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I'm learning Directional Derivative on Khan Academy. Here is the definition of Directional Derivative:

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So the formula for calculating directional derivative is:

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But as I knew, the dot product should be:

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I don't understand this point. Please explain for me why the formula for calculating directional derivative doesn't have "cos($\alpha$)".

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  • $\begingroup$ That’s not the only way to define the dot product. $\endgroup$ – amd Oct 10 '17 at 8:50
  • $\begingroup$ so you means we have 2 ways for calculating dot product. And as I guess, by using those 2 ways, I can calculate cos(alpha) between those 2 vectors. For example, vector (a1,b1) and vector (a2,b2). cos(alpha) = (a1*a2+ b1*b2) / (square_root(a1^2+b1^2) * square_root(a2^2+b2^2)) $\endgroup$ – hqt Oct 10 '17 at 8:58
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    $\begingroup$ The dot product should be $$a \cdot b$$ $\endgroup$ – user14972 Oct 10 '17 at 9:23
  • $\begingroup$ @hqt Just so. This works in higher-dimensional spaces, too. In fact, you can take that equivalence as the definition of angle in any vector space equipped with an inner product. $\endgroup$ – amd Oct 10 '17 at 20:40
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The Dot Product of two vectors $x=[x_1,x_2\cdots, x_n], a=[a_1,a_2\cdots, a_n]$ is algebraically defined as $$x\cdot a=\sum_{i=1}^nx_ia_i=x_1a_1+x_2a_2+\cdots+x_na_n$$ In your case we have the vectors $v=[v_1, v_2, v_3]$ and $df=\Large [\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\frac{\partial f}{\partial z}]$.

I hope it is clear now why the formula described is correct.


In the Euclidean $2D$-space indeed we do have an equivalent geometric form involving $\cos$ as you mention in your question.

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  • $\begingroup$ so cos(alpha) version only appeared in Euclidean 2-D space? I asked this because there is a proof why "gradient descent of multivariable function is a steepest ascent" using this formula for prof. $\endgroup$ – hqt Oct 10 '17 at 15:26
  • $\begingroup$ For example here: math.stackexchange.com/questions/223252/… $\endgroup$ – hqt Oct 10 '17 at 15:27
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    $\begingroup$ The “geometric form” is valid in a Euclidean space of any dimension. Indeed, it can be taken as the definition of the angle between two vectors in an inner product space. $\endgroup$ – amd Oct 10 '17 at 17:35
  • $\begingroup$ @amd Thank you for the clarifications. I did not mean to imply that the geometric form is somehow restricted to $2D$ space but I see that my wording was less than perfect.. $\endgroup$ – MathematicianByMistake Oct 10 '17 at 21:46

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