Math in Medicine

One of the more advanced ways that medical professionals use mathematics is in the use of CAT scans. A CAT scan is a special type of x-ray called a Computerized Axial Tomography Scan. A regular x-ray can only provide a two-dimensional view of a particular part of the body. Then, if a smaller bone is hidden between the x-ray machine and a larger bone, the smaller bone cannot be seen. It is like a shadow. It is much more beneficial to see a three dimensional representation of the body’s organs, particularly the brain. CAT scans allow doctors to see inside the brain, or another body organ, with a three dimensional image. In a CAT scan, the x-ray machine moves around the body scanning the brain (or whichever body part is being scanned) from hundreds of different angles. Then, a computer takes all the scans together and creates a three dimensional image. Each time the x-ray machine makes a full revolution around the brain, the machine is producing an image of a thin slice of the brain, starting at the top of the head and moving down toward the neck. The three-dimensional view created by the CAT scan provides much more information to doctors that a simple two-dimensional x-ray. 

Mathematics plays a crucial role in medicine and because people’s lives are involved, it is very important for nurses and doctors to be very accurate in their mathematical calculations. Numbers provide information for doctors, nurses, and even patients. Numbers are a way of communicating information, which is very important in the medical field.

Week 12, Day 2

Today we learned about graphs of polar equations. As a quick review, the polar coordinate system is very similar to that of the rectangular coordinate system. In a polar coordinate grid, as shown below, there will be a series of circles extending out from the pole (or origin) and five different lines passing through the pole to represent the angles at which the extact values are known for the trigonometric functions. Graphing a polar equation is accomplished in pretty much the same manner as rectangular equations are graphed. They can be graphed by point-plotting, using the trigonometric functions period, and using the equation's symmetry (if any). When graphimg rectangular equations by point-plotting you should pick values for x and then evaluate the equation to determine its corresponding y value. For a polar equation, you would pick angle measurements for theta and then evaluate the equation to determine its corresponding r value. 

Symmetry Tests for Polar Coordinates

(1) Replace theta with -theta. If an equivalent equation results, the graph is symmetric with respect to the polar axis (x-axis). (2) Replace theta with -theta and r with -r. If an equivalent equation results, the graph is symmetric with respect to theta = pi/2 (y-axis). (3) Replace r with -r. If an equivalent equation results, the graph is symmetric with respect to the pole (origin). It is possible for a polar equation to fail  test and still exhibit that type of symmetry when you finish graphing the function over a full period.

Polar equations have some general types of equations. 

Learning to recognize the formulas of these equations will help in sketching the graphs. 

Circles in Polar Form

Limacons

1. r = a +/- b sin theta, where a > 0 and b < 0

2. r = a +/- b cos theta, where a > 0 and b < 0

The limacons containing sine will be above the horizontal axis if the sign between a and b is plus or below the horizontal axis if the sign between a and b is minus. If the limacon contains the function cosine then the graph will be either to the right of the vertical axis if the sign is plus or to the left of the vertical axis if the sign is minus. The ratio of a/b will determine the exact shape of the limacon. 

Rose Curves

A rose curve is a graph that is produced from a polar equation in the form of:

r = a sin n theta or r = a cos n theta, where a does not equal 0 and n is an interger > 1

They are called rose curves because the loops that are formed resemble petals. The number of petals that are present will depend on the value of n. The value of a will determine the length of the petals.

Lemniscates

The last type of polar equation that we will cover in this section is lemniscates, which has the shape of a figure-8 or propeller. Lemniscates have the general polar equation of:

r^2 = a^2 sin 2 theta or r^2 = a^2 cos 2 theta, where a does not equal 0

A lemniscate containing the sine function will be symmetric to the pole while the lemniscate containing the cosine function will be symmetric to the polar axis, to theta = pi/2, and the pole. 

Example 1: 

Graph the polar equation r = 1 - 2 cos theta

Identify the type of polar equation. 

The polar equation is in the form of a limacon, r = a - b cos theta.

Find the ratio of a/b to determine the equation's general shape: a/b = 1/2

Since the ratio is less than 1, it will have both an interger and outer loop. The loops will be along the polar axis since the function is cosine and will loop to the left since the sign btwn a and b is minus.

Test for symmetry

Evaluate r at different values of theta. 

Since the equation passes the test for symmetry to the polar axis, we only need to evaluate the equation over the interval [0, theta] and then reflect the graph about the polar axis.

Plot the points

Use the symmetry to complete the graph

Week 12, Day 1

Today we learned about 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. 

Mathematics in Music

Mathematics is applicable to nearly all aspects of life, as Galileo Galilei once said that the universe "is written in the language of mathematics." It is remarkable the extent to which science and society are governed by mathematical ideas. It is perhaps even more surprising that music, with all its passion and emotion, is also based upon mathematical relationships. Such musical notions as octaves, chords, scales, and keys can be demystified and understood logically using simple mathematics. Among the various components of music, is pitch and wave frequencies. This especially intertwines with math, and only further demonstrates the prevalence of math in the music industry.

Music appears to be transmitted by magic, escaping from your expensive stereo, car radio, or guitar, and accosting your eardrums on one fell swoop. In fact, sound progresses as a wave through the air, and sound cannot be produced without an atmosphere. A sound wave creates minute pockets of higher and lower air pressure, and all the sounds we hear are caused by these pressure changes. With music, the frequency at which these pockets strike your ear controls the pitch that you hear. 

For example, consider the note called "Middle C," typically the first note learned in piano lessons. This note has a frequency of about 262 Hertz. That means that when Middle C is played, 262 pockets of higher air pressure pound against your ear each second. Equivalently, the pockets of air arrive so quickly that one pocket strikes your ear every 0.00382 seconds. We can draw a graph by putting an X at every time a pocket of air arrives: 

This graph provides a representation of Middle C. By itself, it does not tell us much. However, such graphs provide a new perspective on the relationship between different musical notes. 

A basic rule is that higher-pitched notes have a higher frequency, corresponding to more frequent air pocket arrivals. For example, the note Middle G, seven semi-tones higher than Middle C, has a frequency of about 392 Hertz, corresponding to 392 air pockets per second, or a time period of 0.00255 of a second between arrivals:

With the higher note (Middle G), the air pockets arrive more frequently, corresponding to a higher frequency, and thus to more X's in the graph. If you listen carefully to an ambulence siren or a train whistle, you will notice that the noise sounds higher while the vehicle is approaching, and lower after the vehicle has passed by. This is because the approaching movement compresses the X's together, making them arrive more frequently and produce a higher pitch, while the departing movement stretches out the X's and produces a lower pitch. This is musical frequency in action.

Week 11, Day 2

Today we learned about rotation and systems of equations. 

Horizontal or Vertices Axes: Ax^2 + Cy^2 + Dx +Ey + F = 0

Equation in an xy-Plane: Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

To eliminate this xy-term, you can use a procedure called rotation of axes. The objective is to rotate the x- and y-axes until they are parallel to the axes of the conic. The rotated axes are denoted as the x'-axis and the y'-axis. After the rotation, the equation of the conic in thw new x'y'-plane will have the form A'(x')^2 + C'(y')^2 + D'x' + E'y' + F' = 0. Because this equation has no xy-term, you can obtain a standard form by completing the square. The following theorem identifies how much to rotate the axes eliminate the xy-term and also the eauations for determining the new coefficients A', C', D', E', and F'. 

Rotation of Axes to Eliminate an xy-Term

The general second-degree equation Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 can be rewritten as:

A'(x')^2 + C'(y')^2 + D'x' + E'y' + F' = 0 by rotating the coordinate axes theough an angle theta, where x = x' cos(theta) - y'sin(theta) and y = x' sin(theta) + y' cos(theta). 

Example 1: Rotation of Axes for a Hyperbola

Write the equation xy -1 = 0 in standard form.

Because A = 0, B = 1, and C = 0, you have

cot2(theta) = A - C / B = 0

2(theta) = pi / 2

Theta = pi / 4

which implies that

x = x' cos (pi /4) - y' sin(pi / 4)

x = x' ((2)^1/2 / 2) - y' ((2)^1/2 / 2) 

x = x' - y' / (2)^1/2 and

y = x' sin (pi / 4) + y' cos (pi / 4)

y = x' ((2)^1/2 / 2) + y' ((2)^1/2 /2)

y = x' + y' / (2)^1/2

The equation in the x'y'-system is obtained by substituting these expressions into the equation xy -1 =  0. (x' - y' / (2)^1/2) x (x' + y' / (2)^1/2) - 1 = 0

(x')^2 - (y')^2 / 2 - 1 = 0

(x')^2 / ((2)^1/2)^2 - (y')^2 / ((2)^1/2)^2 = 1

In the x'y'-system, this is a hyperbola centered at the origin with vertices at (+/- (2)^2, 0). To find the coordinates of the vertices in the xy-system, substitute the coordinates (+/- (2)^1/2, 0) into the equations x = x' - y' /(2)^1/2 and y = x' + y' / (2)^1/2. This substitution yields the vertices (1,1) and (-1,-1) in the xy-system. Note also that the asymptotes of the hyperbola have equations y' = +/-x', which correspond to the original x- and y- axes.

Example 2: Rotation of Axes for an Ellipse

Sketch the graph of 7x^2 - 6(3)^1/2xy + 13y^2 - 16 = 0

Because A = 7, B = -6(3)^1/2, and C = 13, you have

cot2(theta) = A - C / B = 7 - 13 / 6(3)^1/2 = 1 / (3)^1/2

which implies that theta = pi / 6. The equation in the x'y'-system is obtained by making the substitutions x = x' cos(pi / 6) - y' sin (pi / 6)

x = x'((3)^1/2 / 2) - y'(1/2)

x = (3)^1/2x' - y' / 2 and

y = x' sin(pi / 6) + y' cos(pi / 6)

y = x'(1/2) + y'((3)^1/2 /2)

y = x' + (3)^1/2y' / 2

into the original equation. Thus, you have

7x^2 - 6(3)^1/2xy + 13y^2 - 16 = 0

7((3)^1/2x' - y' / 2)^2 - 6(3)^1/2((3)^1/2x' - y' / 2)(x' + (3)^1/2y' / 2) + 13(x' + (3)^1/2y'/ 2)^2 - 16= 0

which simplifies to: 4(x')^2 + 16(y')^2 - 16 = 0

4(x')^2 + 16(y')^2 = 16

(x')^2 / 4 + (y')^2 / 1 = 1.

This is the equation of an ellipse centered at the origin with vertices (+/- 2, 0) in the x'y'-system.

Example 3: Rotation of Axes for a Parabola

Sketch the graph of x^2 - 4xy + 4y^2 + 5(5)^1/2y + 1 = 0

Because A = 1, B = -4, and C = 4, you have cot2(theta) = A - C / B = 1 - 4 / -4 = 3/4

Using the identity cot2(theta) = (cot^2(theta) - 1) / (2cot(theta)) produces

cot2(theta) = 3/4 = cot^2 - 1 / 2cot(theta) from which you obtain the equation: 

4cot^2(theta) - 4 = 6cot(theta)

4cot^2(theta) - 6cot(theta) - 4 = 0

(2cot(theta) - 4)(2cot(theta) + 1) = 0

Considering 0 < theta < pi / 2, you have 2cot(theta) = 4.

Thus, cot(theta) = 2 ---> theta = 26.6 degrees.

From the triangle above, you obtain sin(theta) = 1 / (5)^1/2 and cos(theta) = 2 / (5)^1/2. 

Consequently, you use the substitutions: 

x = x' cos(theta) - y' sin(theta) = x'(2 / (5)^1/2) - y' (1 / (5)^1/2) = 2x' - y' / (5)^1/2

y = x' sin(theta) + y' cos(theta) = x' (1 / (5)^1/2) + y' (2 / (5)^1/2) = x' + 2y' / (5)^1/2.

Substituting these expressions into the original equation, you have x^2 - 4xy + 4y^2 + 5(5)^1/2 + 1=0

(2x' - y' / (5)^1/2) - 4(2x' - y' / (5)^1/2)(x' + 2y' / (5)^1/2) + 4(x' + 2y' / (5)^1/2)^2 + 5(5)^1/2(x' + 2y' / (5)^1/2) + 1 = 0, which simplifies as follows. 5(y')^2 + 5x' + 10y' + 1 = 0 ---> 5(y' + 1)^2 = -5x' + 4 

---> (y' + 1)^2 = (-1)(x' - 4/5). The graph of this equation is a parabola with vertex at (4/5, -1). Its axis is parallel to the x'- axis in the x'y'-system.

Invariants Under Rotation:

In the rotation of axes theorem listed at the beginning of this section, note that the constant term is the same in both equations---that is, F' = F. Such quantities are invariant under rotation. 

The next theorem lists other rotation invariants:

The rotation of the coordinate axes through an angle that transforms the equation:

Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 has the following rotation invariants:

1. F = F'

2. A + C = A' + C'

3. B^2 - 4AC = (B')^2 - 4A'C'

You can use the results of this theorem to classify the graph of a second-degree equation with an xy-term in much the same way you do for a second-degree equation without an xy-term. Note that because B' = 0, the invariant B^2 - 4AC reduces to B^2 - 4AC = -4A'C'. This quantity is called the discriminant of the equation Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0. Now, from the classification procedure given in the previous lesson, you know that the sign of A'C' determines the type of graph for the eauation A'(x')^2 + C'(y')^2 + D'x' + E'y' + F' = 0. Consequently, the sign of B^2 - 4AC will determine the type of graph for the original equation, as given in the following classification.

1. Ellipse or circle: B^2 - 4AC < 0

2. Parabola: B^2 - 4AC = 0

3. Hyperbola: B^2 - 4AC > 0

Week 11, Day 1

Today we learned about hyperbolas, the set of all points (x,y) the difference of whose distances from two distinct fixed points (foci) is a positive constant. The definition of  hyperbola paralells that of an ellipse. The difference is that for an ellipse the sum of the distances between the foci and a point on the ellipse is foxed, whereas for a hyperbola the difference of these distances is fixed. Every hyperbola has two disconnected branches. The line through the two foci intersects a hyperbola at its two vertices. The line segment connecting the vertices is called the transverse axis, and the midpoint of the transverse axis is called the center of the hyperbola. The development of the standard form of the equation of a hyperbola is similar to that of an ellipse. The following are standard equations of a hyperbola, in which the center is at (h,k): 

The following diagrams illustrate the horizontal and vertical orientations of a hyperbola:

Example 1: Finding the Standard Equation of a Hyperbola

Find the standard form of the equation of the hyperbola with foci at (-1,2) and (5,2) and vertices at (0,2) and (4,2). By the Midpoint Formula, the center of the hyperbola occurs at the point (2,2). Furthermore, c = 3 and a = 2, and it follows that b^2 = c^2 - a^2 = 3^2 - 2^2 = 9 - 4 = 5. Thus, the equation of the hyperbola is [(x - 2)^2 / 4] - [(y - 2)^2 / 5] = 1. 

Asymptotes of a Hyperbola:

Each hyperbola has two asymptotes that intersect at the center of the hyperbola. The asymptotes pass through the vertices of a rectangle of dimensions 2a and 2b, with kts center at (h,k).

The asymptote for a horizontal transverse axis is: y = k +/- b/a(x - h). 

The asymptote for a vertical transverse axis is: y = k +/- a/b(x - h). 

Example 2: Finding the Asymptotes of a Hyperbola

Sketch the hyperbola given by 4x^2 - 3y^2 + 8x + 16 = 0 and find the equations of its asymptotes. 

4x^2 - 3y^2 + 8x + 16 = 0

4(x^2 + 2x) - 3y^2 = -16

4(x^2 + 2x + 1) - 3y^2 = -16 + 4

4(x + 1)^2 - 3y^2 = -12

[y^2 / 4] - [(x + 1)^2 / 3] = 1

From this equation you can conclude that the hyperbola is centered at (-1,0), has vertices at (-1,2) and (-1,-2), and the ends of the conjugate axis occur at (-1 - (3)^1/2, 0) and (-1 + (3)^1/2, 0). To sketch the hyperbola, draw a rectangle through these four points. The asymptotes are the lines passing through the corners of the rectangle. Finally, using a = 2 and b = (3)^1/2, you can conclude that the equations of the asymptotes are: y = [2/(3)^1/2] (x + 1)] and y = [-2/(3)^1/2] (x + 1)]. 

Example 3: Using Asymptotes to Find the Standard Equation

Find the standard form of the equation of the hyperbola having vertices at (3,-5) and (3,1) and with asymptotes y = 2x - 8 and y = -2x + 4. By the Midpoint Formula, the center of the hyperbola is at (3,-2). Furthermore, the hyperbola has a vertical transverse axis with a = 3. From the given equations, you can determine the slopes of the asymptotes to be: m(1) = 2 = a/b and m(2) = -a/b and because a = 3, you can conclude that b = 3/2. Thus, the standard equation is [(y + 2)^2] / 9 - [(x - 3)^2] / 9/4 = 1. 

Eccentricity:

As with ellipsese, the eccentricity of a hyperbola is e = c / a and because c > a, it follows that e > 1. If the eccentricity is large, the branches of the hyperbola are nearly flat. If the eccentricity is close to 1, the branches of the hyperbola are more pointed. 

Classifying a Conic from its General Equation:

The graph of Ax^2 + Cy^2 + Dx + Ey + F = 0 is one of the following.

1. Circle: A = C

2. Parabola: AC = 0 (A = 0 or C = 0, but not both)

3. Ellipse: AC > 0 (A and C have like signs)

4. Hyperbola: AC < 0 (A and C have unlike signs) 

Pi Day!

Have you ever wondered why we dedicate an entire day to a number? 

Well, I am here to inform you why pi is such a special number by listing the following facts:

1. Pi is the most recognized mathematical constant in the world. 

2. Comedian John Evans once quipped: "What do you get if you divide the circumference of a jack-o-latern by its diameter? Pumpkin Pi."

3. Scientists in Carl Sagan's novel Contact are able to find hidden messages from the creators of the human race, allowing humans to access universal understanding.

4. We can never truly measure the circumference or the area of a circle because we can never know the value of pi, as it is an irrational number whose digits continue in a random sequence.

5. In 1995, Hiroyoki Gotu memorized 42,195 places of pi and is considered the current pi champion. Scholars speculate that Japanese is better suited than other languages for memorizing numbers.

6. In 2002, a Japanese scientist found 1.24 trillion digits of pi using a computer called Hitachi SR.

7. "Pi Day" is celebrated on March 14, which was chosen because it resembles 3.14. The official celebration begins at 1:59 p.m. to make an appropriate 3.14159 when combined with the date. Albert Einstein was born on Pi Day (3/14/1879) in Ulm Wurttemberg, Germany.

8. The Bible alludes to pi in 1 Kings 7:23 where it describes the altar inside Solomon's temple: "And he made a molten sea of ten cubits from brim to brim...and a line of thirty cubits did compass it round about." These measurements procure the following equation: 333/106 = 3.141509. 

9. Leonardo da Vinci (1452-1519) and artist Albrecht Durer (1471-1528) both briefly worked on "squaring the circle" or approximating pi.

10. The first six digits of pi appear at least six times among the first 10 million decimal places of pi. 

Week 10, Day 2

Today we learned about another type of conic: ellipses, a set of all points (x,y) the sum of whose distances from two distinct fixed points (foci) is constant. Its components are illustrated below:

The line through the foci intersects the ellipse at two points (vertices). The chord joining the vertices is the major axis, and its midpoint is the center of the ellipse. The chord perpendicular to the major axis at the center is the minor axis of the ellipse. To derive the standard form of the equation of an ellipse, consider the figure shown below with these points: center (h,k); vertices (h + a, k); foci, (h +/- c, k). The sum of the distances from any point on the ellipse to the two foci is constant. Using a vertex point, this constant sum is (a + c) + (a - c) = 2a (length of major axis) or simply the length of the major axis. Now, if you let (x,y) be any point on the ellipse, the sum of the distances between (x,y) and the two foci must also be 2a. That is, [(x - (h - c))^2 + (y - k)^2]^1/2 + [(x - (h + c))^2 + (y - k)^2]^1/2 = 2a. Finally, using the figure below and b^2 = a^2 - c^2, you obtain the following equation of the ellipse. b^2(x - h)^2 + a^2(y - k)^2 = a^2b^2; (x - h)^2/a^2 + (y - k)^2/b^2 = 1. 

Standard Equations of an Ellipse:

Vertical and Horizontal Orientations of an Ellipse:

Eccentricity:

To measure the ovalness of an ellipse, you can use the concept of eccentricity. 

The eccentricity (e) of an ellipse is given by the ratio: e = c / a. 

Example 1: 

x^2 + 4y^2 + 6x - 8y + 9 = 0

Find the standard form, center, foci, vertex, and eccentricity of this equation. 

x^2 + 6x + 4y^2 - 8y = -9

x^2 + 6x + 9 + 4(y^2 - 2y) = -9 + 9

x^2 + 6x + 9 + 4(y^2 - 2y + 1) = -9 + 9 + 4

(x + 3)^2 + 4(y - 1)^2 = 4

(x + 3)^2 / 4 + (y - 1)^2 / 1 = 1

Center = (h, k)

Center = (-3, 1)

b = 1, a = 2

c^2 = a^2 - b^2

c^2 = 4 - 1

c^2 = 3

c = (3)^1/2

Foci = (-3 + (3)^1/2, 1) (-3 - (3)^1/2, 1)

Vertex = (-1, 1) (-5, 1) 

Eccentricity = e = c / a = (3)^1/2 / 2

Week 10, Day 1

Today we introduced conics, a term that encompasses all shapes formed by an intersection of a plane and a double-napped cone, including circles, ellipses, parabolas, and hyperbolas. This lesson focuses specifically on porabolas, a set of all points (x,y) that are equidistant from a fixed line (directrix) and a fixed  point (focus) not on the line. The midpoint between the focus and the directrix is called the vertex, and the line passing through the focus and the vertex is called the axis of the parabola. Using the definition of a parabola, you can derive the following standard form of the equation of a parabola whose directrix is parallel to the x-axis or to the y-axis. The vertex at (h,k), the standard form is: (x-h)^2 = 4p(y-k): vertical axis; directrix: y = k-p. (y-k)^2 = 4p(x-h): horizontal axis; directrix: x = h-p. The focus lies on the axis p units (directed distance) from the vertex. If the vertex is at the origin (0,0), the equation takes one of the following forms. x^2 = 4py: vertical axis. y^2 = 4px: horizontal axis. 

Example 1: Finding the Standard Equation of a Parabola

Find the standard form of the equation of the parabola with vertex (2,1) and focus (2,4).

Because the axis of the parabola is vertical, consider the equation (x-h)^2 = 4p(y-k) where h = 2, k = 1, and p = 4-1 = 3. Thus, the standard form is (x-2)^2 = 12(y-1). 

Example 2: Finding the Focus of a Parabola

Find the focus of the parabola given by y = -1/2(x)^2 - x + 1/2. To find the focus, convert to standard form by completing the square. The steps are listed as follows:

y = -1/2(x)^2 - x + 1/2 (Original equation)

-2y = x^2 + 2x - 1 (Multiply both sides by -2)

1 - 2y = x^2 + 2x (Group the terms)

2 - 2y = x^2 + 2x + 1 (Complete the square)

-2(y - 1) = (x + 1)^2 (Standard form)

Comparing this equation with (x - h)^2 = 4p(y - k), you can conclude that h = -1, k = 1, and p = -1/2. Because p is negative, the parabola opens downward. The focus is (h, k + p) = (-1, 1/2). 

Example 3: Vertex at the Origin

Find the standard equation of the parabola with vertex at the origin and focus (2,0). The axis  of the parabola is horizontal, passing through (0,0) and (2,0). Thus, the standard form is y^2 = 4px where h = k = 0 and p = 2. The equation is y^2 = 8x. 

The Probability of Faith

Thomas Baynes was the minister of the Presbyterian Chapel in the English spa town of Tunbridge Wells from around late 1733. By all accounts he was a better mathematician than he was a church minister. As a non-conformist, he was prevented from attending the English universities, so he began to study in 1719 at Edinburgh University in Scotland. A number of years later, he published a work entitled "Divine Benevolence," and in 1742, he was made a fellow of the Royal Society after defending Isaac Newton against Bishop Berkeley in a 'pamphlet war.' Bayes did work with fluxions, infinite series, and probability. In the late 1740s, Bayes set out his theory of probability that eventually bore his name. His theorem was discovered after his death. It had no practical applications in his lifetime, and was therefore never published. It was his fellow mathematician and church minister Richard Price who discovered it among Bates' effects aftee his death and submitted it for publication as "An Essay Towards Solving a Problem in the Doctrine of Chances." A man named Laplace later rediscovered and modified it into a form we largely recognize today. McGrayne Bayes writes "In discovering its value for science, many supporters underwent a near-religious conversion yet had to conceal their use of Bayes' rule and pretend they employed something else. It was not until the 21st-century that the method lost its stigma and was widely and enthusiastically embraced." Today it is used among other things to forecast weather, identify e-mail spam, improve lowres images on computers, and determine forgeries. The formula known today is: P(A I B) = [P(B I A) x P(A)] / P(B) where P(A) denotes the probability of A and P(A I B) is the probability of A given that B has occurred. 

Interestingly, Bayes' theorem has been used by philosophers of religion such as Richard Swinburne to try and prove God's existence and by other such as philosopher John Mackie and the atheist evangelist Richard Dawkins in an attempt to disprove God's existence. 

Week 9, Day 2

Today we explored data. It is often helpful to describe data by a single number that is most representative of an entire collection of numbers. Such a number is called a measure of central tendency. The most commonly used measures are as follows: Mean, or average, of n numbers is the sum of the numbers divided by n. The median of n numbers is the middle number when the numbers are written in order. If n is even, the median is the average of the two middle numbers. The mode of n numbers is the number that occurs most frequently. If two numbers tie for most frequent occurrence, the collection has two modes and is called bimodal. Quartiles are another concept we learned, and can be divided into two types: upper and lower. An upper quartile is the median of the numbers before m (the median of all numbers) and a lower quartile is the median of the numbers before m (the median of all numbers). Finally, we were taught measures of dispersion. The following are equations relating to variance and standard deviation: 

Example 1:

Given a set of numbers: 42. 62, 40, 29, 32, and 70, find the measures of central tendency. 

Then, using those values, determine the variance and standard of deviation. 

Week 9, Day 1

Today we learned about probability in its various forms.

Independent events are when the occurrence of one event does not affect the occurrence of the 2nd event. The complement of an event is the probability an event does not occur. A happening whose result is uncertain is called an experiment. The possible results of the experiment are outcomes, and the set of all possible outcomes of the experiment is the sample space. Any subcollection of a sample space is an event. The measure of the likelihood that an event will occur based on chance is called the probability of an event. If an event (E) has n(E) equally likely outcomes and its sample space (S) has n(S) equally likely outcomes, the probability of event (E) is P(E) = n(E) / n(S). The probability of an event must be between 0 and 1. If P(E) = 0, the event cannot occur, and (E) is called an impossible event. If P(E) = 1, the event (E) must occur, and (E) is called a certain event. 

Example 1:

A box contains 3 red marbles, 5 balck marbles, and 2 yellow marbles. If a marble is selected at random from the box, what is the probability that it is yellow?

n(E) / n(S) = 2 / 10 = 1 / 5 = 20%

Mutually Exclusive Events:

The probability of an event must be between 0 and 1. If P(E) = 0, the event E cannot occur, and E is called an impossible event. If P(E) = 1, the event E must occur, and E is called a certain event. Two events A and B, from the same sample space, are mutually exclusive if A and B have no outcomes in common. If A and B are events in the same sample space, the probability of A or B occurring is given by P(A U B) = P(A) + P(B) - P(A inverted U B). U refers to union, and inverted U refers to intersection.

Example 2: 

A box contains 3 red marbles, 5 black marbles, and 2 yellow marbles. If a marble is selected at random from the box, what is the probability that it is either red or black.

3 / 10 + 5 / 10 = 8 / 10 = 4 / 5 = 80%

Independent Events: 

If A and B are independent are independent, the probability that both A and B will occur is P(A and B) = P(A) + P(B). That is, to find the probability that two independent events will occur, multiply the probabilities. Let A be an event and let A' be its complement. If the probability of A is P(A), the probability of the complement is P(A') = 1 - P(A). "With relplacement" implies independent events.

Example 3:

A box contains 3 red marbles, 5 black marbles, and 2 yellow marbles. If two marbles are randomly selected with replacement, what is the probability that both marbles are yellow? 

2 / 10 x 2 / 10 = 4 / 100 = 2 / 50 = 4%