Week 17, Day 3

As we continued our review, Harrison presented on lesson 7.6: Linear Programming. In this lesson, we learned how to perform linear programming. A 2D linear programming problem consists of a linear objective function and a system of linear inequalities called constraints. The objective function gives the quantity that is to be maximized or minimized, and the constraints determine the set of feasible solutions.

The following are steps to solving a linear programming problem:

1) Sketch the region corresponding to the system of constraints.

2) Find the vertices of the region.

3) Test the objective function, z = ax + by, at each of the vertices and select the values of the variables that optimize the objective function. For a bounded region, both a maximum and minimum will exist. For an unbound region, if an optimized solution exists, it will occur at a vertex. 

Week 17, Day 2

Today, Grace presented on section 10.6: Polar Coordinates. As opposed to the traditional x-y coordinate system, we use the polar coordinate system. To form this system in the plane, fix a point O, called the pole or origin, and construct from O an initial ray called the polar axis. Then each point P in the plane can be assigned polar coordinates (r, theta) as follows. (1) r = directed distance from O to P (2) Theta = directed angle, counterclockwise from polar axis to segment OP. 

Example 1: Plotting Points in the Polar Coordinate System

a) The point (r, theta) = (2, pi/3) lies two units from the pole on the terminal side 

of the angle theta = pi/3, as shown in the following image. 

b) The point (r, theta) = (3, -pi/6) lies three units from the pole on the terminal side 

of the angle theta = -pi/6, as shown in the following image. 

c) The point (r, theta) = (3, 11pi/6) coincides with the point (3, -pi/6), as shown in the following image. 

In rectangular coordinates, each point (x,y) has a unique representation. This is not true for polar coordinates. For instance, the coordinate (r, theta) and (r, 2pi + theta) represent the same point. Another way to obtain multiple reoresentations of a point is to use negative values for r. Because r is a directed distance, the coordinates (r, theta) and (-r, theta + pi) represent the same point. In general, the point (r, theta) can be represented as (r, theta) = (r, theta +/- 2npi) or (r, theta) = (-r, theta +/- (2n+1)pi) where n os any interger. Moreover, the pole is represented by (0, theta), where theta is any angle. 

Example 2: Multiple Representation of Points

Plot the point (3, -3pi/4) and find three additional polar representations of this point, using -2pi < theta < 2pi. The point is shown in the following image. 

Three other representations are as follows.

(3, -3pi/4 + 2pi) = (3, 5pi/4): Add 2pi to theta.

(-3, -3pi/4 - pi) = (-3, -7pi/4): Replace r by -r, then subtract pi from theta.

(-3, -3pi/4 + pi) = (-3, pi/4): Replace r by -r, then add pi to theta.

Coordination Conversion

To establish the relationship between polar and rectangular coordinates, let the polar axis coincide with the positive x-axis and the pole with the origin. Because (x,y) lies on a circle of radius r, it follows that r^2 = x^2 + y^2. Moreover, for r > 0, the defintions of the trignometric functions imply that tan theta = y/x, cos theta = x/r, and sin theta = y/r. You can show that the same relationships hold for r > 0. 

The polar coordinates (r, theta) are related to the rectangular coordinates (x,y) as follows:

x = rcos theta and tan theta = y/x

y = rsin theta 

r^2 = x^2 + y^2

Example 3: Polar-Rectangular Conversion

Convert the points (a) (2, pi) and (b) ( (3)^1/2, pi/6) to rectangular coordinates.

(a) For the point (r, theta) = (2, pi), you have

x = rcos theta = 2 cos pi = -2 and y = rsin theta = 2 sin pi = 0.

The rectangular coordinates are (x,y) = (-2,0).

(b) For the point (r, theta) = ( (3)^1/2, pi/6), you have

x = (3)^1/2 cos pi/6 = (3)^1/2 ( (3)^1/2 / 2) = 3/2 and

y = (3)^1/2 sin pi/6 = (3)^1/2 (1/2) = (3)^1/2 / 2.

The rectangular coordinates are (x,y) = ( 3/3, (3)^1/2).

Example 5: Converting Polar Equations to Rectangular Form

Describe the graph of each polar equation and find the corresponding rectangular equation.

(a) r = 2 

The graph of the polar equation r = 2 consists of all points that are two units from the pole. In other words, this graph is a circle centered at the origin with a radius of 2. You can confirm this by converting to rectangular form, using the relationship r^2 = x^2 + y^2. 

r = 2 is the polar equation. 

r ^2 = 2^2 

x^2 + y^2 = 2^2 is the rectangular equation. 

Week 17, Day 1

This week we continued our review of topics that will be tested on our final. 
Cristian Vera presented lesson 9.5: Binomial Theorem and Expansion.
When you write out the coefficients for a binomial that is raised to a power, you are expanding a binomial. There are two methods of expansion, both of which provide binomial coefficients. The first is Binomial Theorem, which states that in the expansion of 

(x + y)^n: (x + y) ^n = x^n + (nx^n-1)y +...+ (n)C(r) (x^n-r)y^r +...+ nxy^n-1 + y^n.The coefficient of (x^n-r)y^r is given by a procedure known as combination: (n)C(r) = n! / (n-r)!r!. The second is Pascal's Triangle, in which the first andlast number in each row is 1. Every other number in each row is formed by adding the two numbers immediately above the number. A basic illustration of Pascal's triangle is:

To expand a binomial, it is easiest to use Pascal's Triangle as shown below in an example I got from my original post on the topic:

Because the exponent of the binomial is 6, you would use the 6th row of the triangle. Using the given coefficients, write each appropriate term, and simplify. 

Week 16, Day 3

Today we reviewed the topic of vectors, as explained in Darron's presentation.

Physical forces and velocities are not confined to the plane, therefore it is natural to extend the concept of vectors from two-dimensional space to three-dimensional space. In space, vectors are denoted by ordered triples v = <v1, v2, v3>, otherwise known as component form. The zero vector is denoted by 0 = < 0, 0, 0 >. Using the unit vectors i = < 1, 0, 0 > in the direction of the positive z-axis, the standard unit vector notation for v is v = v1i + v2j + v3k, otherwise known as unit vector form. If v is represented by the directed line segment from P(p1, p2, p3) to Q(q1, q2, q3), the component form of v is produced by subtracting the coordinates of the initial point from the coordinates of the terminal point, v ='<v1, v2, v3> = <q1-p1, q2-p2, q3-p3>. 

Vectors in Space:

1. Two vectors are equal if and only if their corresponding components are equal. 

2. The length of u = <u1, u2, u3> is ||u|| = (u1^2 + u2^2 + u3^2)^1/2 = vector length. 

3. A unit vector u in the direction of v is given by u = v / ||v|| = unit vector.

4. The sum of u = <u1, u2, u3> and v = <v1, v2, v3> is u + v = <u1 + v1, u2 + v2, u3 + ||v3>.

5. The scalar multiple of the real number c and u = <u1, u2, u3> is cu = <cu1, cu2, cu3>.

6. The dot product of u = <u1, u2, u3> and v = <v1, v2, v3> is u x v = u1v1 + u2v2 + u3v3.

Example 1: Finding the Component Form of a Vector

Find the component form and length of the vector v having initial point (3,4,2) amd terminal point (3,6,4). Then find a unit vector in the direction of v. 

The component form of v is:

v = <3-3, 6-4, 4-2> = <0,2,2>

This implies that the vector's length is 

||v|| = (0^2 + 2^2 + 2^2)^1/2 = (8)^1/2 = 2(2)^1/2.

The unit vector in the direction of v is:

u = v / ||v|| = 1 / 2(2)^1/2 <0,2,2> = <0, 1/(2)^1/2, 1/(2)^1/2>.

Example 2: Finding the Dot Product of Two Vectors

Find the dot product of <0,3,-2> and <4,-2,3>. 

<0,3,-2> x <4,-2,3> = 0(4) + 3(-2) + (-2)(3) = 0-6-6 = -12.

The angle between two nonzero vectors is the angle theta, 0 <(=) theta <(=) pi, between its respective standard position vectors. Thus, if theta is the angle between two nonzero vectors u and v, then cos(theta) =u x v / ||u|| ||v||. If the dot product of two nonzero vectors is zero, the angle between the vectors is 90 degrees. Such vectors are called orthogonal. For instance, the standard unit vectors i,j, and k are orthogonal to each other. 

Example 3: Finding the Angle Between Two Vectors

Find the angle between u = <1,0,2> and v = <3,1,0>.

cos(theta) = u x v / ||u|| ||v|| = <1,0,2> x <3,1,0> / ||<1,0,2>|| ||<3,1,0>|| = 3 / (50)^1/2

This implies that the angle between the two vectors is 

theta = arccos 3 / (50)^1/2 = 64.9 degrees. 

Parallel Vectors:

In general two nonzero vectors u and v are parallel if there is some scalar c such that u = cv. 

For example, the vectors shown below u, v, and w are parallel because u = 2v and w = -v. 

Example 4: Parallel Vectors

Vector w has initial point (1,-2,0) and terminal point (3,2,1). 

Which of the following vectors is parallel to w? 

a) u = <4,8,2> 

b) v = <4,8,4> 

Begin by writing w in component form.

w = <3,-1, 2 -(-2), 1-0> = <2,4,1>

a) The vector u is parallel to w because

u = <4,8,2> = 2<2,4,1> = 2w

b) In this case, you need to find a scalar c such that <4,8,4> = c<2,4,1>.

However, equating corresponding components produces c = 2 for the first two components and c = 4 for the third. Hence, the equation has no solution, and the vectors are not parallel. 

Week 16, Day 2

As we continued our review of the semester, we dove into the concept of matrices.
In her presentation, Jasmine explained how to perform inverse matrices.

A^-1 is called the inverse of A.

Show that B is the inverse of A where, 

To show that B is the inverse of A, show that AB = I = BA, as follows.

If a matrix A has an inverse, A is called invertible or nonsingular; otherwise, A is called singular. A nonsquare matrix cannot have an inverse. The following are example problems I took from my original post on the topic demonstrating how to solve for inverse matrices (one is singular and one is not). 

Week 16, Day 1

Today we began our review of this semester's material.
One of our topics, as Axie explained in her presentation, was partial fractions.

In this lesson, we learned how to perform partial fraction decomposition, which involves writing a rational expression as the sum of two or more rational expressions. For instance, (x+7) / (x^2-x-6) can be written as the sum of two fractions with linear denominators. That is, (x+7) / (x^2-x-6) = (2 / x-3) + (-1 / x+2). Each fraction on the right side of the equation is a partial fraction, and together they make up the partial fraction decomposition of the left side. The following are steps to carry out partial fraction decomposition: 

1. Multiply by LC

2. Distribute

3. Collect like terms

4. Factor out the variable

5. Equate the coefficients

6. Solve the system of equations

7. Write the answer as a partial fraction

When the denominator of a partial fraction is, for example, (x+1)^2, it is called a repeated factor.

When the denominator of a partial fraction is, for example, (x^2+1), it is called a quadratic factor. 

Here are a few examples:

Example 1: Distinct Linear Factors

5/(x^2+x-6)--> 5/(x+3)(x-2) = A/(x+3) + B/(x-2)

1. 5 = A(x-2) + B(x+3)

2. 5 = Ax - 2A + Bx + 3B

3. 5 = Ax + Bx - 2A + 3B 

4. 5 = (A + B)x - 2A + 3B 

5. 0 = A + B, 5 = -2A + 3B

6. 5 = 5B--> B = 1, A = -1

7. -1/(x+3) + 1/(x-2)

Example 2:Repeated Linear Factors

2x - 3/(x-1)^2 

In this case, a factor (x-1) repeats, and thus the steps to solve it are as follows: 

1. Multiply by LCD

2. Distribute

3. Collect like terms

4. Equate the coefficients

5. Solve the system of equations

6. Write the answer as a partial fraction

2x - 3/(x-1)^2 = A/(x-1) + B/(x-1)^2 Every exponent gets a factor!

1. 2x-3 = A(x-1) + B

2. 2x-3 = Ax - A + B

3. 2x-3 = Ax + (-A + B)

4. A = 2, -3 = -A + B

5. -3 + 2 = B = -1

6. 2/(x-1) - 1/(x-1)^2

Example 3: Distinct Linear Factors

4x^2 + 2x - 1 / x^2(x +1) = A/x + B/x^2 + C/x+1

1. 4x^2 + 2x - 1 = Ax(x+1) + B(x+1) + Cx^2

2. 4x^2 + 2x - 1 = Ax^2 + Ax + Bx + B + Cx^2

3. 4x^2 + 2x - 1 = Ax^2 + Cx^2 + Ax + Bx + B

4. 4x^2 + 2x - 1 = (A + C)x^2 + (A + B)x + B

5. A + C = 4, A + B = 2, B = -1

6. A = 3, B = -1, C = 1

7. 3/x -1/x^2 + 1/x+1

Example 4: Distinct Linear and Quadratic Factors

x^2 - 1 / x(x^2 + 1) 

In this case, the denominator contains an irreducible quadratic factor, so the set up is as follows:

x^2 - 1 / x(x^2 + 1) = A/x + Bx + C / x^2 + 1

1. x^2 - 1 = A(x^2 + 1) + (Bx + C)x

2. x^2 - 1 = Ax^2 + A + Bx^2 + Cx

3. x^2 - 1 = Ax^2 + Bx^2 + Cx + A

4. x^2 - 1 = (A + B)x^2 + Cx + A

5. A + B = 1, C = 0, A = -1

6. A = -1, B = 2, C = 0

7. -1/x + 2x/ x^2 + 1

When the exponent of the numerator is greater than or equal to the exponent of the denominator, use long division. Once you have found the quotient, take the remainder and put it over the original denominator, using it as your new partial fraction. 

Mathematical Imagery

The connection between mathematics and art goes back thousands of years. Mathematics has been used in the design of Gothic cathedrals, Rose windows, oriental rugs, mosaics, and tilings. Geometric forms were fundamental to the cubists and many abstract expressionists, and award-winning sculptors have used topology as the basis for their pieces. Dutch artist M.C. Escher represented infinity, Möbius bands, tessellations, deformations, reflections, Platonic solids, spirals, symmetry, and the hyperbolic plane in his works. Mathematicians and artists continue to create stunning works in all media. To further explore the visualization of mathematics, refer to origami, computer-generated landscapes, tesselations, fractals, and anamorphic art, which I have shown below. Hope you enjoy!