partial R }{\partial z}\frac { \partial z }{\partial x}\frac{ \partial z }{\partial y}+ R \frac { \partial ^2z }{\partial x \partial y} [/tex]
I guess they got the above anwser after doing the following, but tell me if not.
[tex] \frac { \partial Q }{\partial x}\frac { \partial x }{\partial x} +\frac { \partial Q }{\partial y}\frac { \partial y }{\partial x}+ \frac { \partial Q }{\partial z} \frac { \partial z }{\partial x} + \frac { \partial R }{\partial x} \frac { \partial x}{\partial x}\frac { \partial z }{\partial y}+ \frac { \partial R }{\partial y} \frac { \partial y}{\partial x}\frac { \partial z }{\partial y}+ \frac { \partial R }{\partial z}\frac { \partial z }{\partial x}\frac{ \partial z }{\partial y}+ R \frac { \partial ^2z }{\partial x \partial y} [/tex]
I see that dx/dx is equal to 1, so it does not appear in the anwser. Does the second term drop, because [tex] \frac {\partial y} { \partial x } =o [/tex].
They show this by rewriting it as follows: [tex] \frac {\partial } { \partial x } y =o [/tex].
I dident realize you could separate the partial operator from the numerator term like that.
Could a possible reason be that y is an independent variable, so a change in x, another independent variable, has no effect on y, which is why it is equal to zero? In other words, the change in y has nothing to do with the change in x, which is why the fraction is zero.
And I guess for the R, term, the same follows, except for the fact that they use two chain rules, one for many variables, and one for the calc2 version.
uv= u'v+v'u
So anyways, here's the question:
Part of stokes theorem has the following in it:
[tex] \frac {\partial } {\partial x} ( Q + R \frac{\partial z}{\partial y} ) [/tex]
Which is written as:
[tex] \frac { \partial Q }{\partial x} + \frac { \partial Q }{\partial z} \frac { \partial z }{\partial x} + \frac { \partial R }{\partial x} \frac { \partial z }{\partial y}+ \frac { \partial R }{\partial z}\frac { \partial z }{\partial x}\frac{ \partial z }{\partial y}+ R \frac { \partial ^2z }{\partial x \partial y} [/tex]
I guess they got the above anwser after doing the following, but tell me if not.
[tex] \frac { \partial Q }{\partial x}\frac { \partial x }{\partial x} +\frac { \partial Q }{\partial y}\frac { \partial y }{\partial x}+ \frac { \partial Q }{\partial z} \frac { \partial z }{\partial x} + \frac { \partial R }{\partial x} \frac { \partial x}{\partial x}\frac { \partial z }{\partial y}+ \frac { \partial R }{\partial y} \frac { \partial y}{\partial x}\frac { \partial z }{\partial y}+ \frac { \partial R }{\partial z}\frac { \partial z }{\partial x}\frac{ \partial z }{\partial y}+ R \frac { \partial ^2z }{\partial x \partial y} [/tex]
I see that dx/dx is equal to 1, so it does not appear in the anwser. Does the second term drop, because [tex] \frac {\partial y} { \partial x } =o [/tex].
They show this by rewriting it as follows: [tex] \frac {\partial } { \partial x } y =o [/tex].
I dident realize you could separate the partial operator from the numerator term like that.
Could a possible reason be that y is an independent variable, so a change in x, another independent variable, has no effect on y, which is why it is equal to zero? In other words, the change in y has nothing to do with the change in x, which is why the fraction is zero.
And I guess for the R, term, the same follows, except for the fact that they use two chain rules, one for many variables, and one for the calc2 version.
uv= u'v+v'u