- Your quizzes are returned.
- slicing and dicing
- slicing in two different directions -- Clairaut requires continuity of the second partials.
- Good: you know that differentiability implies continuity, but not vice versa.
- Existence and continuity of first partials implies
differentiability.
- You've got a homework due today. Matthew stopped by and asked for some Mathematica advice. Feel free to do that, or to email me with questions.
- We're generalizing the univariate chain rule: we start with a
function $y=f(x)$, where, in addition, $x(t)$. So $y$ is an implicit
function of $t$. If we want to know the rate of change of $y$ with
respect to $t$, we can do it in two steps:
\[
\frac{dy}{dt}=\frac{df}{dx}\frac{dx}{dt}
\]
or
\[
\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}
\]
- In complete analogy, if we start with a
function $z=f(x,y)$, where, in addition, $y(t)$ and $x(t)$ are
univariate functions of $t$, then if we
want to know the rate of change of $z$ with
respect to $t$, we can do it in steps:
\[
\frac{dz}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}
\]
as mentioned in the text, we often write
\[
\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}
\]
instead.
The proof of this is actually easy to follow, and illustrates the use of the definition of differentiability. Let's take a look at it (p. 948).
Here's a start on a visualization of the situation.... (I just started butchering my slice program -- sorry I've not finished it yet.)
For those of you in discrete math, why might we prefer a graph to a tree (e.g.
- There is a multivariate version of this: suppose that $x(s,t)$ and
$y(s,t)$. Then, when we look for a partial derivative of $z$
with respect to either $s$ or $t$ (since $z$ is a function of both):
\[ \frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t} \]
and symmetrically for $\frac{\partial z}{\partial s}$.
And of course we can do this for functions of any number of independent variables (the "General version", p. 951).
- We can gain some insight into implicit functions from our
univariate days using the ideas in this section. For example, an
implicit function, e.g.
\[
x^2+y^2=1
\]
which we recognize as the unit circle, can be reformulated as
\[
F(x,y)=x^2+y^2-1=0
\]
That is, we can think of it as a particular level curve (=0) of the
multivariate function \(F(x,y)=x^2+y^2-1\).
Applying the chain rule, we get \[ \frac{dy}{dx}=-\frac{F_x}{F_y} \]
Let's have a look at page 953, and think about the conditions under which we can do this (the Implicit Function Theorem). Why do they make sense, given our discussion of differentiability from last time?
Think about the linearization....
- Let's try a few simpler
exercises, and then move along to some of the more complicated
ones:
- #2, p. 954
- #7
- #15
- #37
- #38
- #49
We shouldn't be a stick in the mud about the orientation of our axes: why $x$ and $y$, and not some other pair of directions which are mutually perpendicular? Perhaps we are interested in the slope of the surface along some direction other than $x$ or $y$: hence the idea behind directional derivatives. At a given point at which a function is differentiable, one natural choice for two directions might be the direction in which the function is increasing fastest, and the direction perpendicular to this.
But any direction will do -- take a look at p. 957, Figure 3.
Our author's pretty good TEC animation of that figure.
This figure is partially what motivated me to create that "slicing" demo in Mathematica.
One of the biggest pieces of news is that we're going to be working with vectors -- e.g. $u$ -- and with vector-valued functions; and that will place another demand upon your visualization skills.
But gradients are just another kind of multivariate function: one where the domain is points in the plane, like many of our other functions, but the range is the set of vectors.
At each point in the plane there is a gradient vector pointing -- indicating what?
Are you comfortable with
- the two properties of a vector?
- unit vectors?
- dot products?
- norms?
- The directional derivative of $f$ at $(x_0,y_0)$ in the direction of
unit vector ${\bf{u}} = \langle a,b \rangle$ is
\[
D_{\bf{u}} f(x_0,y_0) =
\lim_{h \to 0}\frac{f(x_0+ha,y_0+hb)-f(x_0,y_0)}{h}
\]
if this limit exists.
- If $f$ is a function of two variables $x$ and $y$, then the gradient of
$f$ is the vector function $\nabla f$ defined by
\[
\nabla f(x,y) = \langle f_x(x,y), f_y(x,y) \rangle =
\frac{\partial f}{\partial x} \hat{i}
+
\frac{\partial f}{\partial y} \hat{j}
\]
- Theorem: \[
D_{\bf{u}} f(x,y) =
\nabla f(x,y) \cdot {\bf{u}}
\]
- More generally (in higher dimensions), we can write
\[
D_{\bf{u}} f({\bf{x}}_0) =
\lim_{h \to 0}\frac{f({\bf{x}}_0+h{\bf{u}} )-f({\bf{x}}_0)}{h}
\]
where $\bf{u}$ is a unit vector. (Isn't that a beautiful analogy with the
limit definition of the univariate derivative -- the most important definition in calculus?)
- Theorem: Suppose $f$ is a differentiable multivariate function. The maximum value of the directional derivative $D_{\bf{u}}f({\bf{x}})$ is $|\nabla f({\bf{x}})|$ and it occurs when ${\bf{u}}$ has the same direction as the gradient vector $\nabla f({\bf{x}})$. A Mathematica example.
- #1, p. 967
- #5
- #7