Week 8 Discussion Notes

Table of Contents

• Directional Derivatives
• Chain Rule

Directional Derivatives

In 1D, if you had a function $f\p{x}$, then the derivative $\deriv{f}{x}$ tells you how much $f$ (the "height") changes when you move a little bit in the positive $x$ direction.

With multiple variables, though, you have a lot more directions to work with, but the interpretation is the same:

$D_{\vec{u}}f\p{a, b}$ tells you how much $f$ (the "height") changes when you move a little bit in the direction specified by the vector $\vec{u}$. Since this is a dot product, you already have another, more geometric formula for the directional derivative:

$\theta$ is the angle between $\nabla f\p{a, b}$ and the vector $\vec{u}$. Recall that $\cos{\theta}$ is between $-1$ and $1$, where $\cos{0} = 1$ and $\cos{\pi} = -1$. So, you immediately get two interesting interpretations of the gradient:

1. The maximum rate of increase is in the direction $\nabla f\p{a, b}$.
2. The maximum rate of decrease is in the direction $-\nabla f\p{a, b}$.

Another interesting case is when $\cos{\theta} = 0$, where $\theta = \frac{\pi}{2}$, i.e., when $\nabla f\p{a, b}$ is orthogonal to $\vec{u}$. In this case, $D_{\vec{u}}f\p{a, b} = 0$, i.e., if you move a little bit in a direction orthogonal to $\nabla f\p{a, b}$, then $f$ (essentially) stays constant. However, if you stay at the same height, that means you're moving along a level curve. This tells you another interpretation of the gradient:

The gradient of $f$ is orthogonal to the level curves of $f$.

These pictures gives a rough summary of the concepts:

Using the interpretation that the gradient is orthogonal to level curves, you'll be able to find normal vectors for tangent planes much more easily.

Example 1.

If a surface is defined by $F\p{x, y, z} = c$, i.e., is a level curve of the function $F$, then you already know that $\nabla F$ will be orthogonal to this surface. Thus, you can use

$$\vec{n} = \nabla F$$

as your normal vector.

Example 2.

As a special case, if you have a surface defined by $z = f\p{x, y}$, then you can view it as a level curve:

$$F\p{x, y, z} = f\p{x, y} - z = 0$$

Then as in the previous example, you can use

$$\vec{n} = \nabla F = \ang{\pderiv{f}{x}, \pderiv{f}{y}, -1},$$

which coincides with the normal vector we've been using for surfaces.

Example 3.

Find a normal vector to the surface $x^2 + y^2 + z^2 = 1$ at the point $\p{a, b, c}$.

Solution.

In this problem, $F\p{x, y, z} = x^2 + y^2 + z^2$ and the surface is a level surface of $F$. Thus, we can use

$$\vec{n} = \nabla F\p{a, b, c} = \boxed{\ang{2a, 2b, 2c}}.$$

Chain Rule

Theorem

Let $f\p{x, y}$ be differentiable, and suppose $x = x\p{s, t}$ and $y = y\p{s, t}$. Then

$$\pderiv{f}{s} = \pderiv{f}{x} \pderiv{x}{s} + \pderiv{f}{y} \pderiv{y}{s}.$$

So to find $\pderiv{f}{s}$, you just need to find all the variables of $f$ that depend on $s$ and multiply out the partial derivatives. This generalizes to more complicated situations:

Example 4.

Let $f\p{x, y, z}$ be differentiable, $x = x\p{s, t}$, and $z = z\p{s}$ (so $y$ is independent of $s, t$). Calculate the partial derivatives

$$\pderiv{f}{s} \quad\text{and}\quad \pderiv{f}{t}.$$

Solution.

For $\pderiv{f}{s}$, only $x$ and $z$ depend on $s$, so we get

$$\boxed{\pderiv{f}{s} = \pderiv{f}{x} \pderiv{x}{s} + \pderiv{f}{z} \pderiv{z}{s}}.$$

Similarly, for $\pderiv{f}{t}$, only $x$ depends on $t$, so

$$\boxed{\pderiv{f}{t} = \pderiv{f}{x} \pderiv{x}{t}}.$$

Example 5.

Derive the formula

$$\deriv{}{t} f\p{\vec{r}\p{t}} = \nabla f\p{\vec{r}\p{t}} \cdot \vec{r}'\p{t}.$$

Solution.

We can write

$$\vec{r}\p{t} = \ang{x\p{t}, y\p{t}},$$

so this becomes finding the derivative $\deriv{}{t} f\p{x, y}$ with $x = x\p{t}$ and $y = y\p{t}$:

$$\begin{aligned} \deriv{}{t} f\p{x, y} &= \pderiv{f}{x} \deriv{x}{t} + \pderiv{f}{y} \deriv{y}{t} \\ &= \ang{\pderiv{f}{x}, \pderiv{f}{y}} \cdot \ang{\deriv{x}{t}, \deriv{y}{t}} \\ &= \nabla f\p{\vec{r}\p{t}} \cdot \vec{r}'\p{t}. \end{aligned}$$