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The directional derivative in the direction <0, 1> is < ∂f/∂x, ∂f, ∂y> ⋅ <0, 1>. This makes perfect sense because this comes out to taking ∂f/∂x and multiplying it to 1, and adding it to (∂f/∂y)*(0), which is zero, coming to an end result of ∂f/∂x. I do not understand how this can be generalized. How can we say, for each and every possible function, that the rate of change at a given point in the direction will be a(∂f/∂x) + b(∂f/ ∂y)? Someone explained the directional derivative as being the sum of the vector's component in the x direction multiplied by a pure step in the x direction, and the vectors component in the y direction multiplied by a pure step in the y direction. For example, how can we say that for every possible function, the directional derivative in the direction <1/sqrt(2), 1/sqrt(2)> is equal to (1/sqrt(2))∂f/∂x + (1/sqrt(2))∂f/∂y? I would think things would vary from function to function. What relationship am I missing here? Please help!

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Let $z=Ax+By+C$ be the tangent plane of the graph $z=f(x,y)$ at the point in question; thus $A=\partial f/\partial x$ and $B=\partial f/\partial y$ at that point. (This assumes that $f$ is differentiable, of course, but so does your question to begin with.)

Then the directional derivative with respect to the unit vector $(a,b)$ is just how much $z$ changes along the tangent plane if you add $a$ to $x$ and add $b$ to $y$. Which is obviously $aA+bB$.

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  • $\begingroup$ What you say is not so obvious to me. I am rather new to calculus. Would you explain the details? What you have already said is making a lot of sense. Thx! $\endgroup$ – King Squirrel Jul 29 '18 at 16:05
  • $\begingroup$ I don't think I have that much more to say... Could you perhaps be more specific about what parts you find unclear? $\endgroup$ – Hans Lundmark Jul 29 '18 at 16:32
  • $\begingroup$ Suppose we want the derivative in the direction of <1,1>. So I’ve been told, we must convert this to a unit vector: <1/sqrt(2), 1/sqrt(2)> in order to use the formula. So I don’t find it intuitive that the partial x must be multiplied by 1/sqrt(2) and the partial y must be multiplied by 1/sqrt(2), take the sum, and that’s your derivative in the direction. <1, 1>. $\endgroup$ – King Squirrel Jul 30 '18 at 0:13
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    $\begingroup$ OK. As far as derivatives are concerned, you may as well replace the graph with its tangent plane (since you only care about the slope at that particular point). So how much does $z=Ax+By+C$ change if you increase both $x$ and $y$ by $1/\sqrt2$? Answer: $z$ increases by $A/\sqrt2+B/\sqrt2$. $\endgroup$ – Hans Lundmark Jul 30 '18 at 7:26
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    $\begingroup$ Haha, you're welcome! $\endgroup$ – Hans Lundmark Jul 31 '18 at 4:55

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