Derivatives and Tangent Vectors

Here’s just a little aside about the pushforward of a unit vector and its relationship to the usual notion of taking a derivative. We’re going to use this idea a lot as the course goes on, so make sure you really “get it!”


Derivatives on the Real Line

So far we’ve been thinking about the differential in a very geometric way: it tells us how to stretch out or push forward tangent vectors as we go from one place to another. In fact, we can apply this geometric viewpoint to pretty much any situation where derivatives are involved. For instance, think about a good old fashioned real-valued function \(\phi(x)\) on the real line. Typically we visualize \(\phi\) by plotting its value as a height over the \(x\)-axis:

In this case, the derivative \(\phi^\prime\) can be interpreted as the slope of the height function, as suggested by the dashed line in the picture above. Alternatively, we can imagine that \(\phi\) stretches out the real line itself, indicated by the change in node spacing in this picture:

Where the derivative is large, nodes are spaced far apart; where the derivative is small, nodes are spaced close together. This picture inspires us to write the derivative of \(\phi\) in terms of the push-forward \(d\phi(X)\) of a unit tangent vector \(X\) pointing along the positive \(x\)-axis:

\[ \phi^\prime = d\phi(X). \]

In other words, the derivative of \(\phi\) is just the “stretch factor” as we go from one copy of \(\mathbb{R}\) to the other. But wait a minute — does this equality even make sense? The thing on the left is a scalar, but the thing on the right is a vector! Of course, any tangent vector on the real line can be represented as just a single value, quantifying its extent in the positive or negative direction. So this expression does make sense — as long as we understand that we’re identifying tangent vectors on \(\mathbb{R}\) with real numbers. Often this kind of “type checking” can help verify that formulas and expressions are correct, similar to the way you might check for matching units in a physical equation.

Here’s another question: how is this interpretation of the derivative any different from our usual interpretation in terms of height functions? Aren’t we also stretching out the real line in that case? Well, yes and no — certainly the real line still gets stretched out into some other curve. But this curve is now a subset of the plane \(\mathbb{R}^2\) — in particular, it’s the curve \(\gamma = (x,\phi(x))\). So for one thing, “type checking” fails in this case: \(\phi^\prime\) is a scalar, but \(d\gamma(X)\) is a 2-vector. But most importantly, the amount of stretching experienced by the curve doesn’t correspond to our usual notion of the derivative of \(\phi\) — for instance, if we look at the magnitude of \(|d\gamma(X)|\) we get \(\sqrt{1+(\phi^\prime)^2}\). (Why is this statement true geometrically? How could you write \(\phi^\prime\) in terms of \(d\gamma(X)\)? Can you come up with an expression where you recover the proper sign?)


Directional Derivatives

So far so good: we can think of the derivative of a real-valued function on \(\mathbb{R}\) as the pushforward of a (positively-oriented) unit tangent vector \(X\). But what does \(d\phi(X)\) mean if \(\phi\) is defined over some other domain, like the plane \(\mathbb{R}^2\)? This question may “stretch” your mind a little, but if you can understand this example then you’re well on your way to understanding derivatives in terms of tangent vectors. Let’s take a look at the geometry of the problem — again, there are two ways we could plot \(\phi\). The usual approach is to draw a height function over the plane:

The derivative has something to do with the slope of this hill, but in which direction? To answer this question, we can introduce the idea of a directional derivative — i.e., we pick a vector \(X\) and see how quickly we travel uphill (or downhill) in that direction. And again we can consider an alternative picture:

Since \(\phi\) is a map from \(\mathbb{R}^2\) to \(\mathbb{R}\), we can imagine that it takes a flat sheet of rubber and stretches it out into a long, skinny, one-dimensional object along the real line. Therefore if we draw an arrow \(X\) on the original sheet, then the “stretched-out” arrow \(d\phi(X)\) gives us the rate of change in \(\phi\) along the direction \(X\), i.e., the directional derivative. What about type checking? As before, everything matches up: \(d\phi(X)\) is a tangent vector on \(\mathbb{R}\), so it can be represented by a single real number. (What if we had continued to work with the height function above? How could we recover the directional derivative in this case?)

By the way, don’t worry if this discussion seems horribly informal! We’ll see a more explicit, algebraic treatment of these ideas when we start talking about exterior calculus. The important thing for now is to build some geometric intuition about derivatives. In particular: a map from any space to any other space can be viewed as some kind of bending and twisting and stretching (or possibly tearing!); derivatives can be understood in terms of what happens to little arrows along the way.

October 20, 2012 | Posted in: Notes | Comments Closed

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