Question

In the definition of a direction derivative, how is the denominator of 'h' obtained?

Answer 1

\[D _{u}f(a,b) = \lim_{h \rightarrow 0} \frac{ f(a+h*\cos \Theta, a+h*\sin \Theta) - f(a,b) }{ h }\]
Where does the 'h' in the denominator come from?

Answer 2

Much like the definition of the normal derivative, I reckon :)

Answer 3

In a normal derivative, the defination is similiar to the traditional algebra's method of finding slope, that is:
\[f \prime () = \frac{ f(x+h) - f(x) }{ (x+h) - x }\]
And the denominator would simplify to 'h'. How is it done with the directional derivative?

Answer 4

A directional derivative is the derivative (rate of change) of a function in a particular direction (courtesy of the unit vector coupled with it)

Answer 5

h is still the difference in the variables.

Answer 6

Or, maybe you can think of it this way.
\[\large D_vf(x,y)=\lim_{h\rightarrow0}\frac{f[(x,y)+h(v_x,v_y)]-f(x,y)}{|(x,y)+h\mathbf{\vec{v}}-(x,y)|}\]

Answer 7

The bottom part has to be a magnitude, since the top part would still be a scalar.

Answer 8

I'm not 100% on it, but I think I'm getting the idea.

Answer 9

I had done something, vaguely similar, though I don't believe my math is sound.

Answer 10

Neither am I, really.
I have always treated the limit
\[\Large \lim_{h\rightarrow0}\frac{f(x+\cos\theta,y+\sin\theta)-f(x,y)}{h}\]as the definition of the directional derivative. And I never used this formula extensively :)

Answer 11

Sorry,
\[\Large \lim_{h\rightarrow0}\frac{f(x+h\cos\theta,y+h\sin\theta)-f(x,y)}{h}\]

Answer 12

for example if \[P _{0}(a,b)\] and \[P(a + h\cos \theta, b +h\sin \theta)\]then roughly the equation is equivalent to:\[\lim_{h \rightarrow 0}\frac{ f(P)-f(P _{0}) }{ P-P _{0} }\]. If I do some arithmetic on the denominator then it should be equal to:\[(a+h\cos \theta -a, b+h\sin \theta - b)\] right?

Answer 13

yeah...

Answer 14

If I simplify this I should get:\[(h*\cos \theta, h*\sin \theta)\] then I should be able to factor 'h' out as a constant right? and get:
\[h(\cos \theta, \sin \theta)\] and I know that in vector terms:\[<\cos \theta, \sin \theta> = i+j\] in other words a unit vector. then I wanted to multiply h by that and get h. but the problem with this is I out of nowhere just used a coordinate as a vector...

Answer 15

there is a one-to-one correspondence between vectors and points.
Also, i+j is a lot of things, but it's DEFINITELY not a unit vector.
\[\huge |\vec i+\vec j|=\sqrt2\]

Answer 16

could you explain the correspondence? And oops, thanks for that.

Answer 17

Correspondence?
We can form a bijection :/
Let's go on 2d vectors... Let V be the set of all 2d vectors.
Define a function
\[\huge \Phi:\mathbb{R}^2 \ \rightarrow \ V\]

Answer 18

\[\huge (x,y) \ \ |\rightarrow <x , y>\]

Answer 19

And it's clearly a bijective function, no? :)

Answer 20

honestly, I don't what a "bijective function is" :(

Answer 21

No worries.
Let f be a function from a domain X to a set Y
\[\huge f:X \rightarrow Y\]
f is called injective if for any two elements x_1 and x_2, in x
if \[\large x_1 \ne x_2 \rightarrow f(x_1)\ne f(x_2)\]In other words, the function is one to one.

Answer 22

If for any element y in the set Y, there exists an x in X such that
\[\large y = f(x)\]then, the function f is called surjective. In other words, for any element of Y, there is an x that is mapped to that element y, by the function f.

f is also called onto (if it's surjective)

Answer 23

A function that is both injective and surjective is called bijective.
It is only bijective functions which have inverse functions.

Answer 24

I get the one to one bit, but my lack of familiarity with terms like "set" is retarding my learning process.

Answer 25

Well, never too late to learn. :)

Answer 26

indeed.