Proving grad operator yields perpendicular vector to contour

[henry wang]

I saw a link on MIT open courseware proving grad operator yields perpendicular vector to contour, but I can't make sense of how dg/dt=0.
Can someone explain to me please.
http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-b-chain-rule-gradient-and-directional-derivatives/session-36-proof/MIT18_02SC_notes_19.pdf

 

[stevendaryl]

I saw a link on MIT open courseware proving grad operator yields perpendicular vector to contour, but I can't make sense of how dg/dt=0.
Can someone explain to me please.
http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-b-chain-rule-gradient-and-directional-derivatives/session-36-proof/MIT18_02SC_notes_19.pdf

If you have a function f(x,y,z), and you have a constant c, then the set of points satisfying f(x,y,z) = c is a two-dimensional surface. So now we let \vec{r}(t) be any path that stays on that surface. In terms of components, if we write \vec{r}(t) = (x(t), y(t), z(t)), then to say that \vec{r}(t) stays on the surface f(x,y,z)=c just means that for all t, we have:

f(x(t), y(t), z(t)) = c

So far, this is just true by assumption. We're assuming that x(t), y(t), z(t) are three functions such that f(x(t), y(t), z(t)) = c.

Since f(x(t), y(t), z(t)) = c, then it immediately follows that:

\frac{d}{dt} f(x(t), y(t), z(t)) = \frac{dc}{dt} = 0

All they're doing is defining f(x(t), y(t), z(t)) = g(t). That's just the definition of g. So it immediately follows that \frac{d}{dt} g(t) = 0.

 

[henry wang]

If you have a function f(x,y,z), and you have a constant c, then the set of points satisfying f(x,y,z) = c is a two-dimensional surface. So now we let \vec{r}(t) be any path that stays on that surface. In terms of components, if we write \vec{r}(t) = (x(t), y(t), z(t)), then to say that \vec{r}(t) stays on the surface f(x,y,z)=c just means that for all t, we have:

f(x(t), y(t), z(t)) = c

So far, this is just true by assumption. We're assuming that x(t), y(t), z(t) are three functions such that f(x(t), y(t), z(t)) = c.

Since f(x(t), y(t), z(t)) = c, then it immediately follows that:

\frac{d}{dt} f(x(t), y(t), z(t)) = \frac{dc}{dt} = 0

All they're doing is defining f(x(t), y(t), z(t)) = g(t). That's just the definition of g. So it immediately follows that \frac{d}{dt} g(t) = 0.

Thank you very much, that make a lot of sense, I understand now!