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If Iam going to derive the DD of a function without using a unit vector i.e.$u=<a,b>$ (not unit vector). I get the same result $Du=(Fx).a+(Fy).b$.So why are we considering unit vector always?

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  • $\begingroup$ write $a$ and $b$ using polar coordinates. Now do you see the problem? $\endgroup$ – The Great Duck Dec 29 '17 at 7:35
  • $\begingroup$ How is this “the same result?” $\nabla f\cdot v \ne \nabla f\cdot {v\over\|v\|}$ if $v$ isn’t a unit vector. $\endgroup$ – amd Dec 29 '17 at 8:44
  • $\begingroup$ Yes true, but i get the same formula , but different results when i change the magnitude of the vector but in the same direction.By same formula i mean Du=(fx)a+(fy)b, whether it may or many not be a unit vector. $\endgroup$ – CHAND Dec 29 '17 at 16:23
  • $\begingroup$ @CHAND Please, if you are ok, you can accept the answer and set it as solved. Thanks! $\endgroup$ – user Feb 1 '18 at 22:48
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If V = < a, b> is not a unit vector then afx + bfy is not the directional derivative of f at the direction of V. It is the directional derivative multiplied by the norm of V. Therefore you do not get the same result unless you normalize V.

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  • $\begingroup$ That depends on your definition of directional derivative. Not all authors require a unit vector. Indeed, if you’re not working in a normed space, then what’s a unit vector in the first place? $\endgroup$ – amd Dec 29 '17 at 8:43
  • $\begingroup$ Solution of direction derivative changes when we use a vector other than a unit vector at a point but in the same direction, but I'm asking why is it changing even though it is in the same direction? $\endgroup$ – CHAND Dec 29 '17 at 12:23
  • $\begingroup$ That is exactly why you need to consider a unit vector so you get the same answer for all vectors in the same direction. $\endgroup$ – Mohammad Riazi-Kermani Dec 29 '17 at 13:07
  • $\begingroup$ The directional derivative is a dot product of the gradient vector and the unit vector in the desired direction. As you know the dot product depends on the norms of vectors so you get a different answer if you do not normalize your vector. $\endgroup$ – Mohammad Riazi-Kermani Dec 29 '17 at 13:13
  • $\begingroup$ The dot product of the gradient and a unit vector happens to equal the directional derivative for differentiable functions (and some others), but that’s not the fundamental definition of a directional derivative. $\endgroup$ – amd Dec 29 '17 at 23:28
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For any $\vec v=(a,b)$ we can define the directional derivative as:

$$\frac{\partial f}{\partial \vec v}=\lim_{h\to 0}\frac{f(x_0+ah,y_0+bh)-f(x_0,y_0)}{h}$$

Often the directional derivative are defined assuming $\vec v$ as a unit vector, indeed it can be shown that:

$$\frac{\partial f}{\partial \lambda \vec v}=\lim_{h\to 0}\lambda\frac{f(x_0+\lambda ah,y_0+\lambda bh)-f(x_0,y_0)}{\lambda h}=\lambda \frac{\partial f}{\partial \vec v}$$

Notably the partial derivatives are the directional derivatives corresponding to the unit vectors $(1,0)$ and $(0,1)$.

Wheter $f$ is differentiable the following holds:

$$\frac{\partial f}{\partial \vec v}=\nabla f\cdot\vec v=v_1\frac{\partial f}{\partial x}+v_2\frac{\partial f}{\partial y}$$

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The formula stated here gives the derivative of f, a real valued function of two variables, with respect to a vector u whose components are 'a' and 'b' at a point where the partial derivatives(denoted in the question by fx and fy) of f are evaluated, in case the function f is differentiable there, as a function of two variables. If we put the additional condition that u is a unit vector we get what is usually referred to as the directional derivative of f,at the point, along u and which is equal to the component of the gradient vector of f along u. Ref: Tom Apostol: Mathematical Analysis.

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  • $\begingroup$ The formula stated where? $\endgroup$ – amd Dec 29 '17 at 7:00
  • $\begingroup$ @amd in the rest of that reference somewhere. beats me. $\endgroup$ – The Great Duck Dec 29 '17 at 7:26
  • $\begingroup$ That’s not very useful. $\endgroup$ – amd Dec 29 '17 at 8:41

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