...Application Activity: Angry Birds – Andrew Fox Consider the following scenario: Red Bird, Yellow Bird, Blue Bird and Black Bird are angry with the pigs who stole the birds’ eggs. The birds want their eggs back and will stop at nothing to get them back. The flight path of the birds can be modeled with a parabola where “x” is the distance and “y” is the height. Use the data below to help answer the following questions: * Red Bird starts his flight from point (10, 0). His flight path reaches a maximum height of 18 yards and lands at point (38, 0). * Yellow Bird’s flight path can be modeled by the quadratic equation y=-x2+14x-24 * Blue Bird’s flight is modeled by the following graph: * The table below contains partial data points of Black Bird’s trajectory: x | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | y | 0 | 7.5 | 14 | 19.5 | 24 | 27.5 | 30 | 31.5 | 32 | 31.5 | | In developing responses to the problems, be sure to show all work: 1. What is the maximum height of each bird’s flight: (6 points) 2. What is the axis of symmetry for each bird’s flight: (6 points) 3. What was the total distance of each bird’s flight: (7 points) 4. Which bird flew the highest? (2 points) 5. Which bird traveled the longest? (2 points) 6. Which bird hit the following pigs: a. King Pig located at point (21, 19.5) (1 point) b. Moustache Pig located at point (9, 21) (1 point) 1. Red max height (24,18) Yellow max height (7,25) ...
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...variable. Show all of your work and all of your steps. (Hint: Use the properties of logarithms.) (4 marks each) a) b) c) d) Question 6) Solve for the variable. Show all of your work and all of your steps. Show the answer to 4 decimal places. (Hint: Use the common logarithm.) (4 marks each) a) b) c) Question 7) Solve for . Show all of your work and all of your steps. Show the answer to 4 decimal places. (Hint: Use the natural logarithm and the definition of a logarithm.) (4 marks each) a) b) c) Question 8) Ms. Mary bought a condo for $145 000. Assuming that the value of the condo will appreciate at most 5% a year, how much will the condo be worth in 5 years? Section 2: Conic Sections Standard forms to Know: * Parabola * Circle * Ellipse * And what does a hyperbola look like? (No formula necessary) Question 1) Write an equation for the circle that satisfies each set of conditions. (2 marks each) a) centre (12, -4), radius 81 units _________________________________________ b) centre (0, 0), radius 3/5 units...
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...Term Paper Mathematics NAME: BIPIN SHARMA ROLL NO: B59 SECTION: C1903 Conics Conic sections are the curves which result from the intersection of a plane with a cone. These curves were studied and revered by the ancient Greeks, and were written about extensively by both Euclid and Appolonius. They remain important today, partly for their many and diverse applications. Although to most people the word “cone” conjures up an image of a solid figure with a round base and a pointed top, to a mathematician a cone is a surface, one which is obtained in a very precise way. Imagine a vertical line, and a second line intersecting it at some angle f (phi). We will call the vertical line the axis, and the second line the generator. The angle f between them is called the vertex angle. Now imagine grasping the axis between thumb and forefinger on either side of its point of intersection with the generator, and twirling it. The generator will sweep out a surface, as shown in the diagram. It is this surface which we call a cone. Notice that a cone has an upper half and a lower half (called the nappes), and that these are joined at a single point, called the vertex. Notice also that the nappes extend indefinitely far both upwards and downwards. A cone is thus completely determined by its vertex angle. Now, in intersecting a flat plane with a cone, we have three choices, depending on the angle the plane makes to the vertical axis of the cone. First, we may...
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...Sydney Harbour Bridge – Directed Investigation – Quadratic functions Introduction: Aim: To find the multiple unknowns in the Sydney Harbour bridge The Sydney Harbour Bridge will be used to investigate a diverse number of points in the structure such as the height and length. Quadratics will be used to solve the height of the bridge at different points on the x axis. A quadratic is an equation constructed from information collected from a graph; this equation can also be used to produce a graph. Quadratics can be used to solve the problem dealt in this investigation as the Sydney Harbour Bridge is identified as a parabola shape. Only basic information is given about the bridge and the answers can be established by solving the additional questions. The quadratic equation will be tested from the graph to prove the accuracy of the quadratic identified. Quadratics are mainly seen in the form ax^2+bx+c=0, where x is a variable of any number. The bridge is mounted on two base pylons on the opposite ends. The highest point of the bridge is 182.5cm above sea level and the longest vertical cable is 135m from the origin. This position is the vertex, which is the highest or lowest point inside a quadratic, and in this case it is highest in this scenario. The axis of symmetry is an imaginary vertical line that cuts through and splits both sides evenly on the graph. The main supporting arc has 2 points that sits on the road situated 50m above sea level. Mathematical Investigations: The...
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...The Ellipse Definition of Ellipse Ellipse is the locus of point that moves such that the sum of its distances from two fixed points called the foci is constant. The constant sum is the length of the major axis, 2a. General Equation of the Ellipse From the general equation of all conic sections, A and C are not equal but of the same sign. Thus,the general equation of the ellipse is Ax2 + Cy2 + Dx + Ey + F = 0 or Standard Equations of Ellipse From the figure above, and From the definition above, Square both sides Square again both sides From triangle OV3F2 (see figure above) Thus, Divide both sides by a2b2 The above equation is the standard equation of the ellipse with center at the origin and major axis on the x-axis as shown in the figure above. Below are the four standard equations of the ellipse. The first equation is the one we derived above. Ellipse with center at the origin Ellipse with center at the origin and major axis on the x-axis. Ellipse with center at the origin and major axis on the y-axis. Ellipse with center at (h, k) Ellipse with center at (h, k) and major axis parallel to the x-axis. Ellipse with center at (h, k) and major axis parallel to the y-axis. The Hyperbola Submitted by Romel Verterra on February 21, 2011 - 1:33pm Definition Hyperbola can be defined as the locus of point that moves such that the difference of its distances from two fixed points called the foci...
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...the cone horizontally, you are left with a circle. If you take a slice roughly at a forty five degree, you will be dealing with an ellipse. If you take a slice that is parallel from one edge of the cone to the other cone, you are dealing with a parabola. If you take a slice from directly off centered but straight down from top to bottom, you give yourself a hyperbola. These are a few terms with definitions you will see while working with conic sections. In a circle, ellipse, and a hyperbola you have a Center. Which is usually at the point of (h,k.) The focus or “Foci” is the point which distances are measured in forming the conic. The directrix is the distance that is measured in forming the conic. The major access is the line that is perpendicular to the directrix that passes through the foci. Half of the major axis between the center and the vertex is called the semi major access. There is a general equation that covers all the conic sections and goes as follows: Ax2+Bxy+Cy2 + Dx+Ey+F=0. From this equation you can create equations for circles, ellipses, parabolas and hyperbolas. There is a test to find out which conic section you are dealing with by just looking at the equations. If both variables not squared then it’s a parabola, if it is you can move on and look to see if the squared...
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...equation: p = -25x2 + 300x quadratic equation Therefore you can find the max profit by finding the value of x of the axis of symmetry and find the vertex with that: Axis of symmetry formula: x = -b/ (2a) In equation; p = -25x2 + 300x a = -25 b = 300 x = -300/ (2)-25 x = -300/ -50 x = 6 clerks will maximize the profit To find the max profit, substitute 6 for x in the original equation: P =-252+300x=0 P = -25(62) + 300(6) Substitute 6 for x. P = -25(36) + 1800 factor both left and right side. P = -900 + 1800 add to solve equation. P = $900 is the actual profit A profit/clerk graph will look like this: The basic shape of the graph in this equation of a parabola that opens downwards (coefficient of x2 is negative) so the maximum value of P will be found at the parabola's vertex. The parabola will cross the x axis at 0 and 6. To maximize profits, the manager should employ 6 clerks. The maximum profit can be found by substituting 6 for x in the original equation for P. The graph represents the maximum number of clerks needed to gain the maximum daily...
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...e^-0.12t = 1.85454/5 = 0.370908 -0.12t = ln 0.370908 t = ln 0.370908/-0.12 = 8.265 ≈ 8 weeks 2. According to the information (bacteria in culture after 3 hours is 100 and 5 hours is 400), the bacteria appears doubling every hour. If this is the case a) the growth rate is 200% every hour . b) Following this same trend to find the initial growth rate we would divide 100 by 2, 3 times. Thus the initial number of bacteria would be 12.5 c) Since the number of bacteria of 5 hours is 400, to find the number of bacteria at 6 hours is going to be double the amount at 5 hours, 800. 3. To solve we need to find the equation of the parabola in vertex form, since we know a point of the vertex, also that is vertically symmetrical from this point. Recall that the equation we’ll use is the quadratic y = a(x-h)² + k. Since we are looking for how wide the parabola is at 10 meters(8 meters) and its vertex’s x coordinate is at 0 and its vertically symmetrical the a point that is passed through at that height(10 meters) is 4 (4,10).Substitute this information into the equation (10=a(4-0)^2) + 12 simplify -2=16a 12/16=a a = -1/8 Now substitute the equation with a=-1/8, leave x and y as these are this is the point we are looking for and the variables we will solve for this equation is y=-1/8x^2 +12 Now find the x-intercepts at this height (0 or ground level) ((-1/8x)^2 ) +12 =y set y = to 0 ((-1/8x)^2 )+12 =0 now simplify (-1/8x)^2 = -12 x^2 = 96 x = ±√96 x=±4√6 Now subtract...
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...* Link Mechanism A link mechanism can be defined as a system of connecting parts/rods that move or work together when a particular action is pre-determined. * Parabola The path traced out by a point, which moves in a plane, so that the ratio of its distance from a fixed point [FOCUS] to any point on the curve, and from the curve to a perpendicular distance on the directrix is always constant and equal to one. This constant is also known as the eccentricity. * Ellipse The path traced out by a point, which moves in a plane, so that the ratio of its distance from a fixed point [FOCUS] to any point on the curve, and from the curve to a perpendicular distance on the directrix is always constant and less than one. This constant is also known as the eccentricity. * Hyperbola: The path traced out by a point, which moves in a plane, so that the ratio of its distance from a fixed point [FOCUS] to any point on the curve, and from the curve to a perpendicular distance on the directrix is always constant and greater than one. This constant is also known as the eccentricity. * Archimedean Spiral The path traced out by a point along a rod, as the rod pivots about a fixed end. The linear movement of the point along the rod is constant with the angular movement of the rod. * Involute The path traced out by a point when an end of a plane figure is wrapped or unwrapped when held firm. * Epicycloid Tracing the path of a point as a circular disc rolls on the outside...
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...Definition of Ellipse Ellipse is the locus of point that moves such that the sum of its distances from two fixed points called the foci is constant. The constant sum is the length of the major axis, 2a. General Equation of the Ellipse From the general equation of all conic sections, [pic] and [pic] are not equal but of the same sign. Thus, the general equation of the ellipse is [pic] or [pic] Standard Equations of Ellipse Elements of the ellipse are shown in the figure above. 1. Center (h, k). At the origin, (h, k) is (0, 0). 2. Semi-major axis = a and semi-minor axis = b. 3. Location of foci c, with respect to the center of ellipse. [pic]. 4. Length latus rectum, LR 5. Consider the right triangle F1QF2: Based on the definition of ellipse: [pic] [pic] [pic] By Pythagorean Theorem [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] You can also find the same formula for the length of latus rectum of ellipse by using the definition of eccentricity. 6. Eccentricity, e DEFINITION: Eccentricity of Conic Eccentricity is a measure of...
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...Parabola Investigation In this task, I will investigate the patterns in the intersections of parabolas and the lines y=x and y=2x, then I will prove my conjectures and to broaden the scope of the investigation to include other lines and other types of polynomials. 1. Consider the parabola y=(x-3)2 +2=x2-6x+11 and the lines y=x and y=2x. The original graph is shown below (graph 1.1) Find the intersections of the parabola with y=x and y=2x, Graph 1.2. Also, label the x-values of the intersections with the line y=x as they appear from left to right on the x-axis as a1 and a2; label the x-values of the intersections with the line y=2x as b1 and b2. Now, I will using the graph and graph calculator, find the values of a1-b1 and b2-a2 and name them respectively SL and SR. SL=a1-b1=2.381966-1.763932=0.618034 SR=b2-a2=6.236068-4.618034=1.618034 Now, calculate the quantity D= │SL-SR│ D= │SL-SR│=│0.618034-1.618034│=1 By algebra calculation, D=│SL-SR│ =│ a1-b1-(b2-a2) │ =│ a1-b1-b2+a2 │ =│ (a1+ a2 )-(b1+b2) │ Now, I will try other parabolas of the form y=ax2+bx+c, a>0, with vertices in quadrant 1, intersected by the lines y=x and y=2x. y=x2+2x+1 [pic] From the graph we can see there is no intersection of the parabola and y=x, y=2x. Using the algebra way: Solve: (a) x2+2x+1=x (b) x2+2x+1=2x ...
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...arrow is 50 feet from where it was released. The target is 120 feet away and the center of the bulls eye is exactly 3 feet above the ground, and 4 inches in diameter. Assuming that Cool guy shoots a perfectly straight shot, will he hit a bulls eye? We need to use the vertex form of the equation y = a(x – h)^2 + k To find the vertical height of the arrow as it approaches the target. Since the arrow has a maximum height of 5feet when it has traveled 50 feet toward the target, we know that the vertex is at (50, 5). Substituting this into the equation, we get: y = a(x – 50)^2 + 5 To find the stretching factor, a, we need to plug in a point other than the vertex. Since the arrow was initially launched at a height of 3 feet, the parabola goes through the point (0,3). Replacing x and y in the equation, we get:3 = a(0 – 50)^2 + 53 = 2500a + 5-2 = 2500aa = -0.0008 This gives a final equation of y = -0.0008(x – 50)^2 + 5We can find the height of the arrow when it hits the target by plugging in 120 feet in for x.y = -0.0008(120 – 50)^2 + 5y = 4.888 feet The bulls eye is at 3 feet above the ground. Subtracting the height from 3, we will find the place on the target that the arrow hits.3 – 4.888 = -1.888 feet below the bulls eye-1.888 ft * 12 in/ft = 22.656 inches below the bulls eye Cool guy missed the bulls eye. In fact he missed the whole target. Maybe next time, Cool...
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...Programming Example: The Length of a Parabola Segment http://www.cs.mtu.edu/~shene/COURSES/cs201/NOTES/chap02/p-lengt... Problem Statement Given base b and height h, the length of a special segment on a parabola can be computed as follows: Write a program to read in the values of base and height, and use the above formula to compute the length of the parabola segment. Note that both base and height values must be positive. Solution ! ----------------------------------------------------------! Calculate the length of a parabola given height and base. ! ----------------------------------------------------------PROGRAM ParabolaLength IMPLICIT NONE REAL REAL :: Height, Base, Length :: temp, t 'Height of a parabola : ' Height 'Base of a parabola Base : ' * WRITE(*,*) READ(*,*) WRITE(*,*) READ(*,*) ! ... temp and t are two temporary variables t = 2.0 * Height temp = SQRT(t**2 + Base**2) Length = temp + Base**2/t*LOG((t + temp)/Base) WRITE(*,*) WRITE(*,*) WRITE(*,*) WRITE(*,*) END PROGRAM 'Height = ', Height 'Base = ', Base 'Length = ', Length ParabolaLength Click here to download this program. Program Output Height of a parabola : 100.0 Base of a parabola 78.5 Height = 100. Base = 78.5 Length = 266.149445 : The input values for Height and Base are 100.0 and 78.5, respectively. The computed length is 266.149445. Discussion 1 of 2 4/5/2014 9:30 PM Programming Example: The Length of a Parabola Segment http://www.cs.mtu.edu/~shene/COURSES/cs201/NOTES/chap02/p-lengt... ...
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...Section 3.1 - Quadratic Functions A quadratic function is a 2nd degree polynomial function There are two forms of the quadratic function: 1. The general form 2. The standard form The graph of every quadratic function is a parabola. The parabola opens up if and opens down if The highest or lowest point on the parabola is known as the vertex . When discussing relative maxima or minima, we write to mean that is a maximum (if the parabola opens down) or that is a minimum (if the parabola opens up). To obtain the vertex from the standard form simply write using the numbers from the equation (notice you are changing the sign of the from the equation). To obtain the vertex from the general form use the formulas , . (an alternate way to find is to evaluate once is known) Graphing A minimum of 3 points must be used; the vertex and one point on each side of the parabola. Therefore, to graph a quadratic function do each of the following: Determine whether the parabola opens up or down Determine the vertex Solve the equation for the x-intercepts Evaluate for the y-intercept Sketch the graph (if there are no x-intercepts, calculate additional points using the equation) Applications 1. You have 200 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the river, find the length and width of the plot that will maximize the area. What is the largest area that can...
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...(Dugopolski, 2012). Problem #26: h(x) =-3x2 The relation is h(x) =-3x2 h (0) =-3×02=0 h (-2)=-3×(-2)2=-12 h (-1) =-3×(-1)2=-3 h(1)=-3(1)2=-3 h(2)=-3(2)2=-12 X | h(x)=-3x2 | 0 | 0 | -2 | -12 | -1 | -3 | 1 | -3 | 2 | -12 | The above represents the points I will plot to graph the function h(x) = 3x2. The function is shaped like a parabola. This parabola opens down. The x-axis interception point of -3x2 :( 0, 0) and the y-axis interception of -3x2 (0,0). The domain is all real number, which can be written in standard notation as D= (-∞, ∞) and R= (-∞, 0). When I plot the parabola I notice it has a vertex of (0,0). From the above I can truly this relation and is also a function as every element of the domain has one and only once value associated with it in the range and passes through the vertical line test. Image below represents my function h(x) =-3x2. My second problem comes from page 711 #34.The problem states h(x) =-√x-1 h(x) =-√1-1=0 h(x) =-√2-1=-1 h(x) =-√3-1=-1.41 x | h(x) =-√x-1 | 1 | 0 | 2 | -1 | 3 | -1.41 | From the results, when I plot my graph it resembles half of a parabola, which opens to the left with a line which falls below the x-axis, except my starting point (1,0). This is functionsy=-x-1. The function has a range [0, -∞) because the line starts at and includes the x-axis, then it continues going down forever. The domain (left/right) would be [1,...
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