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Apr 7 2007, 01:28 AM
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Super Member ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 397 Joined: 9-June 06 From: Wisconsin Member No.: 24,924 |
If you remember stuff from your algreba class, you should recall that an equation that is used on a graph is usually in the form of y = x, or x = y, or something that includes both x and y. Parametric equations invovle a third variable, T, which is generally used to represent time. It takes two equations to graph a parametric equation, x = T and y = T. The x equation defines the x coordinate of the point you are plotting, and the y equation defines the y coordinate of the point.
The variable T is often referred to as time because parametric equations make it very easy to graph things like motion. It also makes it very easy to to graph lines at an angle. With y = x, you can kind of graph things at an angle, for example y = 2x will producer a steeper line than y = x. But with parametric equations you can graph and exact angle like so. This makes parametric equations somewhat helpful in trigonometry, or the math of finding the missing side or angle in a triangle. x = cos(Angle) * T y = sin(Angle) * T After about 30 minutes of really thinking about it, I have found a mathematical proof explaining why the above equation will generate a line with the angle that is specified by Angle. Every equation will have a starting point. For simplicity in the proof, lets call this point (0,0). This is the point where the line starts, and also is where the angle is formed. With a graphing calculator, you can have T start and end on any value, but in this case we will have it start on zero. We are drawing a line with a 45 degree angle thus our equations would be: x = cos(45) * T y = sin(45) * T Each parametric equation can be modeled by a right triangle, where the line being made is the hypotenuse of the triangle, x is the length of the side that runs horizontally, and y is the length of the line that runs vertically. I've drawn a picture to help illustrate my point. I appolgize for this being probably the worst drawn picture in existance. Im using a program for making icons. [If you once again remember back to your algebra days, you will remember three ratios for right triangles: sine (sin) opposite side length/ hypotenuse length cosine (cos) adajencent side length/ hypotenuse length tanget (tan) opposite side length/ adajcent side length I'd draw you another pretty picture but hopefully you already get the idea. Remember that whatever the angle of the triangle, the ratios between the sides will always be the same. Lets assume now that T is equal to one and we are plotting our next point. Go back to our original equations: x = cos(45) * T y = sin(45) * T Since T is equal to 1 x = cos(45) y = sin(45) Those are out points for x and y. However, you can also get the length of the legs based of of this. You already know that the angle between the horizontal line and the vertical line is going to a be 90 degrees. You also know the other angle measure, 45 degrees. The last angle measure can be determined by subtracting both of those from 180, because there are 180 degrees in a triangle. The length of the horizontal leg is equal to Tcos(45) - 0 (0 is our x starting point). The length of the vertical leg is equal to Tsin(45) - 0 (0 is our y starting point). The Law Of Sines states that any angle divided by its opposite side will be the same for every angle in the triangle. I learned this is Honors Geometry last year and Honors Advanced Algebra last year. I'll have a proof for you soon. However using this you can find out the angle between the the horizontal line and the line that you are plotting. We are solving for our angle. sin a (our angle) sin((180-90-a)) --------------------------- = ------------------------- length of vertical side length of horizontal side Using proportions, multiply this out to get: sin A * length of horizontal side = sin 45 * length of vertical side Multiply this done further, and you will find that the sin of our angle (45 degrees), is what you should get. Lastly, to use parametric equations on a graphing calculator (a TI one), press mode, and instead of func choose PAR or PARAM. |
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Apr 7 2007, 04:32 PM
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Premium Member ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 193 Joined: 26-April 06 From: NYC BABY Member No.: 22,568 |
Question: What Are Parametric Equations?
Answer: I HAVE NO IDEA Background Info: I am not good at math, I got a B on my report card for it. |
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Sep 29 2007, 11:44 AM
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#3
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Privileged Member ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: [HOSTED] Posts: 786 Joined: 13-April 07 From: mreža Member No.: 41,558 |
Question: What Are Parametric Equations? Answer: I HAVE NO IDEA Background Info: I am not good at math, I got a B on my report card for it. To put it simple for you man. Parametric equations are those equations that you can solve however one of the letters or as they are kjnown in the mathematical vocabulary are considered as any number and in fact any number can replace such parameter and according to that parameter other variables get their values. Also sometimes in certain cases people like to consider just specific case that is the reasons why they write that parameter is from certain numbers set for example (natural numbers or real numbers or any other data structure). Hopefully I helped to you. |
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