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

[Byron Chen]

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?

 

[micromass]

These are so-called directional derivatives.

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

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}

First of all, note that if \mathbb{v}=(1,0,0), 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 f(x,y,z)=x^5yz^4+\log(x)\sin(yz). We wish to find the partial derivative with respect to the line x=y=-z. A unit vector on this line is given by \mathbb{v}=(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}). (of course, -\mathbb{v} is another unit vector but this vector will give us the same directional derivative up to a sign). So, we have

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

So

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))

 

[micromass]

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

f:\mathbb{R}^3\rightarrow \mathbb{R}

We want to find the derivative when this is restricted to a line.
First, take a point \mathbb{a} 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 \mathbb{v} which is parallel to L. Then we can write all points on L in the form \mathbb{a}+\alpha\mathbb{v}.

So f restricted on the line gives us the function

g(\alpha)=f(\mathbb{a}+\alpha\mathbb{v})

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

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}

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

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

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

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

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!

 

[GarageDweller]

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