The Quick Calculus Tutorial

This text is a quick introduction into Calculus ideas and techniques. Itis designed to help you if you take the Calculus based course Physics 211at the same time with Calculus I, but you do not yet have any Calculusbackground. But we will assume that you have some algebra/trigonometryskills, of course you can refresh while you read. This guide is accompaniedby six short You-tube lectures going through the six sections (links see atthe 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 firstyourself! Stop and think and calculate all Examples yourselves while you goalong. 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 howthe 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/Calculuscourses. You can also read the sections out of order.

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Lecture 1. What is the derivative?

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

Three real numbers x(t), y(t), z(t) tell usan object’s position at time t and thus tell ushow the position is changing with time. Cal-culus begins with the question to describehow fast the position is changing in timeand to describe the change in detail. Thisleads to the notion of velocity. For examplex(1) = 2, y(1) = 2, z(1) = −1 means thatat time t = 1 we reach the object from theorigin by going two steps in the direction ofthe 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. Oftenthis may be a vertical motion like when you throw a ball vertically up and tellthe 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) formotions that are not vertical like a car moving along a straight line.

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

z(t) = 64 − 16t2, 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 ⋅ 22 = 0 and the ball hits theground.

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

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In Calculus books the notation is often different. But this is just a matterof taste. (When we say y = 64− 16x2 for 0 ≤ x ≤ 2 and tell that x is time andy is the height above ground, we have given the same information, variablenames don’t matter but try to remember at all times what the variable namesstand for.)

Don’t confuse the picture of the orbit ofa moving object with the position graph.To the left you see the graph of the posi-tion graph of the object falling according toz(t) = 64 − 16t2 during the interval [0,2].

Now the graph of the position functionsloping down means that the object getsfaster and faster on its way down, the speedis not constant. If the speed would be con-stant the object would fall equal distancesin equal time intervals. If we set up a tableshowing the height z at times t we get

t 0.0 0.5 1.0 1.5 2.0z 64 60 48 28 0

and see that while the object only falls 4 feet within the first half second itfalls 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 andaccelerates within the first 10 sec-onds. The following picture showsthe graph of x(t). How fast is thecar at t = 5? Imagine you take yourfeet from the pedal for a short mo-ment ∆t at t = 5 and see how faryou go. Note that if we stop accel-erating we move with the constantvelocity, 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) = 6t2

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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 averagevelocity in this interval, which is for ∆t = 0.1:

vav =∆x

∆t= x(5.1) − x(5)

5.1 − 5.

Let us calculate this average velocity (note that you use the binomial formula)

6 ⋅ 5.12 − 6 ⋅ 52

0.1= 6 ⋅ (5 + 0.1)2 − 52

0.1= 6 ⋅ 5

2 + 2 ⋅ 5 ⋅ 0.1 + 0.01 − 52

0.1= 6 ⋅ (10 + 0.1)

If we replace 0.1 by ∆t we get 6 ⋅ (10+∆t). Of course, when ∆t gets smallerand 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 casewe have the notation:

v(t) = dxdt

= lim∆t→0

∆x

∆t

You just have to be careful with the notation for ∆x because it does notspecify for which t we consider the average velocity.

∆x

∆t= 6 ⋅ (t +∆t)2 − 6t2

∆t= 6 ⋅ t

2 + 2t ⋅∆t + (∆t)2 − t2∆t

= 12t +∆t

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

v(t) = dxdt

= 12t

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

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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) = t3. Calculate ∆x∆t for t = 1 and find from this

dxdt∣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 sothat x(t) is the coordinate of the center of the train, measured from someorigin O. You measure how far to the right you are from the center of thetrain and call it y(t). Then your coordinate at time time is x(t)+y(t), whichis the distance to the right from the origin O.

What is your velocity with respect to ground: It obviously is the velocityof the train added to your velocity with respect to the train. In Calculuslanguage this means:

d

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

dt+ dydt.

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

d

dt(kx(t)) = kdx

dt.

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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 turnsaround.II. Power Rule:

d

dttn = ntn−1,

where n can be any real number. In particular ddtk = d

dt(k ⋅ 1) = ddtt

0 = 0 fora constant k. (Get a feeling for the rule: First take the exponent as a factordown, then subtract one from the exponent.)Examples: The following examples are calculated using linearity and thepower rule.

1.

d

dt(6t4 + 8t2 + 12t + 4) = 6 ⋅ 4t3 + 8 + 8 ⋅ 3t2 + 12 ⋅ 1 + 0 = 24t3 + 24t2 + 12

So, for example the slope of x(t) = 6t4 + 8t3 + 12t + 4 (do you have anidea 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 12 + t− 1

3) = 1

2t12−1 + (−1

3)t− 1

3−1 = 1

2t−

12 − 1

3t−

43

= 1

2⋅ 1√

t− 1

3⋅ 1

3√t4

3. Suppose you throw up a ball such that the height at time t is given by

z(t) = 4 − (t − 2)2 = 4 − (t2 − 4t + 4) = 4t − t2 = t(4 − t).

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

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We calculate:dz

dt= dtdt

(4t − t2) = 4 − 2t,

and sodz

dt∣t=1

= 4 − 2 = 2.

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

There a few more important derivatives.

d

dtet = et ;

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 slopesof sin(t) with the values of cos(t):

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III. Product Rule and Quotient Rule: This tells how a product of twofunctions 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 sidesof a rectangle so that f(t) ⋅ g(t) is the area. Note that for ∆t small ∆f ⋅∆gis 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(t2 sin t) = ( d

dtt2) ⋅ sin t + t2 (⋅ d

dtsin t) = 2t sin t + t2 cos t

2.d

dt(et sin t) = et sin t + et cos t = et(sin t + cos t)

Note that x(t) ⋅ x(t)−1 = x(t) ⋅ 1x(t) = 1 and the derivative of a constant is

0. So an application of the product rule gives

0 = dt(x(t)x(t)−1) = dx

dt⋅ x(t)−1 + x(t) ⋅ d

dtx(t)−1,

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and thus by solving for ddtx(t)−1 we get

d

dtx(t)−1 = −

dxdt

x(t)2

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

d

dt

x

y= x

′ ⋅ y − x ⋅ y′y2

,

where we have abbreviated x′ = dxdt and y′ = dy

dt .

Example:

d

dttan t = d

dt

sin t

cos t=

d sin tdt cos t − sin td cos t

dt

cos2 t

= cos t cos t − sin t(− sin t)cos2 t

= cos2 t + sin2 t

cos2 t

= 1

cos2 t= sec2 t

What is the derivative of e−t?Let’s look at the graphs of et and e−t:

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

d

dte−t = −e−t

(Note that this means in particular: ddte

−t∣t=−1

= −e−(−1) = −e1 = − ddte

t∣t=1)

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In general for each constant k:

d

dtekt = kekt,

for example ddte

3t = 3e3t.

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 ekt? 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) = eu.

Then the chain rule is:

df

dt= dxdt

= dxdu⋅ dudt.

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 = ddue

u = eu and dudt = d

dt (kt) =k and the result follows.

Examples:

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1. Let x = ecos t. Then x(u) = eu and u(t) = cos t. Thus ddue

u = eu andddx cos t = − sin t

dx

dt= ecos 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

⋅ ddt

sin t = − sin(sin t) ⋅ cos(t)

3. Suppose a particle moves along the z-axis and its position at time t isgiven by x(t) = e−t2+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) =eu with u(t) = −t2 + 4t. Note that for t = 1 we have u = −12 + 4 ⋅ 1 = 3.Then dx

du = eu and dudt = −2t + 4. Thus

dx

dt∣t=1

= dx

du∣u=3⋅ dudt

∣t=1

= e3 ⋅ 2 = 2e3

Exercises:

1. Find the derivatives dxdt :

� x = t5 +√t + 1√

t(Hint:

√t = t1/2)

� x = (t2 + 6)e3t

� x = esin 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 isgiven by y(t) = t2 cos(et). Find the velocity at time t = ln(π2 ). (Hint:

You have to calculate ddt (t2 ⋅ cos(et))∣

t=ln(π2). The calculation of the

derivative requires the product rule, the chain rule, the power rule andthe derivatives of et and cos t).

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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 velocityalone won’t be able to tell us the position. But since the velocity is thederivative of the position, the position has to be an anti-derivative of thevelocity. Note that such an anti-derivative can only be determined up to aconstant. We write

∫ v(t)dt

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

∫ v(t)dt = ∫ tdt = 1

2t2 +C.

Here C stands for an arbitrary constant. Note that

d

dt(1

2t2 +C) = 1

2

d

dtt2 + d

dtC = 1

2⋅ 2t = t

Let’s say that for a motion along the x-axis we have given v(t) = 2t2 and aninitial position x(0) = 2. We first find the anti-derivative of v(t):

∫ 2t2dt = 2

3t3 +C,

so x(t) = 23t

3 + C for some constant C. This can be calculated from theposition at time 0:

x(0) = 2

3⋅ 0 +C = 2

and so C = 2 and the position is

x(t) = 2

3t3 + 2.

But how did we know the anti-derivative of 2t2. Well we can check that

d

dt(2

3t3 +C) 2

3⋅ 3t2 = t2.

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In general taking anti-derivatives re-quires to turn around the rules we learned inSection 2. This is easy for some of the rulesbut hard for others. It also requires you toget used to some conceptual understanding.If you take a derivative you actually performan operation with input a function, let’s sayx, 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 sayv(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 lateron).

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.

∫ tndt = 1

n + 1tn+1 +C for n ≠ −1

(Get a feeling for the rule: First add 1 to the exponent, then take the reciprocalof 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 theexceptional rule:

∫dt

t= ln ∣t∣ +C

Examples:

∫√tdt = ∫ t1/2dt = 2

3t3/2 +C

∫1

t+ t5dt = ln ∣t∣ + 1

6t6 +C

Of course we have for constants k ≠ 0:

∫ ektdt = 1

kekt +C

∫ sin(kt)dt = −1

kcos(kt) +C ;∫ cos(kt)dt = 1

ksin(kt) +C

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Note that you will never need the above rules for k = 0 because e0 = 1,sin(0) = 0 and cos(0) = 1. You will need the special case of the ReversePower 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! Weknow that the position function is an anti-derivative for a specif valueof the constant:

∫ 8 cos(2t)dt = 8∫ cos(2t)dt = 8 ⋅ 12

sin(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 dxdt = e−t and x(1) = 1. Find x(t)! We first find the

anti-derivative (k = −1):

∫ e−tdt = −e−t +C.

Then x(1) = −e−1 +C = 1 shows C = 1 + 1e and we get

x(t) = 1 + 1

e− e−t

Check your answer: ddt

(1 + 1e − 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:

� ∫ t2 + 2t1/2 + t−1dt

� ∫ 2 cos t sin tdt (Hint: ddt sin2(t) = 2 cos t sin(t) from the chain rule.)

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2. Consider the motion given by the velocity v(t) = dydt = v0 + at for a, v0

constants and let y(0) = y0. What is the position function y(t)? Whatis the meaning of a and v0? Explain why d

dt(v(t)) = ddt

(dydt) = a. The

notation for the last function is d2ydt2 and is called the second order

derivative.

3. Find d2xdt2 for x(t) = t4 + 3t3 + 6t2 + 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 hoursalong the x-axis. How far did we get within those two hours? Of course, 100miles.

The velocity graph in this is a horizontalline at height 50. Note that we can interpretthe distance covered in this case as the areaof the rectangle bounded by the lines x =0, x = 50, t = 0 and t = 2. The units are infact correct because 50 mi/hr⋅2 hr = 50miles.The number though can be interpreted as anarea. Of course this works in general: If wemove with constant velocity k the positionat time t will be kt, where we assume thatthe 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) = atalong the x-axis. We know that if x(0) = 0. By taking the anti-derivative weknow that:

x(t) = 1

2at2.

How is this number related with the graph of the velocity function? Is thenumber x(t) again the area under the graph of x(t) > 0 and above the t-axisbetween 0 and t. In this case we the area is that of a triangle with baselength t and height at so is 1

2at2. By subtracting areas of rectangles we see

that in this case the distance covered within the interval [t1, t2] is 12a(t22− t21).

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This is the motivation for the notation

∫t2

t1v(t)dt = ∫ v(t)dt∣

t2

t1

= x(t)∣t2t1 = x(t2) − x(t1)

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

∫b

af(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 calledindefinite integral), which is a class of functions. Net area means that weactually subtract the area above the graph and below the t-axis. This makessense because when the velocity is negative we have to subtract from ourposition. This is indicated in the picture on the below.

We have that

∫t

0v(t)dt = A1 −A2 = x(t) − x(0)

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

∫b

af(t)dt = F (t2) − F (t1)

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

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A related statement is that

d

dt ∫t

af(u)du = f(t)

Note that ∫t

a f(t)dt is a specific anti-derivative of f(t), namely the one withf(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 backthe function. Note that taking the integral of a derivative of a function givesback the function up to a constant. In fact, when we take the derivativewe lose the initial condition and the integral won’t know this information.In the picture below you get the visual representation of the FundamentalTheorem assertions stated above.

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

∫c

af(t)dt = ∫

b

af(t)dt + ∫

c

bf(t)dt

for any three numbers a, b, c. Also:

∫a

bf(t)dt = −∫

b

af(t)dt

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

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

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For example5

∑k=1k = 1 + 2 + 3 + 4 + 5 = 15

or6

∑k=2k3 = 23 + 33 + 43 + 53 + 63 = 8 + 27 + 64 + 125 + 216 = 440

I believe you can guess how the symbol is used in general. This integralsymbol is supposed to remind you of the fact that integrals are approximatedby sums in the following way: Suppose we divide an interval [a, b] into nsmall equal length sub-intervals of length ∆t and pick values f(ti) in the i-thinterval. Then

∫b

af(t)dt ≈

n

∑k=1

f(tk)∆t

Note that on the right hand side you are summing about areas of n rectangleswith base lengths ∆t and heights f(tk).Examples:

1.

∫2

0t2dt = 1

3t3∣

2

0

= 1

3⋅ 23 − 1

3⋅ 03 = 8

3.

Note that this number is the area bounded by the t-axis, the graph oft2 and the line t = 2.

2.

∫1

0t3+ sinπtdt = 1

4t4 − 1

πcosπt∣

1

0

= 1

4⋅1− 1

π⋅(−1)−(1

4⋅0− 1

π⋅0) = 1

4+ 2

π

3.

∫1

0t5 + t1/2 + et − 4dt = t6

6+ 2

3t3/2 + et − 4t∣

1

0

= e − 25

6

4. What is ∑5k=0 k2?

5

∑k=0k2 = 02 + 12 + 22 + 32 + 42 + 52 = 1 + 4 + 9 + 16 + 25 = 55

18

Exercises:

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

� ∫π

0 sin(t)dt� ∫

1

0 1 + t + t2 + t3dt� ∫

1

0 e−tdt

� ∫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 dydt = cos(πt)+1

and initial position y(0) = 0. Determine y(1) by calculating a definiteintegral. Interpret the resulting number as net area for the graph of afunction.

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 = d2xdt2

is the acceleration. Here we assume that F is a force acting on an objectmoving along the x-axis.

Suppose that a force F (t) = cos t is the force and m = 1. How do wefind the position function x(t) of the particle if we know that x(0) = 1 andx′(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.

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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 wecalculate C = 2 and thus

x(t) = 2t − cos t + 2.

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

dx

dt(t) = 2 + sin t ,

d2x

dt2= cos t,

andx(0) = − cos 0 + 2 = 1 , x′(0) = 2 + sin 0 = 2.

The following coordinate system shows the graphs of x(t), dxdt and d2xdt2 . Do

you understand the relations between those graphs in terms of slope and netarea?

Exercise: Suppose an object with mass 2 kg moves along the y-axis. Aforce of F (t) = 2 − e−2t Newtons acts on the object. We have y(0) = −1 anddydt∣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 time0 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∆t2 +∆t3 − 1

∆t= 3 + 3∆t +∆t2

When ∆t gets smaller and smaller, ∆t2 will be even smaller. Thus

dx

dt∣t=1

= lim∆t→0

∆x

∆t= 3

Exercises 2.1.:

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� x = t5 +√t + 1√

t= t5 + t1/2 + t−1/2, and so by linearity and power rule:

dx

dt= 5t4 + 1

2t−1/2 − 1

2t−3/2

� x = (t2 + 6)e3t, and by using the product rule, linearity, the power ruleand the derivative of the exponential function (including a constant inthe exponent):

dx

dt= ( d

dt(t2 + 6)) ⋅ e3t + (t2 + 6) ⋅ d

dte3t = 2te3t + (t2 + 6) ⋅ 3e3t

= e3t(3t2 + 2t + 18)

� x = esin t cos t, and so by using product rule and chain rule we get:

dx

dt= ( d

dtesin t) ⋅ cos t + esin t ⋅ d

dtcos t

= (esin t cos t) cos t + esin t(− sin t)= esin t(cos2 t − sin t)

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

d

dtcos(cos(cos t)) = d

ducosu∣

u=cos(cos t)⋅ ddt

cos(cos t)

= − sin(cos(cos t)) d

ducosu∣

u=cos t

⋅ ddt

cos 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 ddt cot t = d

dtcos tsin t using the quotient rule:

d

dt

cos t

sin t= − sin t ⋅ sin t − cos t ⋅ cos t

sin2 t= − 1

sin2 t= − csc2 t

Exercise 2.2: y(t) = t2 cos(et). We want to calculate

dy

dt∣t=lnπ/2

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We use product rule and chain rule to get

dy

dt= 2t cos(et) + t2(− sin(et))et = 2t cos(et) − t2et sin(et)

and by evaluating at lnπ/2 and using the identity elnu = u for all u we get

dy

dt∣t=lnπ/2

= 2 lnπ

2cos(π

2) − (ln π

2)2π

2sin(π

2) = −π

2(ln π

2)2 ≈ −0.320

Exercise 3.1:

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

∫ t2 + 2t1/2 + t−1dt = 1

3t3 + 2 ⋅ 2

3t3/2 + ln ∣t∣ +C = t

3

3+ 4

3t2/3 + ln ∣t∣ +C

� Since from the chain rule ddt sin2 t = 2 sin t cos t we know that the deriva-

tives of sin2 t is our integrand and thus

∫ 2 cos t sin tdt = sin2 t +C

Exercise 3.2:Since v(t) = dy

dt we know that y(t) is an anti-derivative of v(t):

∫ vdt = ∫ v0 + atdt = v0t +1

2at2 +C

The constant can be determined by evaluating at t = 0:

v0 ⋅ 0 +1

2a ⋅ 0 +C = C = y0,

so the constant is y0. So we have:

y(t) = y0 + v0t +1

2at2.

The velocity is not constant but changing uniformly according to dvdt = a,

where a is the constant acceleration. Also v(0) = v0 is the initial velocity.This is the uniformly accelerated motion.

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Exercise 3.3: From x = t4 + 3t3 + 6t2 + 5 we get

dx

dt= 4t3 + 9t2 + 12t

andd2x

dt2= d

dt(dxdt

) = d

dt(4t3 + 9t2 + 12t) = 12t2 + 18t + 12

Exercise 4.1:

� ∫π

0 sin tdt = − cos t∣π0 = − cosπ + cos 0 = −(−1) + 1 = 2

� ∫1

0 1+ t+ t2 + t3dt = t + t2

2 + t3

3 + t4

4 ∣1

0= 1+ 1

2 + 13 + 1

4 = 12+6+4+312 = 25

12 ≈ 2.083

� ∫1

0 e−tdt = −e−t∣10 = −e−1 − (−e0) = 1 − 1

e ≈ 0.632

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

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

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

y(τ) = ∫τ

0v(t)dt

is the anti-derivative with y(0) = 0. This is because ∫0

0 g(t)dt = 0 for everyfunction g(t). Thus

y(1) = ∫1

0cos(πt) + 1dt

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

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 =md2ydt2 we get

d2y

dt2= 1 − 1

2e−2t.

Recall that we have also given y(0) = −1 and dydt∣t=0 = 0. By taking the

anti-derivative we get

∫d2y

dt2dt = t + 1

4e−2t +C

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and using the Fundamental theorem and dydt∣t=0 = 0 we get 1

4 +C = 0 and thus

C = −14 . Thus

dy

dt= t + 1

4e−2t − 1

4Again 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 −18 +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+ 02

2− 1

8e−2⋅0 = −7

8− 1

8= −1

and by taking derivatives

dy

dt= −1

4+ t + 1

4e−2t ,

d2y

dt2= 1 − 1

2e−2t

The object is moving because even though the velocity at t = 0 vanishes theacceleration does not. This is like dropping a ball from rest, it starts withvelocity 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 givenforce?

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

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