A brief Overview, Review and
Summary of the basic Calculus
needed for AP Physics C
Animation borrowed from: http://www.calculusinmotion.com/gsp4.html
Joe Stieve 2010
An Overview of Calculus
The Calculus involves the solution of Three types of mathematical problems.
I. The
slope problem.
Differential Calculus
II. The “un-slope” problem.
Integral Calculus
III. The area problem.
3
y
Here we have a parabolic curve.
The first problem of calculus is to
assign a quantity at each point to
describe the rate of change of the
function at that point.
For a graph, that can be interpreted
as “find the slope”.
If Charlie Brown starts a trip up the
hill clearly he will find the slope
becoming steeper as he climbs.
We can think of the slope at each
point as if it were the slope of a
line that is aligned parallel to the
direction of his motion.
x
y
You can also think of the instantaneous
slope as the direction of travel of a car
or motorcycle as it drives along the
curve as indicated by the headlight
beams.
x
But how do we get a mathematical value
for the slope, say at the point (x,y) for a
specific curve? Let’s look at a relatively
simple example.
y
(x2,y2)
y = x2
The idea is that we write a
general expression for the slope
of a straight line drawn from the
given point let us call it (x,y) to
any other point (x2,y2) on the line.
For our specific function this becomes:
Slope = 2x +Δx
(x,y)
x
So for
y  x2
= 2x  x
we found a general expression for the slope
There is no reason the second x must be to
the right of the first x.
y
(x2,y2)
As the (x2, y2) moves toward the
point (x,y) the slope of the red line
connecting the two points
becomes less.
As the (x2, y2) moves toward the
point (x,y) the slope of the blue
line connecting the two points
becomes greater. (note that ∆x is
negative.)
In either case mathematically observe
what happens for our function if the
second x gets closer to the first, that is if
Δx gets small.
The slope gets close to being simply 2x.
We say the slope at x is 2x
Slope = 2x + Δx
(x,y)
Slope = 2x + Δx
(x2,y2)
x
We have derived from the original function, x2, a new function 2x. We say that the second
function is the derivative of the first and we call the process “taking the derivative”.
The slope of the two lines we have draw get y
closer to being the same as we approach from
above or below x.
(x2,y2)
The line drawn through the point (x,y) with
the slope equal to the derivative at that
point is said to be tangent to the curve at
that point and represents the instantaneous
slope at that point.
Slope = 2x
(x,y)
(x2,y2)
x
y
To put numbers on our given example
we can choose the point (1.5,2.25) and
the slope comes out 3 at that point.
6
5
4
For the next slide I have borrowed an
animated gif to illustrate how the slope of
chords, or secant lines approach the slope
of a tangent line as the two points come
together.
Slope = 2x = 3
3
(1.5, 2.25)
2
1
0
0
1
2
x
Notice this is NOT the same function y = x2 that we used earlier
but rather y = 0.75 x2, but the principle is the same.
Here the tangent at the point x = 0.5 is shown in green, and the secant line
draw to the yellow moving dot is shown in purple.
Source of animation http://www.ima.umn.edu/~arnold/graphics.html

Just a word about notation before we go on to generalize what we have learned.
Frequently, instead of the y notation we substitute the function notation
and write f(x) instead of y, that is f(x) = x2 instead of y =x2 .
Because the derivative represents the instantaneous slope we frequently
use the notation: dy
dx
Or to remind us of its ancestry we might just put a prime on the old symbol and
write f’(x)

So for the example we have been working with we can write y = f(x)=x2 and
dy
= f’(x) = 2x .
dx
Now that the idea and notation of the derived function, or derivative, is known,
we find that we “luck out” and that the rules for finding the derivatives of most
functions are relatively easy. As an example we have been using a monomial.
For the general monomial f(x) = axn the derivative is found by multiplying
the coefficient by the exponent and reducing the exponent by one. The
derivative is then f’(x) =naxn-1
What if we have a polynomial? It turns out that the derivative of a sum is
simply the sum of the derivatives.
If f (x)  4 x 3  2x 2  3x  2

we get
f '(x)  12x 2  4 x  3
Although I have not listed it as a rule ,you can see that the derivative
of a constant term is zero because it doesn’t change.
Caution to students: Though this rule for sums (and differences) works
for all derivatives it cannot be extended to things like products and
quotients. The derivative of a product/quotient is NOT the
product/quotient of the derivatives.

Some important applications of the derivative to kinematics.
Because velocity is the rate of change of position and acceleration is the rate of
change of velocity we can use the calculus notation to write:
r
r ds
v
where " s" is a generalized space coordinate. In a particular problem
dt
it might be x, y, z, r etc., and
r
(The vector notation is not absolutely necessary and can be
r dv
a=
eliminated for one dimensional equations.)
dt
If then the position of an object is given as x = 25t2 -16t +4, (where the units:
25 m/s2, -16 m/s and 4 m for the numbers have been left off for convenience)
we can find its velocity as a function of time by taking the derivative…
v = 50t -16, and the acceleration by taking the derivative again to get…
a = 50 m/s2. Note that in the case of a quadratic position function the
acceleration will always come out constant.
Other examples and applications will appear in the lectures for the C exam.
A note about notation. The derivative of a derivative is referred to as

2
dx
d
d
x

dt 
the second derivative and is often written as f" (x) or
for

dt 2
dt




Part II: the “unslope” problem
Now to address the second “problem” of calculus: The “unslope”, usually
referred to as the anti-derivative. The situations is that you are given the slope at
all points and you are to find the original function. Inspection of the following
graphs will illustrate the problem.
At any value of x the slopes for all
three curves shown, and indeed for
a whole family of curves, have the
same value. How do we distinguish
which curve?
y
Here are the three equations for the graphs and
their derivatives.
x

f (x) = x 2  2 ,
f '(x)  2x
f (x)  x 2
,
f '(x)  2x
f (x)  x 2  2 ,
f '(x)  2x
To find the function from its derivative you must be
given some more information.
First lets use the rules for the derivative in reverse for f’(x) = 2x. We now increase
the exponent by one and divide by the new exponent. Here is this rule for taking
the anti-derivative of a monomial.
ax n 1
n
If f '(x)  ax we have f (x) 
C
n 1
y
In the other notation we have :

dy
ax n 1
n
If
 ax we have y 
C
dx
n 1

x
We need to add the “C” because we
don’t know what constant might have
vanished when the derivative was taken.
The “hint” we need is just one point on the graph. Here
for example we found the equation is y = x2+C. If we are
told that y =2 when x = 0 we can substitute in and solve
for C= 2 and we have our answer: y = x2 +2.
Because, in the symbol dy/dx, the dy and dx are called differentials,
equations that contain them are called differential equations.
Near the earth’s surface the acceleration of gravity is about 9.8 m/s2. This
information could be written as the differential equation:
dv
 9.8 m/s 2
dt
Taking the anti-derivative gives us v = 9.8 m/s2 t + C

So for any free object near the earth’s surface we can calculate the
velocity at any time if we are given some initial information such as at
1s the object is going 20 m/s the we can do the arithmetic and find
the general equation for all times is v = 9.8 m/s2 t + 10.2 m/s
This difficulty of being stuck with a constant might seem like a major pitfall,
however it is a wonder in disguise.
With a simple algebraic equation we get one solution. With a
differential equation we get a family of related solutions. For this
reason many of the laws and principles of classical physics and
engineering are written as differential equations. These equations
then apply to the solution of an infinite family of related problems.
We need only insert the specific parameters. In this way we get
general laws. By putting in specific data we can “pick” our curve and
predict the future behavior as the path follows that curve or rule.
This technique will be applied many time over in AP C and we will
illustrate that as we move through the course.
Part III: The “Area” Problem
y
Here is our old friend the parabola f(x) = x2
Suppose, for some unknown reason, we want
to find the area bounded by the curve on top,
the x axis on the bottom and the vertical lines
at x = 1 and x = 2.
1
2
x
Now imagine a
in the region.
inscribed
Its height would be the value of the function where it
intersects the curve, f(x), and its width would be the difference
in the x coordinate at its left side and that at its right side, Δx.
Its area would be length times width or simply f(x)Δx.
Stretch your imagination one more time and image that Δx is on the
verge of being “infinitely” small. We represent such a quantity by a
differential symbol dx to distinguish it from the finite Δx
f(x) = x2
y
The impossibly minute area of this little
section can be written as dA and we have
dA = f(x)dx
Now all we have to do is add up the
infinite number of little rectangles that
it would take to cover the area we
described. That is…
1
2
The Area from x = 1 to 2 is equal to the
sum of the areas dA from x = 1 to 2.
x
Using some notation to “spell this out” we have…
Eventually, and officially the Sum began to be written as a distorted S.
S becomes
becomes
And we write:
becomes
f(x) = x2
y
This notation is all well and good but how in
the world do we evaluate such a thing?
We have to leave the details of the derivation
to Calculus but the great thing is we can do
this without knowing those details.
A “miracle” occurs.
1
2
x
It turns out that the solution of our problem
connects with the solution of the “unslope” or
anti-derivative problem.
Here the F(x) is simply any ant-derivative for the function and you
need not worry about the constant since it would subtract out.
In our example the anti-derivative of x2 is x3/3 + C. Dropping the constant
then our area comes out exactly (2)3/3 - (1)3/3 = 7/3
In general then for any function f(x) that has a primitive (anti-derivative) F(x) you
can find the area by using the expression:
There are some “tricks”. If b>a and the area or part of an area is below the xaxis that area or part of the area will come out negative. So for example the
area from x = 0 to x = 2 for f(x)=x -1 comes out zero.
y
1
+0.5
-0.5
-1
1
2
x
The graph makes the reason for a total of zero
area clear. In other words you need to be
concerned about signed areas.
Now that we know the integral symbol means sum we can do a brief review
of the line and surface integrals of a vector field.
For our purposes in this presentation I will simply use the gravitational field of
the earth. For the AP the ideas are more heavily used for the electric and
magnetic fields.
Consider the gravitational field near the earth’s surface.
The work done in moving an
object from a to b is the force
times the displacement and
could be represented by any
of the following expressions.
(The scalar or “dot” product notation is not required on the AP)
This is the line integral of the force vector taken along the line a to b.
Now that we know the integral symbol means sum we can do a
brief review of the line and surface integrals of a vector field.
In the more general case the notation will be a little different but
the result is the same. Here ds is a minute displacement vector
and at each point we are taking the product of the magnitude of
the force with the component of the displacement in that
direction, that is we are calculating mgdscosθ.
The work done in moving an
object from a to b is the force
times the displacement that
could be represented as:
If we start and end in the same spot since gravitation is a
conservative field we know the work will be zero. That is not
always the case, but the important thing for the future is the
notation.
ds
Here the circle on the
integral sign merely means
we have integrated around
a closed loop and it is
called reasonably enough
a closed integral.
Lets just “jump in” to the surface integral. In physics we most are
often interested to the flow of something through a surface.
For example the flow of water in a pipe.
A
v
If you multiply the velocity of an incompressible fluid by the cross
sectional area you get the volume flow rate. For example 3 m/s
times an area of 4 m3 gives a flow rate of 12 m3/s.
Now if the flow in general is not perpendicular to the area in
question such as water passing through the frame of a net you need
to take the component of the velocity that is perpendicular to the
area. This is most easily done and represented if you invent a
vector of magnitude equal to the area and drawn perpendicular to
the surface. This is called surface vector.
If the angle between and the net were say 30o then the
flow rate for an area of 0.50 m2 with a velocity of water
at 0.20 m/s would give a flow rate of (0.2m/s)(0.5m2)cos
30o =0.087 m3/s
For a complex surface with different velocities at different points
clearly the problem becomes very complicated and we don’t have to
do it. The notation is however very important.
The rate of flow becomes:
(Here the “S” beneath the integral just means to take the sum over the entire surface.)
Lastly we extend this notation to a closed surface (indicated by the
circle on the integral sign):
The convention is that the direction of dA is taken outward for a
closed surface, that is from the finite volume out into the infinite
volume of space.
Thus for an incompressible fluid flowing through a volume of space
the flow into the volume (-) will be equal to the flow out of the
volume (+) by the conservation of the incompressible fluid. In that
case we can state categorically that the total surface integral is zero.
As would be indicated by a net flow of zero.
Closed line and surface integrals will be revisited when we study
Electricity and Magnetism, but don’t be concerned. The notation and
the associated concepts are what will be important not the calculus
solution of these integrals.
When we get there it will be “Look Ma, no calculus.”