- You have a homework due Friday.
- Return of your homework: graded #1, 5, 63
- #1: you have to know that longitude increases to the
west to deduce that \(\frac{\partial T}{\partial x} > 0\):
Also that, as hot air blows in from the west, the temperature will be increasing (i.e. \(\frac{\partial T}{\partial t}\ge 0\))
- #5: I modelled the function given. You don't have to, but
it's encouraged.
You need to justify answers. Just writing down exactly what's in the back of the book is insufficient.
You might draw a slice in each direction, for example, or simply describe a slice.
- #63: pretty straightforward!
- #1: you have to know that longitude increases to the
west to deduce that \(\frac{\partial T}{\partial x} > 0\):
The point is to illustrate the univariate functions we get when we take a slice of a function in a given direction.
(PS: I hope that my constant introduction of Mathematica code will encourage everyone to get a copy for themselves!)
- 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).
By the way, if the function is twice differentiable (and we can fit a "tangent bowl", rather than just a tangent plane), then the form of
- 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 $\frac{\partial z}{\partial t}$.
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, and
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