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Partial Derivatives with Respect To Lines That Are Not In The Direction of Axis

  1. Aug 11, 2012 #1
    A 3-dimensional graph has infinite number of derivatives (in different directions) at a single point. I've learned how to find the partial derivative with respect to x and y, simply taking y and x to be constant respectively. But what do I do if I want to take the partial derivative with respect to a line that is not in the direction of an axis, but rather a diagonal line, like a line cutting through the x-axis at 45 degrees? In this case you can't simply take one of the axis to be a constant, right?
     
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  3. Aug 11, 2012 #2

    micromass

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    These are so-called directional derivatives.

    So, let's say we are given a function [itex]f:\mathbb{R}^3\rightarrow \mathbb{R}[/itex] and a unit vector [itex]\mathbf{v}=(v_1,v_2,v_3)[/itex]. The the directional derivative of f in v is given by

    [tex]D_\mathbb{v} f = v_1\frac{\partial f}{\partial x}+v_2\frac{\partial f}{\partial y}+v_2\frac{\partial f}{\partial z}[/tex]

    First of all, note that if [itex]\mathbb{v}=(1,0,0)[/itex], then we just obtain the partial derivative with respect to x. So this is clearly the case where you differentiate with respect to the x-axis.

    As an example, given [itex]f(x,y,z)=x^5yz^4+\log(x)\sin(yz)[/itex]. We wish to find the partial derivative with respect to the line x=y=-z. A unit vector on this line is given by [itex]\mathbb{v}=(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}})[/itex]. (of course, [itex]-\mathbb{v}[/itex] is another unit vector but this vector will give us the same directional derivative up to a sign). So, we have

    [tex]\frac{\partial f}{\partial x}=5x^4yz^4 + \frac{\sin(yz)}{x}[/tex]
    [tex]\frac{\partial f}{\partial y}=x^5z^4 + z\log(x)\cos(yz)[/tex]
    [tex]\frac{\partial f}{\partial z}= 4x^5yz^3 + y\log(x)\cos(yz)[/tex]

    So

    [tex]D_\mathbb{v} f = \frac{1}{\sqrt{3}}(5x^4yz^4 + \frac{\sin(yz)}{x})+\frac{1}{\sqrt{3}}(x^5z^4 + z\log(x)\cos(yz)) -\frac{1}{\sqrt{3}}(4x^5yz^3 + y\log(x)\cos(yz))[/tex]
     
  4. Aug 11, 2012 #3

    micromass

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    The previous post gave information on how to calculate it, but not on why this is the derivative with respect to a line. To see how to define it, take a function

    [tex]f:\mathbb{R}^3\rightarrow \mathbb{R}[/tex]

    We want to find the derivative when this is restricted to a line.
    First, take a point [itex]\mathbb{a}[/itex] in which we wish to find the directional derivative. We wish to find the directional derivative with respect to a line L.
    First, take a unit vector [itex]\mathbb{v}[/itex] which is parallel to L. Then we can write all points on L in the form [itex]\mathbb{a}+\alpha\mathbb{v}[/itex].

    So f restricted on the line gives us the function

    [tex]g(\alpha)=f(\mathbb{a}+\alpha\mathbb{v})[/tex]

    Note that [itex]g(0)=f(\mathbb{a})[/itex]. So we wish to find the derivative of g with respect to [itex]\alpha[/itex], and we wish to find the derivative of this in 0. This is

    [tex]g^\prime(0)=\lim_{h\rightarrow 0}\frac{g(h)-g(0)}{h} = \lim_{h\rightarrow 0} \frac{ f(\mathbb{a}+h\mathbb{v})-f(\mathbb{a})}{h}[/tex]

    This expression is called the directional derivative of f in [itex]\mathbb{a}[/itex]. So

    [tex]D_\mathbb{v} f = \lim_{h\rightarrow 0} \frac{ f(\mathbb{a}+h\mathbb{v})-f(\mathbb{a})}{h}[/tex]

    Now, you may wonder, how are the things in this post and my previous post equal. Why is

    [tex]g^\prime(0)=v_1\frac{\partial f}{\partial x}+v_2\frac{\partial f}{\partial y}+v_3\frac{\partial f}{\partial z}[/tex]

    The equality of these two things can very easily be shown by the chain rule. Try it yourself once you've seen the chain rule!
     
  5. Aug 11, 2012 #4
    If you have a parametric curve, r(t), then the directional derivative of a scalar function along that curve is
    Df/dt=r'(t) O f'(t)

    O is the dot product, f'(t) is a vector with components equal to the respective derivatives of f
     
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