No one email and no one review session
can cover every possible test topic, but
below I give a summary that touches on many
topics and includes several practice problems.

For a day-by-day list of topics covered,
see the attached retrospective schedule.
When you see a topic you want to study
in more detail, look at the board photos
for that day.

I've also attached my tables of rules for
differentials and antidifferentials.

(1) DIFFERENTIALS AND DERIVATIVES.
The differential approximates the
change in the output of a function
given starting input(s) and small change(s)
of those input(s).

(1a) For a function f(x) with one input,
with starting input x=a and small input change dx,
f(a+dx) is approximately f(x)+df where
df can be computed by first using the differential rules
and then plugging in the given values for x and dx.
Alternatively, you can compute the derivative f'(x)
using derivative rules (or your calculator) and
then use df=f'(x)*dx with the given x and dx plugged in.

For practice you could compute the approximate
value of f(1.1) where f(x)=ln(x)*exp(1-x^3)
using x=1 and dx=0.1.

(1b) For a function f(x,y) with two inputs,
with starting inputs a and b and small input changes dx and dy,
f(a+dx,b+dy) is approximately f(a,b)+df where
df is the total differential which can be computed
just like any other differential by first using the differential
rules and then plugging in the given values for x, y, dx, and dy.
Alternatively, you can compute the partial derivatives
f_x(x,y) and f_y(x,y) using derivative rules
(or your calculator) and then use df=f_x(x,y)*dx+f_y(x,y)*dy
with the given x, y, dx and dy plugged in.
A partial derivative of a function with respect to an
input variable is computed like a derivative but with
all the other input variables held constant.

For practice you could compute the approximate
value of f(3.1,3.8) where f(x,y)=(x+y)/(x^2+y^2)
using x=3, dx=0.1, y=4, and dy=-0.2.

(1c) If dx and/or dy are not given because formula for a
tangent line or plane is desired, then compute
f(a)+df or f(a,b)+df as before except plug in x-a for dx
and, in the two-input case, y-b for dy.

(1d) Be careful not confuse relative and absolute change.
For example, y=4 and dy=-0.2 can be described both
as y decreasing from 4 by 0.2 and
as y decreasing from 4 by 5%.
(because dy/y=0.2/4=0.05=5%).

(2) APPLICATIONS OF DIFFERENTIALS AND DERIVATIVES.

(2a) For a function g of time t, the derivative g'=dg/dt
can be interpreted as an (instanteous, absolute) rate of change.

For practice, if f(x,y)=(x+y)/(x^2+y^2), x=3, y=4,
x is currently changing at +0.5 per minute, and
f is currently changing at -0.3 per minute,
then what is the current rate of change of y?
(Find df, divide by dt to find df/dt, then
solve for dy/dt.)

(2b) The relative rate of change of f is f'/f=(1/f)*(df/dt).
When the relative rate of change of f is a constant r,
then f grows or decays exponentially: f(t)=f(0)*exp(r*t).

(2c) To maximize of minimize f(x), look for critical points,
i.e., points x=c where f'(c)=0. Usually the max or min
you seek is f(c). When in doubt, use the second-derivative
test: if f''(c)>0, f(c) is a local min; if f''(c)<0,
f(c) is a local max.

For practice, if production cost is C=100+10x
and demand is given by 2*p+(x/1000)=60, then find
the maximum possible profit P=p*x-C.
(Start by expressing P as a function
of just p or of just x.)

(2d) To maximize of minimize f(x,y), look for critical points,
i.e., points (x,y)=(a,b) where f_x(a,b)=f_y(a,b)=0.
Usually the max or min you seek is f(a,b).
There is also a second-partial-derivative test:
if f_xx(a,b)*f_yy(a,b)<f_xy(a,b)^2, then (a,b) is a saddle point
and f(a,b) is not a local min nor a local max;
if f_xx(a,b)*f_yy(a,b)>f_xy(a,b)^2 and f_xx(a,b)>0,
then f(a,b) is a local min;
if f_xx(a,b)*f_yy(a,b)>f_xy(a,b)^2 and f_xx(a,b)<0,
then f(a,b) is a local max;

For practice, find and classify the four critical points of
f(x,y)=x+3*y+(1/x)+(2/y).

(3) INTEGRALS AND APPLICATIONS

(3a) An integral is a total change. For example,
the integral of d(x^3/3) from x=1 to x=2 is
2^3/3 - 1^3/3 = 7/3. The catch is that you might
be asked to integrate x^2*dx from x=1 to x=2,
in which case it is up to you (possibly with
your calculator's help) to recognize that x^2*dx
equals d(x^3/3). The table of antidifferential
rules may help.

(3b) u-substitutions are very powerful technique
for finding antidifferentials. An important special
case: if f(x)*dx=d(g(x)) and A, B are constants,
then f(Ax+B)*dx=d(g(Ax+B)/A) because, setting u=Ax+B,
we obtain du=A*dx and then f(u)*dx=f(u)*du/A=d(g(u)/A).

For practice, how much time it take to travel
at speed dx/dt=50+x/10 from x=0 to x=8?
(Get a formula for dt and integrate it;
a u-substitution will help with the integral.)

For very similar practice, if your savings S of
initially 180 (thousands of dollars) is earning 4%
interest per year but you're withdrawing money
at a constant rate of 10 (thousands of dollars per year),
then how long until your savings run out?
(Solve dS/dt=0.04S-10 for dt and integrate
dt from S=180 to S=0.)

(3c) An integral can also be interpreted as an area.
In particular, the area between curves
y=g(x) and y=f(x) and between x=a and x=b
is the integral of (g(x)-f(x))*dx from x=a to x=b,
provided y=g(x) is above y=f(x).

For practice, find the producer surplus, that is,
the area above the supply curve
p=S(x)=5+(x^2)/2000 and below the horizontal line
p=bar(p) where bar(p) is equilibrium price
S(bar(x)) where bar(x) is the equilbrium quantity
x>0 satisfying S(x)=D(x)=30-(x^2)/1000.
(The intended x-range is from 0 to bar(x).)

(3d) The average value of a function f(x)
from x=a to x=b is the integral of
f(x)*dx/(b-a) from x=a to x=b.

For practice, what is the average value
of 5*x*exp(-x/2) from x=0 to x=10?
(Integration by parts will help!)

(3e) Using integrals of integrals, we can
compute the average of a two-input function
f(x,y) over a rectangle of inputs [a,b]x[c,d]
as well as the volume between two surfaces
z=g(x,y) and z=f(x,y), again with x and y
restricted to a rectangle. In the first case,
we integrate f(x,y)*dx/(b-a)*dy/(d-c), first
with respect to one variable, say, from y=c to y=d,
treating x and dx as constants so as to produce
some formula h(x)*dx, which we then integrate
from x=a to a=b. The volume case is the same,
except that we integrate (g(x,y)-f(x,y))*dx*dy,
provided g(x,y)>f(x,y).

For practice, what is the average value
of (x+y)^7 on [0,1]x[-1,1]?
(u-substitutions will help!)
