In Algebra 2 we learned to solve systems of equations with more than two variables. Generally the technique used was elimination. The goal was to get a system that looked like: X + Y+ Z = 6 X - Y + Z = 2 2X - Y + Z = 3 Into a format where we had a triangular row of zeros such that the system was modified to look like: X + Y + Z = 6 Y + Z = 5 Z = 3 (where the coefficients for the missing variables were 0) This allows us to use the value of Z to solve for Y and then to use Y and Z to solve for X. We can also represent this format in matrix format which allows for a shorthand way to write the system as well as various shorter methods to solve systems of equations using matrices. The first assignments and videos below review the solving of a system of equations using elimination without matrices. The next section than represents a way to solve similar systems of equations in matrix format. 9.3 Several Variables pp.657 – 58: 6, 8, 10, 16, 20 9.3 (cont-d) p. 658: 15, 17, 19, 25, 31 9.3 Dependent p.658: 18, 22, 24, 26, 30 These problems should be completed by Thursday, May 7. System of equations with 3 variables part 1 www.youtube.com/watch?v=wIE8KSpb-E8 System of equations with 3 variables - part 2 www.youtube.com/watch?v=5FvY8XLrqmM Examples solving systems using elimination: www.youtube.com/watch?v=3RbVSvvRyeI&feature=youtu.be www.youtube.com/watch?v=EytTXf8_KYA&feature=youtu.be No solution Case www.youtube.com/watch?v=ryNQsWrUoJw&feature=youtu.be Infinite Solution Case www.youtube.com/watch?v=mThiwW8nYAU&feature=youtu.be The notes and answer keys for these problems are at the very end of this post. Representing Systems using Matrices A matrix (plural matrices) is a format to represent data. A matrix represents data with rows and columns of numbers. Rows are horizontal and columns are vertical. A matrix can be used to represent a system of equations. Example: 2X + 3Y = 25 4X - 7Y = 22 can be represented in matrix form as _____ _____ | 2 3 25 | The elements of a matrix are enclosed in the | 4 -7 22 | [ ] symbols. ____ ____ Note that the rows represents each of the equations. The columns represent the various variables and the constants in the equation. X's are the first column. Y's are the second column. The constants are the 3rd column. In this first lesson, we will discuss syntax and notation of matrices, representing systems of equations using matrices, and reducing matrices/solving systems of equations using the method of Gauss -Jordan elimination / row-reduction. Homework: Text page 673: #'s 2, 4, 6, 16, 18, 20 Due: Monday 5/11/20 Homework: Text page 674: #'s 30, 34, 38, 42 Due: Tuesday 5/12/20. (More practice on matrices.) Videos: Basic notation and description of the order or size of a matrix. www.youtube.com/watch?v=ilFJYjfKYjk&feature=youtu.be Changing an equation into matrix format and then reducing the matrix to row-echelon form using row/reduction or Gauss Jordan elimination. www.youtube.com/watch?v=BWBckWPjfpw Solving a system of equations (2 x 2) using an augmented matrix www.youtube.com/watch?v=V8mb5BFmJO0&feature=youtu.be Solving a system of equations with no solutions using matrices www.youtube.com/watch?v=NgXXKmQHFDg&feature=youtu.be Solving a system of equations with infinite solutions using matrices www.youtube.com/watch?v=9_wGz6zcZ6s&feature=youtu.be Solving a system of three equations using a system of matrices www.youtube.com/watch?v=WiVeiVIu_SM&feature=youtu.be Solving a system of equations using matrices and REDUCED row-echelon form. www.youtube.com/watch?v=-bPPDq0Y8s4 Using the TI-84 matrix menu to enter a Matrix and show row operations on a Matrix www.youtube.com/watch?v=ABY5rYu3Rmc&feature=youtu.be Using the TI-84 to show Matrix operations on the home screen of the calculator www.youtube.com/watch?v=syx9LY7qMgo&feature=youtu.be There are more videos at the page www.mathispower4u.com/alg-2.php if you wish to see more examples. On the page, press the control key and the f key to open up a search box and type matrix to see the list of videos. lesson_plan_-_section_9.4_-_matrices_-_day_1.pdf Download File answer_key_-_section_9.4_-_matrices_day_1.pdf Download File answer_key_-_section_9.4_matrices_day_2.pdf Download File
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Good luck on your test. Let me know if you have any questions.
Remember, once you submit your responses, you will not be able to go back and change them. Starting next week we will be working on Matrices. docs.google.com/forms/d/e/1FAIpQLScp6fZ8cUqcPf3iI0FU3JTqgoXxkEM-lf1T0xvSnU5fohyOCA/viewform The definition of a hyperbola is the set of all points in a plane, the DIFFERENCE of whose distance from two fixed points (foci) is constant. The definition is similar to an ellipse, but instead of adding the distances from a point on the hyperbola to each of the two foci, we subtract those distances. For any point that you pick on the hyperbola, when you take the two distances from that point to each of the foci and subtract them, you will get the same value. The general format of the equation for a hyperbola is similar to that of an ellipse. The difference is that you have a subtraction sign between the x squared and y squared terms. The general form of the equation of a hyperbola centered at (0,0) is below. Unlike an ellipse, where the orientation was determined based on whether the bigger denominator was under the x or the y, with a hyperbola, the orientation is determined based on which coordinate is being subtracted, the x or the y. For the hyperbola with the horizontal transverse axis (y^2 is being subtracted), the basic properties are as follows: Vertices are at (a,0) and (-a,0) Equations of asymptotes are y= - b/a and y = b/a (slope m= (change in y) / (change in x) Note In this case the b is under the y and the a is under the x which is why we use b/a (y/x). Foci are at (c,0) and (-c,0) where c is defined as c^2 = a^2 + b^2. For a parabola with a vertical transverse axis (x^2 is being subtracted), you are pretty much swapping the coordinates. Vertices are at (0,a) and (0,-a) Equations of asymptotes are y= - a/b and y = a/b (slope m= (change in y) / (change in x) Note In this case the a is under the y and the b is under the x which is why we use a/b (y/x). Foci are at (0,c) and (0,-c) where c is defined as c^2 = a^2 + b^2. If you have a horizontal and/or vertical shifts, the format of the equations looks as follows: Note that to find the asymptotes for a hyperbola with a shifted center, use the point slope formula with (h,k) as your point, and the slope still being +/- ( a/b) or (b/a) depending on the orientation of the hyperbola. Included at the end of this post are some more notes on hyperbolas with examples worked out, the homework answers worked out and some good websites with videos on hyperbolas. Videos - basics of a hyperbola www.youtube.com/watch?v=i6vM82SNAUk Part I www.youtube.com/watch?v=6Xahrwp6LkI Part II Graphing hyperbolas centered at the origin. www.youtube.com/watch?v=W0IXOdsna9A&feature=youtu.be www.youtube.com/watch?v=vT-gnaAG51o&feature=youtu.be Graphing a hyperbola not centered at the origin. www.youtube.com/watch?v=_ssQfu8N6XE&feature=youtu.be www.youtube.com/watch?v=0qAB_5v_2G4&feature=youtu.be Finding the equation of a hyperbola given center, focus, vertex www.youtube.com/watch?v=ZwqNou_Z7oI&feature=youtu.be Not a video site, but goes into nice detail on how the calculations are derived. www.ck12.org/book/ck-12-algebra-ii-with-trigonometry-concepts/section/10.7/ www.ck12.org/book/ck-12-algebra-ii-with-trigonometry-concepts/section/10.8/ www.ck12.org/book/ck-12-algebra-ii-with-trigonometry-concepts/section/10.9/ Homework Assignment: pages 768-69: 2, 4, 6, 16, 18, 20, 28 pages 781-82: 10, 12 Due Wednesday - 4/29/20 There will be a short assessment on this unit at the end of this week on Friday 5/1/20. As always, shoot me an email with any specific questions you may have on the problems. Stay Healthy!!!!
This next unit is called conic sections and deals with the equations of parabolas, ellipses and hyperbolas. It has a number of formulas in it that are very similar so it's easy to mix them up. Pretty much, the problems in this section consist of two main problem types: 1) getting the formula into the standard form for an ellipse, hyperbola or parabola 2) Given some information about the conic section, use that information to figure out the equation Think back to the equations of a line. You know the formula is y=mx+b, but if you are not given m and b, you can use information given such as two points or the slope and a point to find the needed information m and b. Then you can fill out the m and b to write the line's equation. The idea is similar here except the equations are a little more involved and often require more algebraic manipulation to get the equation to standard form. Attached below are the notes, and answer keys for the home works as well as links to some good videos explaining in more detail. Look through the work and let me know what questions you have if you get stuck on something. This post will contain some more explanations on parabolas and ellipses. A second post will be out soon discussing hyperbolas in more detail. The formulas that you should have for handy reference are on pages 745, 747, 755, 757, 763, and 776 (776 deals with horizontal and vertical shifts of the different conic sections). First Section: Parabolas The first conic section to be discussed will be the parabola. The two parts that define the path of a parabola are the Focus ( a point) and the directrix (a line). The definition of a parabola is the set of all points equally distant from a focus and the directrix. If some point (x,y) on a given parabola is a distance of 5 from the focus, it is also 5 from the directrix. If another point(x2,y2) is a distance of 10 from the focus on the this same parabola, it is also a distance of 10 from the directrix. Every point on the parabola meets this requirement. The focus will be somewhere on the axis of symmetry of the parabola (the line that bisects the parabola). The directrix line and the focus point are on the opposite sides of the turning point or vertex of the parabola and the vertex will be halfway between the focus and directrix. General format of equations: Given a parabola with a vertical axis of symmetry. (points up and down) and a vertex at (0,0) Vertex = (0, 0) Focus point is (0,P) (P = focal point, is on the y axis) Directrix is the equation y = - p General format of the equation is: x^2 = 4py Given a parabola with a horizontal axis of symmetry (points left and right) and a vertex at (0,0) Vertex = (0,0) Focus point is (P,0) (P = focal point on the x axis) Directrix is the equation x = -p General format of the equation is y^2 = 4px If the parabola has been shifted either horizontally, vertically or both, the general format of the equations are: vertical parabola = ( x-h)^2 = 4p(y-k) where the vertex has been shifted to the point (h, k) Note how the sign of h and k change when going between equation and vertex. horizontal parabola = (y - k)^2 = 4p(x-h) where the vertex has been shifted to the point (h,k) Note in the formulas above that: The x coordinate of the vertex, (h) is always with the x in the equation The y coordinate of the vertex ,(k) is always with the y in the equation. When trying to find the vertex, focus and directrix of a parabola that does not have its vertex at the origin (0,0), it is often helps to figure out what these would be without the shift and then applying the shift. For example: Equation = (x-2)^2 = 12(y-3) This is the formula for a parabola that is vertical The parenthesis tell us the shift is 2 right on the x axis and 3 up on the y (Shift is always in the opposite direction of the sign) Shift = (2,3) Totally ignore the shift in parenthesis and the equation looks like x^2 = 12 y This is the general form of x^2 = 4p y 1) Solve for p. 4p = 12 in this example (The coefficient in front of y always = 4p). So p must = 3. Therefore from the formulas above: Pre - shift Shift(2,3) Post Shift Vertex = (0,0) x+2, y+3 Vertex = (2,3) Focal point = (0,3) Focal Point = (2,6) Directrix = y= - 3 Directrix y = 0 (added 3 to the y from the shift) Attached below are notes explaining in more detail and examples worked out as well as some helpful links. The answers to the homework questions are posted again as well. Homework problems: p. 751 #'s 2,4,6,12,28,40 pp. 781-782 #'s 6,8,14,22 General description of various conic sections. www.youtube.com/watch?v=iJOcn9C9y4w Description of parabola conic section and basic equations www.youtube.com/watch?v=k7wSPisQQYs Parabola part II www.youtube.com/watch?v=CKepZr52G6Y There are a number of specific examples on this page for graphing parabolas with and without shifts. Once you click on the link below, click Ctrl f and a search box opens up. Type parabola and you will be taken to the section of the page with many different videos. www.mathispower4u.com/alg-2.php Section on ellipses Geometric definition of an ellipse - The set of all points in a plane whose sum of distances from 2 fixed points called foci (plural of focus) is constant. The shape of an ellipse is like a circle that got squashed so it resembles a football. Inside the ellipse are two fixed points called foci. Pick any point on the ellipse and find the distance from it to focus 1, call it d1. From that same point, find the distance from it to focus 2, call it d2. The sum of d1 + d2 will always add to the same number no matter what point you chose to pick. Just like with the parabolas, these problems rely on knowing the general format of the equations and applying them based on the given information. x^2/a^2 + y^2/b^2 = 1. Equation of an ellipse centered at the origin. The general formats for the equations of ellipses centered at the origin and with shifts are in the word document below. The orientation of the ellipse, whether elongated more along the vertical axis or the horizontal axis is determined by which is the bigger denominator, a^2 or b^2. If the bigger denominator is under the x, then the main axis (longest) is the horizontal. If the biggest denominator is under the y, then the main axis (longest) is the vertical. So when graphing, the important thing is which coordinate is over the larger denominator, x or y. Below are some links showing the basic graph of an ellipse, and to solve some problems. Also included are notes with problems worked out and the answer key with all the homework problems worked out in detail. As always, please email me with any specific questions. Some time by the end of this week or early next week, you will have a short assessment on these problems. Homework problems: pages 759-60: 2, 4, 6, 10, 14, 20, 30 pages 781-82: 2, 4, 16 ( Hint for when you have difficulty getting the coefficient in front of the x^2 or y^2 to 1 when the right side is already 1 e:g: 9x^2 / 1 + 10y^2 / 1 = 1) Change the coefficient in front of the x^2 or y^2 to 1 and represent the denominator as 1/Coefficient of what had been in front of the x or y. The above equation is equal to: x^2 / (1/9) + y^2 / 1/10 = 1 This is now in standard form. Link for ellipses. Basic concepts. Parts 1 and 2 www.youtube.com/watch?v=LVumLCx3fQo www.youtube.com/watch?v=oZB69DY0q9A Graphing ellipses centered at the origin with major axis along x and along y www.youtube.com/watch?v=azI5kALyiXs&feature=youtu.be (horizontal axis) www.youtube.com/watch?v=3qckea8OuN8&feature=youtu.be (major axis vertical) Graphs not centered at the origin: www.youtube.com/watch?v=tJmp1PJD9o8&feature=youtu.be www.youtube.com/watch?v=oWGyVpq94CM&feature=youtu.be Given an equation, rewrite in standard form and graph. www.youtube.com/watch?v=-i48L0WQ-2I&feature=youtu.be www.youtube.com/watch?v=CL7SNu-5riw&feature=youtu.be Given some information about the ellipse (e.g focus, vertex, etc), find the equation. www.youtube.com/watch?v=k5gMLDjMfvI&feature=youtu.be www.youtube.com/watch?v=RWaEIJOlHlw&feature=youtu.be There are a number of videos on this website for ellipses: www.mathispower4u.com/alg-2.php in addition to the ones posted above. As far as a timeline goes, I would aim to have the questions in this post done by Monday of next week - 4/27/20. As always, email me with any questions
Click on the link below to take the test on the Polar and parametric equations unit.
Good luck! docs.google.com/forms/d/e/1FAIpQLScD4MKe5lJEXZ3StxqvQNmqirfiDjWox7hk5CK7qCAwRS-ZvA/viewform In addition to graphing and converting polar equations to rectangular equations from the previous post, there is one other type of equation you need to understand.
The last type of equations to review are parametric equations. A parametric equation is similar to the normal rectangular equations with x's and y's, except they have an extra parameter (variable) for time - (t). It is generally used to track the course of an object along some kind of path (this is an equation in terms of x and y coordinates) but the t tells where the object is located on the path at a give time t: - Think of tracking the path of an asteroid or comet as it passes through the solar system. The entire track of the comet is generally described by the equation of an ellipse, but the t component tells where the comet is on this elliptical path at time t. When converting from parametric to rectangular equations, you want to be able to get rid of all the t's in the equations. The main techniques are to either 1) Rewrite one of the equations to get t by itself on one side with all the x's and y's on the other. Then then use that value for t in terms of x and y as a substitute in the second equation to replace all ts. 2) If you have sin t and cos t and you want to get a t by itself, you will often need to write in terms of the identity (sin t)^2 + (cos t)^2 = 1. And then do a suitable substitution for t. Some helpful videos: Graphing parametric equations with the TI 84 www.youtube.com/watch?v=4Y14XhPD7Os Converting parametric equations to rectangular form www.youtube.com/watch?v=tW6N7DFTvrM Converting parametric equations of a line into a rectangular equation www.youtube.com/watch?v=BGWX0TOthsY&feature=youtu.be The files included with the beginning posts of this unit have some examples worked out. After you go over these types of problems, try the questions on the review sheet. The answer key was posted as well. Let me know what questions you have. If you are ready for it by Friday, I will give you a link to an assessment. You will have 24 hours to complete the test once I post the link. I would prefer to get it done by this weekend so we can start fresh with the next unit on Monday.. Otherwise, I will assign the test to be due by next Tuesday. Email any questions you may have. Per your mom, you also have a phone contact to which you may text messages. Don't be shy about asking questions and stay healthy.. Now that you have some understanding of converting single points between polar and rectangular form, we are going to move to the next step which is changing equations back and forth between polar and rectangular form.
You already have the notes and homework answer keys from the previous posts. Below are additional notes and links to some videos that demonstrate some examples. Please email me or send me a text if you have questions. I would like to give an assessment by the end of this week. Essentially you are doing identities where you are converting the trig into polar form of an equation or vice versa. To convert from polar to rectangular you want to get rid of all the r's and theta's in the equation and change to terms of x and y. To summarize: There are three main types of techniques as listed below to convert between polar and rectangular forms: 1) You can multiply both sides by r to get: r squared - converts to x squared + y squared or to get: r cos theta - converts to x or to get: r sin theta - converts to y 2) Sometimes you only have r on 1 side of the equation and you don't want to multiply r to the other side because it is already in x and y terms. In this case, SQUARE both sides of the equation so that one side becomes r squared which is now = x squared + y squared. The other side is already in x and y terms, but now squared. So it is still in rectangular form. 3) If you have a theta by itself on one side, take the tangent of both sides of the equation. Tan(theta) = Tan(right side) Since tan x = sin x/cos x (Identity), which is exactly the same as r sin x / r cos x, the left side can now be written as y/x. Then finish the algebra to solve the equation. Some helpful videos. Graphing polar equations in the TI 84 https://www.youtube.com/watch?v=PZwiiZQhM0c&feature=youtu.be Example of a conversion from rectangular (circle) to polar form. https://www.youtube.com/watch?v=tUH6tUiJbH8&feature=youtu.be Example converting a parabola in rectangular form to polar form. https://www.youtube.com/watch?v=XkZAtV8jNb4&feature=youtu.be Video showing how to convert equations from polar back to rectangular coordinates form. https://www.youtube.com/watch?v=IKbRiU7kL2w Email me with your questions. The first video below explains how to map a point on a polar coordinate grid instead of the usual (x,y) Cartesian coordinate system we normally use. The format is (r,theta) theta = the angle
Where r represents how far out from the center pole you are (distance) and theta is the angle you are moving away from the central pole (distance at pole itself is r = 0) https://www.youtube.com/watch?v=-tZR3ggdoIU The next video shows how to give different equivalent representations in polar form for the same exact point. https://www.youtube.com/watch?v=ds-YLa7Yd0k The next video below gives some examples of polar and rectangular coordinates being graphed. https://www.youtube.com/watch?v=A-AEuohaP4o The next video shows how to convert from (x,y) rectangular coordinates to the equivalent in polar coordinates (r, theta). https://www.youtube.com/watch?v=Vg88CWOlDTY The next video shows how to convert from polar coordinates( r, theta) back to rectangular coordinates (x,y). https://www.youtube.com/watch?v=Txx9rvLnuTA Note that polar form is really just a short form of writing the trig form of a vector. If the trig form of a vector is 5(cos 90i +sin90j) then polar form is written as (5, 90) The 5 is the r and the 90 is the theta. You do distribution to go to (x,y) form. cos 90 = 0, and sin 90 is 1 so trig form of 5(cos90i + sin90j) is exactly the same as polar form of (5,90) = (5x0, 5x1) = (0,5) in rectangular or (x,y) form. To write the rectangular form in polar form. r is found by doing the pythagorean theorem on the x,y values r = the square root of x squared + y squared. theta is found the same way as finding the angle with vectors that 3 step process we talked about 1) sketch the vector, 2) arctan of y/x to find the reference angle 3) put the reference angle into the sketch to find the actual angle. Attached are the notes, homework assignments and answer keys for converting between polar and parametric equations. I believe that you received this packet at our last session even though we did not go over much (if any) of it at our last session. You need not print it out if you just wish to view it from the computer. By tomorrow afternoon, I would like you to look at, (not necessarily complete), but open up the following pdf files below: 1) the assignments doc, 2) the first lesson plan section 8.1 below (polar coordinates) 3) The answer key to the assignment for section 8.1 polar coordinates 4) Day 2 handout - worksheet - , polar coordinates 5) Answer key - detailed - handout on worksheet on polar coordinates. This is the first lesson in the unit. Polar coordinates are very similar to trig form of a vector or complex number. I want you to look at these files and compare the problems in them to your vector notes. After you look that over, send me an email with your thoughts, questions, etc and we will continue from there.
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