## Uses for the Gradient: Line Integrals.

### May 19, 2010

I was all set to post about lagrange multipliers when I realized that they are pretty dull.  Since this is my blog, after all, I thought I’d take a slight detour and talk about why we’d ever care about taking the gradient of functions, besides for all of that max-min stuff I talked about before.  We haven’t formally talked about line integrals yet, but they are not horrible creatures: if you have not seen them yet, just think about integrals in one dimension.

As it turns out, and you can probably imagine why, the gradient acts as a kind of “universal derivative” in multivariable calculus.  Whereas we had, in single variable calculus,

$\displaystyle \int_{a}^{b} f'(x)dx = f(b) - f(a)$

we don’t have a nice “anti-derivative” in vector calculus such that this equality should hold!  After all, we could ask “the anti-derivative…in what direction?”

In general, the integral of $\int_{\gamma}f\cdot ds$ where $\gamma$ is our path and $f(x,y)$ is some function, is not easy to compute.  In fact, it may be extremely tedious and near impossible to do by hand!  At the very least, they are not easy.

But, for a special type of function, they are trivial.  Specifically, if $\gamma(t)$ is a curve where $0\leq t\leq 1$, then

$\displaystyle \int_{\gamma} \nabla f(x,y)\cdot ds = f(\gamma(1)) - f(\gamma(0))$.

Yes!  It’s that simple!  It’s almost like the fundamental theorem of calculus!  Yes, if we have the gradient of $f(x,y)$, then we just have to recover $f(x,y)$ in order to integrate!

Let’s just give an example of this.

Example:

Let $\gamma(t) = (t, t^2)$ for $0\leq t\leq 1$.

We want to evaluate $\displaystyle \int_{\gamma} (2xy, x^2)\cdot ds$

So our function is, say, $g(x,y) = (2xy, x^2)$, but let’s note that for $f(x,y) = x^{2}y$, we have that $\nabla f(x,y) = (2xy, x^2) = g(x,y)$!  That’s cool.  That means that, according to our theorem, we have

$\displaystyle \int_{\gamma} (2xy, x^2)\cdot ds$

$= f(\gamma(1)) - f(\gamma(0)) = f((1,1)) - f((0,0)) = 1 - 0 = 1$.

That was pretty easy!  So, that’s another use for the gradient of a function.  Next time, we’ll take another small detour to talk about the other ways to use our $\nabla$ symbol if we consider it as the formal symbol and (slight abuse of notation) $\nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial dy}, \frac{\partial}{\partial z})$; we’ll meet the brothers Div, Grad, Curl.