Directional Derivative, Gradient

1. Suppose we have a scalar function of several variables
(1)\displaystyle  f\colon{\mathbb R}^{3}\rightarrow{\mathbb R}
Let {\widehat{u}} be some unit vector. How does {f} change in the {\widehat{u}} direction?

We can consider this quantity as a function
(2)\displaystyle  g(\vec{x}) = \lim_{h\rightarrow0}\frac{f(\vec{x}+h\widehat{u})-f(\vec{x})}{h}
What does this look like?

2. Lets restrict our attention to the smallest non-boring case: {f\colon{\mathbb R}^2\rightarrow{\mathbb R}}. Then we write {\widehat{u} = \langle p,q\rangle}. We have
(3)\displaystyle  f(\vec{x}+h\widehat{u}) = f(x+hp,y+hq).
Expanding this to first order in {h} lets us write
(4)\displaystyle  \begin{aligned} f(x+hp,y+hq) &= f(x,y+hq) + hp\partial_{x}f(x,y+hq)+\mathcal{O}(h^{2})\\ &= \bigl(f(x,y)+hq\partial_{y}f(x,y)+\mathcal{O}(h^{2})\bigr)\\ &\quad+hp\partial_{x}\bigl(f(x,y)+hq\partial_{y}f(x,y)+\mathcal{O}(h^{2})\bigr)\\ &\quad+\mathcal{O}(h^{2})\\ &= f(x,y) + h\bigl(q\partial_{y}f(x,y) + p\partial_{x}f(x,y)\bigr) +\mathcal{O}(h^{2}). \end{aligned}
But what does this look like? It’s simply
(5)\displaystyle  f(\vec{x}+h\widehat{u}) = f(\vec{x}) + h\widehat{u}\cdot\langle\partial_{x},\partial_{y}\rangle f(\vec{x}) + \mathcal{O}(h^{2}).

Example 1. What is the derivative of {f(x,y)=x/y} in the direction of {\vec{v}=\langle 1,3\rangle} at {(5,3)}?

Solution: We find the directional derivative is
(6)\displaystyle  v_{1}\partial_{x}f + v_{2}\partial_{y}f
We compute
(7)\displaystyle  \partial_{x}f = 1/y
(8)\displaystyle  \partial_{y}f = -x/y^{2}
Thus we have
(9)\displaystyle  v_{1}\partial_{x}f + v_{2}\partial_{y}f = v_{1}(1/y) + v_{2}(-x/y^{2}).
We plug in {v_{1}=1}, {v_{2}=3}
(10)\displaystyle  v_{1}\partial_{x}f + v_{2}\partial_{y}f = (1/y) + 3(-x/y^{2}).
Then we evaluate {(x,y)=(5,3)} to get
(11)\displaystyle  v_{1}\partial_{x}f + v_{2}\partial_{y}f = (1/3) + 3(-5/3^{2}) =-4/3.
This gives us the directional derivative of {f}.

3. We denote
(12)\displaystyle  \vec{\nabla} = \langle\partial_{1},\dots,\partial_{n}\rangle
and call it the “Gradient”. Note we will write {\nabla} interchangeably with the vector arrow {\vec{\nabla}}, and they mean the same thing. The vector arrow doesn’t add anything semantically, it’s just different syntax.

The directional derivative is then
(13)\displaystyle  \widehat{u}\cdot\vec{\nabla}f(\vec{x}) = \widehat{u}\cdot\langle \partial_{1}f,\dots,\partial_{n}f\rangle.
Note that the gradient acting on a scalar function produces a vector-valued function of several variables, but we can also take the dot product of the gradient with such a monstrosity.

4. Question: What is a vector-valued function of several variables?

For us, in practice, we think of this as a Vector Field: a “function” which assigns to each point a vector. Each vector-component is a function, usually smooth (i.e., infinitely differentiable).

(Again, just as we warned the reader with vectors, this too is a lie. A vector field is a bit more than just a function {{\mathbb R}^{n}\rightarrow{\mathbb R}^{n}}, it’s a more complicated beast which is studied further in differential geometry.)

5. Meaning of Gradient. Consider a family of level curves {f(\vec{x})=c}. The gradient points towards the direction of increasing {c}. How can we see this? Well, consider the function
(14)\displaystyle  f(x,y)=x^{2}-y^{2}.
We see its gradient is
(15)\displaystyle  \vec{\nabla}f(x,y) = \langle 2x, -2y\rangle.
Lets draw a few level-curves and see what the vectors point to:

We see the vectors point towards {(x,y)\rightarrow(\pm\infty,0)}.

Exercise 1. Consider the function {f\colon{\mathbb R}^{2}\rightarrow{\mathbb R}} defined by {f(x,y)=\ln(x^{2}+y^{2})}. What is its gradient?

Exercise 2. Let {g\colon{\mathbb R}^{3}\rightarrow{\mathbb R}} be defined by {g(x,y,z) = (x^{2}+y^{2}+z^{2})^{-1/2}}. Find its gradient.

Exercise 3. Let {f(x,y,z)=x/(y+z)}. Find its derivative in the direction {\vec{u}=\langle 1,1,1\rangle} at the point {(1,6,2)}.

Exercise 4. Let {f(x,y)=x\exp(-y) + 3y}. Find its gradient.


About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Directional Derivative, Gradient, Vector Calculus and tagged . Bookmark the permalink.

One Response to Directional Derivative, Gradient

  1. Pingback: Lagrange Multipliers | My Math Blog

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