Calculus I Blog

The relationship between mathematics and art dates way back.  Art has been around for what seems like forever, and if you think about it, so has mathematics. So it only makes sense that ancient artists as well as more modern day artists would use math in their works. 

In the ancient Egyptians’ artwork, math was necessary in order to make pyramids possible. However, math is used in all forms of art, including architecture, paintings and sculptures as well. It is important for artists to understand certain details of math in order to make their final masterpiece ‘work.’ If a sculptor were to begin making a piece without understanding proportions, it would be likely that the sculpture may not balance, and it definitely would not be visually pleasing.

In painting, proportions are also important, and as the audience we tend to find things to look ‘strange’ if they are not depicted proportionately. Take, for example, a painting like “Madonna of the Long Neck,” by Parmagianino. Because the painter did not understand proportions, the painting looks distorted and visually unpleasing.

Besides only proportions, mathematics can be used in various other ways in art. For example, symmetry is often found in paintings. Symmetry refers to “the correspondence in size, shape, and relative position of elements within the picture plane.” A work of art that is symmetrical appears balanced and neat, while one that does not contain symmetry can look heavier on one side than the other. Oftentimes, a work is perfectly symmetrical, making it more like a pattern than a picture. However, a painting can also be somewhat symmetrical, meaning that it is almost completely balanced although it does not show a mirror image on either side.

Abstract art is a type of art requiring many math skills to make. Many of these works, such as optical illusions, require geometry as well as numerical patterns. Things we look at on a daily basis include math that we would never even realize.

Look at a simple equation for a ball moving at the velocity: v(t)=6t-2

From this we can find the derivative, which is acceleration and a general equation for position.

To find the derivative, we can use the power rule where you bring down the exponent and multiply it by the constant.

Thus, the derivative of the equation for velocity is v'(t)=6, which is also acceleration.

To take the integral of the velocity function, you can apply the same rules of taking derivatives and work backwards to find the integral. Therefore, we can divide 6 by 2 and -2 by 1 to find the equation V(t)=3t^2-2t+C, where C is a constant that is unknown. V(t)=s(t)

From this information, we can plug in any number and find how fast or slow the ball is moving and fow fast or slow it is accelerating at a given time. If we were given the position of the ball at a certain time, then we would be able to figure out the constant. From there we would be able to find the position of the ball at any given time.

Let’s do just that. At 3 seconds, the position is 18. Just solve for C

s(3)=3(3)^2-2(3)+C=18, and after we bring everything, but C to the other side, we find that C=6

So s(t)=3t^2-2t+6

From this we can find all of the local maximums and local minimums by setting the velocity function (derivative of s(t)) equal to zero. This is because the derivative will give the slope of the tangent line to any point on the graph of s(t). Whenever the slope is 0, the tangent line is horizontal and is a local minimum or maximum. From the left and right sides of where the slope is 0 we can find whether the slope is increasing or decreasing on either side of the 0. It will increase when v(t) is positive and it will decrease when v(t) is negative.

Concavity can also be calculated by taking a(t) (the second derivative of s(t)). Any zeroes are inflection points if concavity changes. To the left and the right of the zero, check if it is positive or negative. If it is positive, then it is concave up. If it is negative it is concave down.

This problem has a local minimum at x=1/3 and no local maximum. It has no inflection points and is concave up.

It sure is interesting how much one little equation can tell you.

32) A poster is to have an area of 180 square inches with 1 inch margins at the bottom and sides and a 2 inch margin at the top. What dimensions will give the largest printed area?

Let x=horizontal width and y=vertical height

Let A= printed area

The two equations are:

xy= 180   and   (x-2)(y-3)=A

We want the equation for A to be in the terms of just one variable so we can optimize it.

Therefore we need to be able to substitute an equation involving only x in place of the y variable.

xy=180 divide both sides by x

y=180/x

Now substitute that in for the y in the other equation.

(x-2)(180/x -3)=A(x)

Simplify A(x)

A(x)=180 -3x -360/x +6

Now find the derivative of A(x)

A'(x)=-3 + 360/x^2

Now set equal to 0 to find critical points

A'(x)=-3 + 360/x^2=0 add 3 to both sides

360/x^2=3 multiply both sides by x^2

360=3x^2 divide both sides by 3

120=x^2 take square root of both sides

x=10.95

Put x back into the equation xy=180 to find y

(10.95)y=180 divide both sides by 10.95

y=16.44

The Millennium Propblems is a set of problems presented by The Clay Mathematics Institute of Cambridge, Massachusetts at a meeting in Paris, France.  There are seven problems in total, each of which is an important mathematical concept that no one has been able to solve in the past.  A fund of seven million dollars was created for the contest, with each problem holding a prize of one million dollars.

The first problem is the Birch and Swinnerton-Dyer Conjecture and deals with elliptical curves.  The second problem is the The Hodge conjecture problem and deals with projective algebraic varieties.  The third problem is the Navier-Stokes Equations problem and deals with the motion of liquids and gases.  The fourth problem is the P vs NP problem and deals with how computers solve problems.  The fifth problem is the Poincaré Conjecture and deals with spheres.  The sixth problem is the Riemann Hypothesis and deals with the Riemann zeta function.  The seventh problem is the Yang-Mills Theory and deals with the Maxwell theory of electromagnetism.

To this date, only the Poincaré Conjecture has been solved, the proof for which was submitted in 2003 and accepted in 2006.

http://www.claymath.org/millennium/

The pigeonhole principle, also known as Dirichlet’s box principle states that, given two natural numbers n and m with n > m, if n items are put into m pigeonholes, then at least one pigeonhole must contain more than one item. Another way of stating this would be that m holes can hold at most m objects with one object to a hole; adding another object will force one to reuse one of the holes, provided that m is finite.

The pigeonhole principle is an example of a counting argument, which can be applied to many formal problems, including ones involving infinite sets that cannot be put into one-to-one correspondence.

An easy example of the pigeonhole principle involves the situation when there are five people who want to play softball (n = 5 objects), but there are only four softball teams (m = 4 holes). This would not be a problem except that each of the five refuses to play on a team with any of the other four. To prove that there is no way for all five people to play softball, the pigeonhole principle says that it is impossible to divide five people among four teams without putting two of the people on the same team. Since they refuse to play on the same team, at most four of the people will be able to play.

The following is another example of the pigeonhole principle:

Assume that in a box there are 10 black socks and 12 blue socks and you need to get one pair of socks of the same color. Supposing you can take socks out of the box only once and only without looking, what is the minimum number of socks you’d have to pull out at the same time in order to guarantee a pair of the same color?

The correct answer is three. To have at least one pair of the same color (m = 2 holes, one per color), using one pigeonhole per color, you need only three socks (n = 3 objects). In this example, if the first and second sock drawn are not of the same color, the very next sock drawn would complete at least one same-color pair (m = 2).

The pigeonhole principle may be used to demonstrate possibly unexpected results. For example, there must be at least two people in London with the same number of hairs on their heads. A typical head of hair has around 150,000 hairs; it is reasonable to assume that no one has more than 1,000,000 hairs on his head (m = 1 million holes). There are more than 1,000,000 people in London (n is bigger than 1 million objects). If we assign a pigeonhole for each number of hairs on a head, and assign people to the pigeonhole with their number of hairs on it, there must be at least two people with the same number of hairs on their heads.

Here’s how someone would mathematically solve a pigeonhole problem:

Suppose a musical group has 11 weeks to prepare for opening night, and they intend to have at least one rehearsal each day. However, they decide not to schedule more than 12 rehearsals in any 7-day period to keep from getting burned out.  Prove that there exists a sequence of successive days during which the band has exactly 21 rehearsals.

This problem can be solved by a double application of the pigeonhole principle.

First, since the band has no more than 12 rehearsals per week, they can’t have more than 132 rehearsals in 11 weeks. (To see this, let x_k denote the number of games played in each of the 11 consecutive weeks, and note that they cannot sum to greater than 132 if they are each no greater than 12.)

Now let x_n denote the total number of rehearsals that have been held after n days. Since the band rehearses at least once per day, we have

1   <=   x_1   <   x_2   <   x_3   <  …  <   x_77   <=  132

Also, we can add 21 to each of these numbers to give the sequence

(x_1 + 21)   <   (x_2 + 21)   < …  <   (x_77 + 21)   <=  153

There are 77 numbers x_n and 77 more numbers x_n + 21 for a total of 154 numbers, all in the range from 1 to 153. Thus, at least two of these 154 numbers must be equal.  But the x_n are all distinct, as are the (x_n + 21), so any “overlap” must be between a number of the form x_n and one of the form (x_m + 21), which implies

                   x_n  =  x_m + 21

This proves that there are indices m, n such that exactly 21 rehearsals were held during the sequence of consecutive days from day n+1 today m.