Directional Derivatives definition

icystrike
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Please help me with the geometric interpretation...

I am wondering why they can define u(directional derivative) as dr/ds instead of r(s) ?

Below are my interpretations, i don't not know whether I'm right... :

The r in dr is <dx,dy,dz> and it refers to the change in position of the graph w(x,y,z)?

The s in ds refers to the change in the distance traveled along the trajectory of u?

Conclusively i can state that the u refers to the change in position of the graph w(x,y,z) when the change in the distance of traveled along the trajectory?

(Del)W dot u gives us the relationship between how the change in position (r) relates to the change in value of W?
 
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icystrike said:
Please help me with the geometric interpretation...

I am wondering why they can define u(directional derivative) as dr/ds instead of r(s) ?
I have no idea what you are talking about. I have never seen a directional derivative defined as "dr/ds". In fact, because you have not told us what "r" or "s" are, it really makes no sense.

Below are my interpretations, i don't not know whether I'm right... :

The r in dr is <dx,dy,dz> and it refers to the change in position of the graph w(x,y,z)?
Okay, "dr" is the differential of position. That makes sense now.

The s in ds refers to the change in the distance traveled along the trajectory of u?
Okay, you are taking s as the arclength measured from some fixed point. But where did "u" come from? And how does it have a "trajectory"?

Conclusively i can state that the u refers to the change in position of the graph w(x,y,z) when the change in the distance of traveled along the trajectory?
Well, I don't know. So far you have used the symbol "u" three times without defining it. It could mean anything. Oh, and the "graph w(x,y,z)" does not "change position" at all.

(Del)W dot u gives us the relationship between how the change in position (r) relates to the change in value of W?
Now, you are taking "u" to be a vector and you had said nothing about vectors before! If W(x,y,z) is some quantity defined at each point, then \nabla W\cdot \vec{u} is the rate of change in the quantity W(x,y,z) as you move in the direction of the (unit) vector \vec{u}. I can see no reason to even mention "ds/dr" nor is there any "trajectory" involved.
 
Sorry HallsofIvy for the loose definition...

Please if u don't mind watch "http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-12-gradient/" from 50.09 minutes onwards... It won't talk u more than 3 minutes to understand what I am talking about... :redface:
 
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I'm not sure I understand your questions either.

50.09 is the end of the video; there is nothing after 50.09.

r is a vector-valued function. Specifically, the values of the function r are position vectors. It takes a real number, s, as its input. Its output is a vector r(s) = <r1(s),r2(s),r3(s)>, one vector for each number s. We can think of position vectors as points. If you're picturing them as arrows, think of the points at the tips of the arrows. The other end of the arrows is at one fixed point, namely <0,0,0>, the zero vector.

The points which are values of the function r form a curve in space.

dr/ds is the derivative of r. It too is a vector-valued function, but of a slightly different kind. For each number s, it outputs a vector dr/ds (s). You can imagine this vector, dr/ds (s), as an arrow with its blunt end not at <0,0,0> but at the point r(s). The vector dr/ds (s) is tangent to the curve at this point, r(s). We call such vectors tangent vectors to the curve, or, synonymously, velocity vectors.

By convention, we use the letter s for the independent variable here when the curve is parameterized by arc length. This means that, if a and b are specific values that s can take, specific real numbers, then the length of the curve between r(a) and r(a+b) is |b - a| (the absolute value of "a minus b"). It happens that when we describe a curve by a function for which this rule about arc length holds, each vector dr/ds (s) has unit length; that is ||dr/ds (s)|| = 1.

*

Why is this? This derivative is, by definition, the limit of the change in r divided by the change in s, as the change in s approaches zero. Intuitively, the curve becomes more and more like a straight line when seen at higher magnitude.

We could have used some other function to describe the same curve, a function for which this rule about the length of the curve doesn't hold. In that case, we'd probably use a different letter for the independent variable, such as t. Then the length of the tangent vector could be any real number, not necessarily 1.

We can make some other, arbitrary parameter, t, a function of s, and vice versa. Then, by the chain rule,

dr/ds = dr/dt * dt/ds.

But dt/ds = (ds/dt)-1, the reciprocal of speed (speed being the magnitude of the velocity vector). So dr/ds is a velocity vector divided by its own magnitude. That is, a unit vector.

*

w is a scalar field. It takes a position vector as its input, and gives as its output a real number. That is, for each point in space (each position vector), <x,y,z>, the function w associates a real number, w(x,y,z).

The gradient of w is a vector field, whose values are gradient vectors \nabla w (x,y,z) for each position vector <x,y,z>. This is a function which takes a point in space, the position vector <x,y,z>, as its input, and gives as its output another kind of vector, the gradient vector, associated with that point. The gradient vectors point in the direction of the greatest rate of increase (the steepest rise) of the values of the scalar field w. The rate of increase in that direction, at the point <x,y,z>, is the magnitude of the gradient vector at <x,y,z>, that is, the magnitude of \nabla w (x,y,z).

What if we want to know the rate of increase of w, at some point <x,y,z>, in some other direction, an arbitrary direction, not necessarily the direction with the greatest rate of increase? Then we can take the dot product of \nabla w (x,y,z) with a unit vector, say u, pointing in the desired direction:

\nabla w (x,y,z) \; \cdot \; \textbf{u} = \left \| \nabla w (x,y,z) \right \| \; \left \| \textbf{u} \right \| \; \cos(\theta)

where \theta (theta) is the angle between these two vectors. This gives us the projection of \nabla w (x,y,z) onto the line through u, in other words, the component of \nabla w (x,y,z) in the direction of u. The right side of this equation denotes multiplication of three real numbers, the magnitude of the gradient vector at <x,y,z>, a unit vector, and the cosine of the angle between them.

Is that any help?
 
Thank you Rasalhague!

Its a great help!

To summarize, we can let any unit vector be dr/ds because of the properties you've stated?
Eg. Magnitude is 1.
 
The derivative dr/ds, evaluated at any particular point, s, of any parameterization, r:R-->R3, will be a vector of unit length (that is, magnitude 1) provided the paramaterization is with respect to arc length. And given a point, and a unit vector in any direction, you can find a curve for which that vector is the derivative of a parameterization of the curve with respect to arc length (i.e. the parameter is arc length).
 
The directional derivative is just the derivative of a function along a curve whose tangent vector points in a given direction. The tangent vector, by convention ,is taken to have length 1 to eliminate the effects of different speeds.

If a curve is parametrized by arc length then its tangent vector always has length 1 but it is only required to have length 1 at a single point to yield the directional derivative at that point.
 
Rasalhague said:
" And given a point, and a unit vector in any direction, you can find a curve for which that vector is the derivative of a parameterization"

I'm wondering if there is a proof to this existence property although it seems to be rather obvious. Does it mean that the curve should not cross one another?
 
Stated as such, the proof is trivial: given the point p and a unit vector e, the curve
\gamma: I\to \mathbb{R}
\gamma(t)=p+te
satisfies \gamma(0)=p and \gamma&#039;(t)=e, so is a curve at p with directional derivative e.

More generally: given a smooth vector field (a direction at every point, ivarying smoothly), there is a corresponding flow. This follows from the existence and uniqueness of ODE's.
 
  • #10
Reviving this dead thread. Thanks to those who posted.

Suppose I have a contour f = f(t). In general, it is difficult to calculate an arc length parameterization of the contour f = f(s).

We stated above df/ds = df/dt * dt/ds, by the chain rule. We cared about df/ds because it has unit length everywhere along the contour.

Perhaps I am missing something obvious, but how do I evaluate dt/ds at any point along the curve? Do I still need an arc length parameterization (or maybe its inverse parameterization) to evaluate the quantity dt/ds as opposed to calculating f=f(s) then directly calculating df/ds?

Thanks for any help!
 
  • #11
We generally say that df/ds and df/dt are both in the same direction; it's just that df/ds has unit magnitude.

I believe you can solve to get ds/dt = |df/dt|. You should be able to integrate this to solve for s = s(t).
 

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