The Quick Calculus Tutorial

This text is a quick introduction into Calculus ideas and techniques. It is designed to help you if you take the Calculus based course Physics 211 at the same time with Calculus I, but you do not yet have any Calculus background. But we will assume that you have some algebra/trigonometry skills, of course you can refresh while you read. This guide is accompanied by six short You-tube lectures going through the six sections (links see at the end of each section). Take a look at the videos and read the text below, it is not always quite the same things you will see. Do all the Exercises first yourself! Stop and think and calculate all Examples yourselves while you go along. Stop and rethink what you have learned all the time.

There is also available the Calculus Concept Companion . These are notes from a course that introduces concepts with a different time-line (hav- ing an eye on what’s going on in your physic’s class) in comparison to how the topics are covered in your section of Calculus I. Fell free to read the com- panion and use its resources while you go through your Physics/Calculus courses. You can also read the sections out of order.

Lecture 1. What is the derivative?

Calculus is the Mathematics of Change. A typical framework to introduce Calculus begins with the following question: How do we describe the motion of a moving object?

Three real numbers x(t), y(t), z(t) tell us an object’s position at time t and thus tell us how the position is changing with time. Cal- culus begins with the question to describe how fast the position is changing in time and to describe the change in detail. This leads to the notion of velocity. For example x(1) = 2, y(1) = 2, z(1) = −1 means that at time t = 1 we reach the object from the origin by going two steps in the direction of the positive x-axis, then two steps in the di- rection of the positive y-axis and finally one step down in the direction of the negative z-axis.

In this tutorial we will only consider things moving along a single axis. Often this may be a vertical motion like when you throw a ball vertically up and tell the height z(t) above ground at each point in time in some interval of time.

There is a corresponding position functions z(t) (or sometimes x(t), y(t) for motions that are not vertical like a car moving along a straight line.

For example imagine dropping a ball vertically down from a building 50 feet above the ground. Suppose we place the origin of the z-axis at ground level. Then

z(t) = 64 − 16t^{2}, 0≤ t ≤ 2

will be the height of the ball above ground for all times t in the time interval
[0, 2]. When t = 2 then we have z(2) = 64 − 16 ⋅ 2^{2} = 0 and the ball hits the
ground.

This is an example of a function. We call t the independent variable and z the dependent variable in this case.

In Calculus books the notation is often different. But this is just a matter
of taste. (When we say y = 64 − 16x^{2} for 0≤ x ≤ 2 and tell that x is time and
y is the height above ground, we have given the same information, variable
names don’t matter but try to remember at all times what the variable names

stand for.)

Don’t confuse the picture of the orbit of a moving object with the position graph.

To the left you see the graph of the posi-
tion graph of the object falling according to
z(t) = 64 − 16t^{2} during the interval [0, 2].

Now the graph of the position function sloping down means that the object gets faster and faster on its way down, the speed is not constant. If the speed would be con- stant the object would fall equal distances in equal time intervals. If we set up a table showing the height z at times t we get t 0.0 0.5 1.0 1.5 2.0

z 64 60 48 28 0

and see that while the object only falls 4 feet within the first half second it falls 28 feet within the fourth half second.

Let us consider another exam- ple. A car is moving along the pos- itive x-axis. It starts from rest and accelerates within the first 10 sec- onds. The following picture shows the graph of x(t). How fast is the car at t= 5? Imagine you take your feet from the pedal for a short mo- ment ∆t at t = 5 and see how far you go. Note that if we stop accel- erating we move with the constant velocity, which is the instantaneous speed that we had at that moment. But how can we determine this velocity?

Suppose the position function is

x(t) = 6t^{2}

If ∆t is small the velocity will not change considerably in the interval 5≤ t ≤ 5+ ∆t. So we can approximate the instantaneous velocity by the average velocity in this interval, which is for ∆t= 0.1:

v_{av} = ∆x

∆t = x(5.1) − x(5) 5.1− 5 .

Let us calculate this average velocity (note that you use the binomial formula)
6⋅ 5.1^{2}− 6 ⋅ 5^{2}

0.1 = 6 ⋅ (5+ 0.1)^{2}− 5^{2}

0.1 = 6 ⋅ 5^{2}+ 2 ⋅ 5 ⋅ 0.1 + 0.01 − 5^{2}

0.1 = 6 ⋅ (10 + 0.1) If we replace 0.1 by ∆t we get 6⋅ (10 + ∆t). Of course, when ∆t gets smaller and smaller these number will get closer and closer to 6⋅ 10 = 60. This is the (instantaneous) velocity at t= 5. We write:

v(5) = dx dt∣

t=5= lim_{∆t→0}∆x

∆t

Now we can do this calculation of the velocity for any time t. In this case we have the notation:

v(t) = dx

dt = lim_{∆t→0}∆x

∆t

You just have to be careful with the notation for ∆x because it does not specify for which t we consider the average velocity.

∆x

∆t = 6⋅ (t + ∆t)^{2}− 6t^{2}

∆t = 6 ⋅ t^{2}+ 2t ⋅ ∆t + (∆t)^{2}− t^{2}

∆t = 12t + ∆t

When ∆t get very small this quantity will be very close to 12t. Thus:

v(t) = dx dt = 12t

is the derivative of x(t). Note that v(5) = 12 ⋅ 5 = 60, which we calculated
before. v(t) is the instantaneous velocity at time t. Now Calculus will be
a collection of algebraic rules how to calculate derivatives without going
through the above tedious procedure of calculating a limit. But don’t forget
the interpretation of the numbers x^{′}(t) for a given function x(t): x^{′}(t) is the
slope of the graph of x(t) for time t. In other words: If x^{′}(t) > 0 the values of
x(t) are increasing for time t: the velocity is positive. If x^{′}(t) = 0 the graph
has a horizontal tangent, it usually changes from sloping up to down or vice

versa: the velocity is zero at this point. If x^{′}(t) < 0 then the values of x(t)
are decreasing at time time: the velocity is negative.

Exercise: Consider x(t) = t^{3}. Calculate ^{∆x}_{∆t} for t = 1 and find from this

dx

dt∣_{t=1}= x^{′}(1).

Watch the video Lecture 1∶ What is the Derivative?

Lecture 2. How to take derivatives?

In this section we will learn the basic rules of taking derivatives.

I. Linearity: Imagine you are walking in a train moving along the x-axis so that x(t) is the coordinate of the center of the train, measured from some origin O. You measure how far to the right you are from the center of the train and call it y(t). Then your coordinate at time time is x(t)+y(t), which is the distance to the right from the origin O.

What is your velocity with respect to ground: It obviously is the velocity of the train added to your velocity with respect to the train. In Calculus language this means:

d

dt(x(t) + y(t)) = dx dt +dy

dt.

Obviously if you replace x(t) by kx(t) for a constant number k then also your velocity will be multiplied by k. Thus

d

dt(kx(t)) = kdx dt.

This should be clear from the following picture, noting that k⋅ x(t) − k ⋅ x(t + ∆t) = k∆x.

Note that if k< 0 you move in the other direction and the velocity also turns around.

II. Power Rule:

d

dtt^{n}= nt^{n−1},

where n can be any real number. In particular _{dt}^{d}k = _{dt}^{d}(k ⋅ 1) = _{dt}^{d}t^{0} = 0 for
a constant k. (Get a feeling for the rule: First take the exponent as a factor
down, then subtract one from the exponent.)

Examples: The following examples are calculated using linearity and the power rule.

1.

d

dt(6t^{4}+ 8t^{2}+ 12t + 4) = 6 ⋅ 4t^{3}+ 8 + 8 ⋅ 3t^{2}+ 12 ⋅ 1 + 0 = 24t^{3}+ 24t^{2}+ 12
So, for example the slope of x(t) = 6t^{4}+ 8t^{3}+ 12t + 4 (do you have an
idea how the graph looks like?) for t= 1 is

dx dt∣

t=1= 24 + 24 + 12 = 60 2.

d dt(√

t+ t^{−1/3}) = d

dt(t^{1}^{2} + t^{−}^{1}^{3}) = 1

2t^{1}^{2}^{−1}+ (−1

3)t^{−}^{1}^{3}^{−1}= 1

2t^{−}^{1}^{2} −1
3t^{−}^{4}^{3}

= 1 2⋅ 1

√t −1 3⋅ 1

√3

t^{4}

3. Suppose you throw up a ball such that the height at time t is given by
z(t) = 4 − (t − 2)^{2}= 4 − (t^{2}− 4t + 4) = 4t − t^{2} = t(4 − t).

What is the velocity after 1 second? (or what is the slope of the graph of the position function shown below).

We calculate:

dz dt = dt

dt(4t − t^{2}) = 4 − 2t,
and so

dz dt∣

t=1= 4 − 2 = 2.

Does this answer look reasonable? Go from the point(1, 3) one unit to
the right and 2 units up. This would be the point you would end up if
gravity could be cut off (which of course it can’t) at t= 1. Note that
z^{′}(2) = 0. This is the high point of the ball.

There a few more important derivatives.

d

dte^{t}= e^{t} ; d

dtln(t) = 1 t d

dtsin(t) = cos(t) ; d

dtcos(t) = − sin(t)

Look at the graphs of the sine and cosine function and compare the slopes of sin(t) with the values of cos(t):

III. Product Rule and Quotient Rule: This tells how a product of two functions f(t) and g(t) is changing when we change the independent variable.

df

dt (f(t)g(t)) = df

dt ⋅ g(t) + f(t)dg dt The idea is to see the necessary quantities

∆(fg) = (∆f) ⋅ g(t) + f(t) ⋅ ∆g + ∆f ⋅ ∆g

in the picture below. Here f(t) and g(t) denote the lengths of the two sides of a rectangle so that f(t) ⋅ g(t) is the area. Note that for ∆t small ∆f ⋅ ∆g is very small so that:

∆(fg)

∆t = ∆f

∆t ⋅ g(t) + f(t) ⋅∆g

∆t By going over to the limit we get the product rule.

Examples:

1. d

dt(t^{2}sin t) = (d

dtt^{2}) ⋅ sin t + t^{2}(⋅d

dtsin t) = 2t sin t + t^{2}cos t

2. d

dt(e^{t}sin t) = e^{t}sin t+ e^{t}cos t= e^{t}(sin t + cos t)

Note that x(t) ⋅ x(t)^{−1} = x(t) ⋅x(t)^{1} = 1 and the derivative of a constant is
0. So an application of the product rule gives

0=d

t (x(t)x(t)^{−1}) = dx

dt ⋅ x(t)^{−1}+ x(t) ⋅ d

dtx(t)^{−1},

and thus by solving for _{dt}^{d}x(t)^{−1} we get
d

dtx(t)^{−1}= −x(t)^{dx}^{dt} ^{2}

Combining this with the product rule a simple calculation gives the quotient

rule d

dt x

y =x^{′}⋅ y − x ⋅ y^{′}
y^{2} ,
where we have abbreviated x^{′}= ^{dx}_{dt} and y^{′}=^{dy}_{dt}.
Example:

d

dttan t= d dt

sin t

cos t = ^{d sin t}^{dt} cos t− sin t^{d cos t}_{dt}
cos^{2}t

=cos t cos t− sin t(− sin t)

cos^{2}t = cos^{2}t+ sin^{2}t
cos^{2}t

= 1

cos^{2}t = sec^{2}t
What is the derivative of e^{−t}?

Let’s look at the graphs of e^{t} and e^{−t}:

Consider for example at the slopes of e^{−t} at t= −1. This is the negative of
the slope of e^{t} at t= 1. In fact in general we have that

d

dte^{−t}= −e^{−t}

(Note that this means in particular: _{dt}^{d}e^{−t}∣_{t=−1}= −e^{−(−1)}= −e^{1} = − dt^{d}e^{t}∣_{t=1})

In general for each constant k:

d

dte^{kt}= ke^{kt},
for example _{dt}^{d}e^{3t}= 3e^{3t}.

The above rule is actually a special case of the

IV. Chain Rule: Consider a composition of two functions f(t) = x(u(t)) (check your Precalculus book if you forgot what this means!)

What has this to do with e^{kt}? Well recall that is is also written exp(kt),
and we can consider this to be the composition of the function which multi-
plies by k and the exponential function, so u(t) = kt and x(u) = e^{u}.

Then the chain rule is:

df dt = dx

dt = dx du⋅du

dt.

The first term on the right hand side is more precisely dx

du∣

u(t),

so you have to take the derivative and then substitute u(t) for the argument.

Let’s apply this to the example above: Then ^{dx}_{du} = du^{d}e^{u}= e^{u} and ^{du}_{dt} = dt^{d} (kt) =
k and the result follows.

Examples:

1. Let x = e^{cos t}. Then x(u) = e^{u} and u(t) = cos t. Thus du^{d}e^{u} = e^{u} and

d

dxcos t= − sin t

dx

dt = e^{cos t}⋅ (− sin t)

2. Let x= cos(sin t). Then x(u) = cos(u) and u(t) = sin t. We get d

dtcos(sin(t)) = d

ducos(u)∣

sin t⋅ d

dtsin t= − sin(sin t) ⋅ cos(t) 3. Suppose a particle moves along the z-axis and its position at time t is

given by x(t) = e^{−t}^{2}^{+4t}. Find the velocity at time t= 1. This is
dx

dt∣

t=1

We note that we have to find the derivative of the composition of x(u) =
e^{u} with u(t) = −t^{2}+ 4t. Note that for t = 1 we have u = −1^{2}+ 4 ⋅ 1 = 3.

Then ^{dx}_{du} = e^{u} and ^{du}_{dt} = −2t + 4. Thus
dx

dt∣

t=1= dx du∣

u=3⋅ du dt∣

t=1= e^{3}⋅ 2 = 2e^{3}

Exercises:

1. Find the derivatives ^{dx}_{dt}:

x= t^{5}+√

t+^{√}^{1}_{t} (Hint: √

t= t^{1/2})

x= (t^{2}+ 6)e^{3t}

x= e^{sin t}⋅ cos(t)

x= cos(cos(cos t))

x= ln cot t

2. Suppose an object moves along the y-axis and its position at time t is
given by y(t) = t^{2}cos(e^{t}). Find the velocity at time t = ln(^{π}2). (Hint:

You have to calculate _{dt}^{d} (t^{2}⋅ cos(e^{t}))∣_{t=ln(}^{π}

2). The calculation of the
derivative requires the product rule, the chain rule, the power rule and
the derivatives of e^{t} and cos t).

Watch the video Lecture 2∶ How to calculate Derivatives?

Lecture 3. How to calculate anti-derivatives?

Given the function of velocities for a motion, how do we find the position?

Well certainly we need to know additionally an initial position, the velocity alone won’t be able to tell us the position. But since the velocity is the derivative of the position, the position has to be an anti-derivative of the velocity. Note that such an anti-derivative can only be determined up to a constant. We write

∫ v(t)dt

to denote the anti-derivative, which is a class of functions. Now what does this mean? Let’s say v(t) = t. Then

∫ v(t)dt = ∫ tdt =1

2t^{2}+ C.

Here C stands for an arbitrary constant. Note that d

dt(1

2t^{2}+ C) = 1
2

d
dtt^{2}+ d

dtC= 1

2⋅ 2t = t

Let’s say that for a motion along the x-axis we have given v(t) = 2t^{2} and an
initial position x(0) = 2. We first find the anti-derivative of v(t):

∫ 2t^{2}dt= 2
3t^{3}+ C,

so x(t) = ^{2}_{3}t^{3} + C for some constant C. This can be calculated from the
position at time 0:

x(0) = 2

3⋅ 0 + C = 2 and so C = 2 and the position is

x(t) = 2
3t^{3}+ 2.

But how did we know the anti-derivative of 2t^{2}. Well we can check that
d

dt(2

3t^{3}+ C)2

3⋅ 3t^{2}= t^{2}.

In general taking anti-derivatives re- quires to turn around the rules we learned in Section 2. This is easy for some of the rules but hard for others. It also requires you to get used to some conceptual understanding.

If you take a derivative you actually perform
an operation with input a function, let’s say
x, and output a function, then denoted ^{dx}_{dt}.
The anti-derivative is an operation somehow
turning this around. The anti-derivative has as input a function, let’s say
v(t), and as output a class of functions (a function plus all possible con-
stants), then denoted ∫ v(t)dt. (We will explain the weird notation later
on).

Here are the basic rules: First linearity holds for integrals:

∫ (f(t) + g(t))dt = ∫ f(t)dt + ∫ g(t)dt ; ∫ kf(t)dt = k ∫ f(t)dt, where k denotes a constant.

∫ t^{n}dt= 1

n+ 1t^{n+1}+ C for n ≠ −1

(Get a feeling for the rule: First add 1 to the exponent, then take the reciprocal of the resulting number down as a factor.) Compare with the Power Rule.

Note that the right hand side is not defined for n = −1. This is the exceptional rule:

∫ dt

t = ln ∣t∣ + C Examples:

∫ √

tdt= ∫ t^{1/2}dt= 2

3t^{3/2}+ C

∫ 1

t + t^{5}dt= ln ∣t∣ + 1
6t^{6}+ C
Of course we have for constants k≠ 0:

∫ e^{kt}dt= 1

ke^{kt}+ C

∫ sin(kt)dt = −1

kcos(kt) + C ; ∫ cos(kt)dt = 1

k sin(kt) + C

Note that you will never need the above rules for k = 0 because e^{0} = 1,
sin(0) = 0 and cos(0) = 1. You will need the special case of the Reverse
Power Rule:

∫ kdt = kt + C Examples:

1. Suppose that the velocity of a moving object is given by v(t) = 8 cos(2t) and the initial position is x(0) = 0. Find the position function! We know that the position function is an anti-derivative for a specif value of the constant:

∫ 8 cos(2t)dt = 8 ∫ cos(2t)dt = 8 ⋅1

2sin(2t) + C = 4 sin(2t) + C Since 0= x(0) = 4 sin(2 ⋅ 0) + C shows C = 0 we have

x(t) = 4 sin(2t)

2. Suppose that ^{dx}_{dt} = e^{−t} and x(1) = 1. Find x(t)! We first find the
anti-derivative (k= −1):

∫ e^{−t}dt= −e^{−t}+ C.

Then x(1) = −e^{−1}+ C = 1 shows C = 1 + ^{1}_{e} and we get
x(t) = 1 +1

e− e^{−t}

Check your answer: _{dt}^{d} (1 +^{1}e − e^{−t}) = −(−e^{−t}) = e^{−t} and x(1) = 1 +^{1}e −
e^{−1}= 1 + e^{−1}− e^{−1}= 1.

Exercises:

1. Find the anti-derivatives:

∫ t^{2}+ 2t^{1/2}+ t^{−1}dt

∫ 2costsintdt (Hint: ^{dt}^{d} sin^{2}(t) = 2 cos t sin(t) from the chain rule.)

2. Consider the motion given by the velocity v(t) = ^{dy}dt = v^{0}+ at for a, v^{0}
constants and let y(0) = y0. What is the position function y(t)? What
is the meaning of a and v0? Explain why _{dt}^{d}(v(t)) = _{dt}^{d} (^{dy}_{dt}) = a. The
notation for the last function is ^{d}_{dt}^{2}2^{y} and is called the second order
derivative.

3. Find ^{d}_{dt}^{2}^{x}2 for x(t) = t^{4}+ 3t^{3}+ 6t^{2}+ 5.

Watch the video Lecture 3∶ How to calculate Anti − derivatives?

Lecture 4. The Fundamental Theorem of Integral Calculus Suppose we are moving with constant velocity 50 miles/hr for two hours along the x-axis. How far did we get within those two hours? Of course, 100 miles.

The velocity graph in this is a horizontal line at height 50. Note that we can interpret the distance covered in this case as the area of the rectangle bounded by the lines x = 0, x = 50, t = 0 and t = 2. The units are in fact correct because 50 mi/hr⋅2 hr = 50miles.

The number though can be interpreted as an area. Of course this works in general: If we move with constant velocity k the position at time t will be kt, where we assume that the position at time 0 is 0. Again kt is the area of the rectangle with lengths t on the x-axis and height the velocity.

Next assume that you are accelerating with constant acceleration a: v(t) = at along the x-axis. We know that if x(0) = 0. By taking the anti-derivative we know that:

x(t) = 1
2at^{2}.

How is this number related with the graph of the velocity function? Is the
number x(t) again the area under the graph of x(t) > 0 and above the t-axis
between 0 and t. In this case we the area is that of a triangle with base
length t and height at so is ^{1}_{2}at^{2}. By subtracting areas of rectangles we see
that in this case the distance covered within the interval[t^{1}, t2] is ^{1}_{2}a(t^{2}2−t^{2}1).

This is the motivation for the notation

∫_{t}^{t}^{2}

1

v(t)dt = ∫ v(t)dt∣^{t}^{2}

t1

= x(t)∣^{t}t^{2}1 = x(t^{2}) − x(t^{1})

This observation is true in general. The change in position is the net area between the graph of the velocity function and the t-axis. The integral

∫_{a}^{b}f(t)dt

is called the definite integral of the function f(t) over the interval [a, b].

This is a number and not to be confused with the anti-derivative (also called indefinite integral), which is a class of functions. Net area means that we actually subtract the area above the graph and below the t-axis. This makes sense because when the velocity is negative we have to subtract from our position. This is indicated in the picture on the below.

We have that

∫_{0}^{t}v(t)dt = A^{1}− A^{2}= x(t) − x(0)

is the geometric interpretation of the anti-derivative. Now recall that the velocity is the derivative of the position and the position function is the anti-derivative. The general statement

∫_{a}^{b}f(t)dt = F(t^{2}) − F(t^{1})

for a reasonably nice function f(t) and its anti-derivative F(t) is often called the Fundamental Theorem of Calculus. It tells the way how we calculate for example position from a given velocity.

A related statement is that d dt ∫

t

a f(u)du = f(t)

Note that ∫^{a}^{t}f(t)dt is a specific anti-derivative of f(t), namely the one with
f(a) = 0. But of course the derivative will not depend on which anti-
derivative we pick. If we take the derivative of the integral we get back
the function. Note that taking the integral of a derivative of a function gives
back the function up to a constant. In fact, when we take the derivative
we lose the initial condition and the integral won’t know this information.

In the picture below you get the visual representation of the Fundamental Theorem assertions stated above.

In calculating definite integrals we use some obvious rules, which are immediate from the net area interpretation or the Fundamental Theorem.

For instance

∫_{a}^{c}f(t)dt = ∫_{a}^{b}f(t)dt + ∫_{b}^{c}f(t)dt
for any three numbers a, b, c. Also:

∫_{b}^{a}f(t)dt = − ∫_{a}^{b}f(t)dt

Finally recall that evaluating a definite integral is usually done by finding the anti-derivative first and then taking the difference of the values of the anti-derivative at the two limits of the integral.

Remark: You may wonder about the strange symbol ∫ for integrals. This symbol is similar to the so called Sigma symbol Σ used to abbreviate sums.

For example

∑5 k=1

k= 1 + 2 + 3 + 4 + 5 = 15

or 6

∑k=2

k^{3} = 2^{3}+ 3^{3}+ 4^{3}+ 5^{3}+ 6^{3}= 8 + 27 + 64 + 125 + 216 = 440

I believe you can guess how the symbol is used in general. This integral
symbol is supposed to remind you of the fact that integrals are approximated
by sums in the following way: Suppose we divide an interval [a, b] into n
small equal length sub-intervals of length ∆t and pick values f(t^{i}) in the i-th
interval. Then

∫_{a}^{b}f(t)dt ≈ ∑^{n}

k=1

f(tk)∆t

Note that on the right hand side you are summing about areas of n rectangles
with base lengths ∆t and heights f(t^{k}).

Examples:

1.

∫_{0}^{2}t^{2}dt= 1
3t^{3}∣^{2}

0= 1

3⋅ 2^{3}−1

3⋅ 0^{3}= 8
3.

Note that this number is the area bounded by the t-axis, the graph of
t^{2} and the line t= 2.

2.

∫_{0}^{1}t^{3}+sin πtdt = 1
4t^{4}−1

πcos πt∣^{1}

0= 1 4⋅1−1

π⋅(−1)−(1 4⋅0−1

π⋅0) = 1 4+2

π 3.

∫_{0}^{1}t^{5}+ t^{1/2}+ e^{t}− 4dt = t^{6}
6 +2

3t^{3/2}+ e^{t}− 4t∣^{1}

0= e −25
6
4. What is ∑^{5}^{k=0}k^{2}?

5 k=0∑

k^{2} = 0^{2}+ 1^{2}+ 2^{2}+ 3^{2}+ 4^{2}+ 5^{2} = 1 + 4 + 9 + 16 + 25 = 55

Exercises:

1. Calculate the following definite integrals and interpret in terms of net areas:

∫^{0}^{π}sin(t)dt

∫^{0}^{1}1+ t + t^{2}+ t^{3}dt

∫^{0}^{1}e^{−t}dt

∫^{−1}^{1} f(t)dt where f(t) is any odd functions (which means f(t) =
f(−t)).

2. Suppose an object moves along the y-axis with velocity ^{dy}_{dt} = cos(πt)+1
and initial position y(0) = 0. Determine y(1) by calculating a definite
integral. Interpret the resulting number as net area for the graph of a
function.

Watch the video Lecture 4∶ The Fundamental Theorem of Integral Calculus Lecture 5. Working with the tool box

Newton’s Law is usually stated as F = ma where m is the mass and a = ^{d}_{dt}^{2}^{x}^{2}
is the acceleration. Here we assume that F is a force acting on an object
moving along the x-axis.

Suppose that a force F(t) = cos t is the force and m = 1. How do we
find the position function x(t) of the particle if we know that x(0) = 1 and
x^{′}(0) = 2. If we let v(t) denote the velocity along the x-axis then

dv

dt = cos t and so by taking the anti-derivative

v(t) = sin t + C
Since v(0) = x^{′}(0) = 2 we get C = 2 and so

v(t) = 2 + sin t.

Then since x^{′}(t) = v(t) we get x(t) by taking the anti-derivative again:

2t− cos t + C,

for another constant C. Then from x(0) = 2 ⋅ 0 − cos 0 + C = −1 + C + 1 we calculate C = 2 and thus

x(t) = 2t − cos t + 2.

It is easy to check whether we have found the correct solution. We find by differentiating

dx

dt(t) = 2 + sin t , d^{2}x

dt^{2} = cos t,
and

x(0) = − cos 0 + 2 = 1 , x^{′}(0) = 2 + sin 0 = 2.

The following coordinate system shows the graphs of x(t),^{dx}dt and ^{d}_{dt}^{2}2^{x}. Do
you understand the relations between those graphs in terms of slope and net
area?

Exercise: Suppose an object with mass 2 kg moves along the y-axis. A
force of F(t) = 2 − e^{−2t} Newtons acts on the object. We have y(0) = −1 and

dy

dt∣_{t=0} = 0. Find the position function of the object and graph the function.

Why is the object moving at all, taking into account that it’s velocity at time 0 vanishes? Discuss the motion for large t.

Watch the video Lecture 5∶ Working with the tool box Lecture 6. Some Solutions for the Exercises Exercise 1.: We calculate

∆x

∆t = x(1 + ∆t) − x(1)

∆t = (1+ ∆t)^{3}− 1

∆t = 1+ 3∆t + 3∆t^{2}+ ∆t^{3}− 1

∆t

= 3 + 3∆t + ∆t^{2}

When ∆t gets smaller and smaller, ∆t^{2} will be even smaller. Thus
dx

dt∣

t=1= lim_{∆t→0}∆x

∆t = 3 Exercises 2.1.:

x= t^{5}+√

t+^{√}^{1}_{t} = t^{5}+ t^{1/2}+ t^{−1/2}, and so by linearity and power rule:

dx

dt = 5t^{4}+1

2t^{−1/2}−1
2t^{−3/2}

x= (t^{2}+ 6)e^{3t}, and by using the product rule, linearity, the power rule
and the derivative of the exponential function (including a constant in
the exponent):

dx dt = (d

dt(t^{2}+ 6)) ⋅ e^{3t}+ (t^{2}+ 6) ⋅ d

dte^{3t}= 2te^{3t}+ (t^{2}+ 6) ⋅ 3e^{3t}

= e^{3t}(3t^{2}+ 2t + 18)

x= e^{sin t}cos t, and so by using product rule and chain rule we get:

dx dt = (d

dte^{sin t}) ⋅ cos t + e^{sin t}⋅ d
dtcos t

= (e^{sin t}cos t) cos t + e^{sin t}(− sin t)

= e^{sin t}(cos^{2}t− sin t)

x= cos(cos(cos(t)). We have to apply the chain rule twice:

d

dtcos(cos(cos t)) = d

ducos u∣

u=cos(cos t)⋅ d

dtcos(cos t)

= − sin(cos(cos t)) d

ducos u∣

u=cos t⋅ d dtcos t

= − sin(cos(cos t)) ⋅ (− sin(cos t)) ⋅ (− sin t)

= − sin(cos(cos t)) ⋅ sin(cos t) ⋅ cos t

x= ln(cot t). We calculate dt^{d} cot t= dt^{d}
cos t

sin t using the quotient rule:

d dt

cos t

sin t = −sin t⋅ sin t − cos t ⋅ cos t

sin^{2}t = − 1

sin^{2}t = − csc^{2}t
Exercise 2.2: y(t) = t^{2}cos(e^{t}). We want to calculate

dy dt∣

t=ln π/2

We use product rule and chain rule to get dy

dt = 2t cos(e^{t}) + t^{2}(− sin(e^{t}))e^{t}= 2t cos(e^{t}) − t^{2}e^{t}sin(e^{t})

and by evaluating at ln π/2 and using the identity e^{ln u}= u for all u we get
dy

dt∣

t=ln π/2= 2 lnπ 2cos(π

2) − (lnπ
2)^{2}π

2sin(π

2) = −π 2(lnπ

2)^{2} ≈ −0.320
Exercise 3.1:

By linearity and the power rule for anti-derivatives (the power rule backwards) we get

∫ t^{2}+ 2t^{1/2}+ t^{−1}dt= 1

3t^{3}+ 2 ⋅2

3t^{3/2}+ ln ∣t∣ + C =t^{3}
3 +4

3t^{2/3}+ ln ∣t∣ + C

Since from the chain rule _{dt}^{d} sin^{2}t= 2 sin t cos t we know that the deriva-
tives of sin^{2}t is our integrand and thus

∫ 2 cos t sin tdt = sin^{2}t+ C
Exercise 3.2:

Since v(t) = ^{dy}dt we know that y(t) is an anti-derivative of v(t):

∫ vdt = ∫ v^{0}+ atdt = v^{0}t+1

2at^{2}+ C
The constant can be determined by evaluating at t= 0:

v0⋅ 0 +1

2a⋅ 0 + C = C = y^{0},
so the constant is y_{0}. So we have:

y(t) = y^{0}+ v^{0}t+1
2at^{2}.

The velocity is not constant but changing uniformly according to ^{dv}_{dt} = a,
where a is the constant acceleration. Also v(0) = v^{0} is the initial velocity.

This is the uniformly accelerated motion.

Exercise 3.3: From x= t^{4}+ 3t^{3}+ 6t^{2}+ 5 we get
dx

dt = 4t^{3}+ 9t^{2}+ 12t

and d^{2}x

dt^{2} = d
dt(dx

dt) = d

dt(4t^{3}+ 9t^{2}+ 12t) = 12t^{2}+ 18t + 12
Exercise 4.1:

∫^{0}^{π}sin tdt= − cos t∣^{π}0 = − cos π + cos 0 = −(−1) + 1 = 2

∫^{0}^{1}1+ t + t^{2}+ t^{3}dt= t +^{t}_{2}^{2} +^{t}_{3}^{3} +^{t}_{4}^{4}∣^{1}

0 = 1 +^{1}_{2}+^{1}_{3}+^{1}_{4} = ^{12+6+4+3}_{12} =^{25}_{12} ≈ 2.083

∫^{0}^{1}e^{−t}dt= −e^{−t}∣^{1}0 = −e^{−1}− (−e^{0}) = 1 −^{1}e ≈ 0.632

The definite integral is zero. This is because the integral is the net area, and counts area below the t-axis negative and the area above the t-axis positive. The two contributions will cancel by the symmetry (see graph below)

In the first three cases the graph of the function is above the t-axis over the corresponding interval. So the definite integral is the blue area. In the last case the contributions on the negative axis cancel with those on the positive axis.

Exercise 4.2: Let v(t) = cos(πt) + 1. For a general number τ the definite integral

y(τ) = ∫_{0}^{τ}v(t)dt

is the anti-derivative with y(0) = 0. This is because ∫^{0}^{0}g(t)dt = 0 for every
function g(t). Thus

y(1) = ∫_{0}^{1}cos(πt) + 1dt

is the definite integral calculating y(1). Using the Fundamental Theorem we get

y(1) = −1

πsin(πt) + t∣^{1}

0= 1

πsin(π) + 1 − (sin(0) + 0) = 1.

Compare this with the shaded area for the graph below:

Exercise 5.1: From Newton’s law m= 2 and F = m^{d}_{dt}^{2}^{y}^{2} we get
d^{2}y

dt^{2} = 1 −1
2e^{−2t}.

Recall that we have also given y(0) = −1 and ^{dy}dt∣_{t=0} = 0. By taking the
anti-derivative we get

∫ d^{2}y

dt^{2}dt= t +1

4e^{−2t}+ C

and using the Fundamental theorem and ^{dy}_{dt}∣_{t=0}= 0 we get ^{1}4+ C = 0 and thus
C= −^{1}4. Thus

dy

dt = t + 1

4e^{−2t}−1
4
Again we take the anti-derivative and get

∫ dy

dtdt= ∫ t + 1

4e^{−2t}−1
4 = t^{2}

2 −1

8e^{−2t}− t
4+ C
From y(0) = −1 we get −^{1}_{8}+ C = −1 and thus C = −^{7}_{8}. This gives

y(t) = −7 8− t

4+t^{2}
2 −1

8e^{−2t}

You may want to check whether this y(t) satisfies the conditions:

y(0) = −7 8−0

4+0^{2}
2 −1

8e^{−2⋅0}= −7
8−1

8 = −1 and by taking derivatives

dy dt = −1

4+ t + 1

4e^{−2t} , d^{2}y

dt^{2} = 1 −1
2e^{−2t}

The object is moving because even though the velocity at t= 0 vanishes the acceleration does not. This is like dropping a ball from rest, it starts with velocity zero but the acceleration due to gravity makes the object move.

Below you the the graph of y(t). Can you relate the graph to the given force?

For large t we have F ≈ 2 is constant and the motion approaches a uni- formly accelerated motion with position graph a parabola (what is the for- mula for the parabola?)

Watch the video Lecture 6∶ Some Solutions for the Exercises