- 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