## Monday, May 12, 2014

### Understanding Parametric Equations

Suppose you wish to track the flight of a ball thrown and know how high the ball is after the ball have traveled 150 feet horizontally if the ball flight is represented by a parabolic curve. The curve can be represented as a function of time using two equations, one related to x (the horizontal distance) and t (representing time) and one related to y (the vertical distance) and t.

For example, suppose a player throws a ball from a height of six feet at an initial speed of100 feet per second at a 45 degree angle to the horizontal axis. After t seconds the location of the ball can be described as x = 100cos(45)t and y = 6 + 100sin(45) - 16t2. The x-component is the ball's horizontal distance in feet and y is the ball's vertical distance in feet. Where is the ball located 1.5 seconds after it is thrown? We substitute 1.5 in for t in both equations to get x to be approximately 106 feet and y to be approximately 40.7 feet. We can graph the movement of the ball by substituting values for t and plotting the values of x and y as the ordered pair (x,y) and connecting the dots.

Suppose we want to eliminate the parameter. We do this by solving for t in one of the two equations and substituting the value in for the other equation. For example, suppose we have the parametric equations x = 2t and y = t2 + 2. We'll solve for t in the first equation since it's easier. Therefore dividing both sides of the equation by 2, we get t = x/2. Now using t = x/2 and y = t2 + 2, we get y = (x/2)2 + 2. Notice we just did a simple substitution, putting x/2 in for y in for t in the second equation. Simplifying the equation, we get y = (1/4)x2 + 2. One familiar with the equations of conic sections knows that this is a parabola. Written in standard form the equation is x2 = 4(y - 2). This is a parabola that opens up with vertex at (0, 2).

Suppose we want to graph a curve represented by the parametric equations x = t + 1 and y = √t. First use y = √t and square both sides to solve for t. Therefore we get y2 = t. Now substitute y2 for t in the first equation to get x = y2 + 1. We Know this is a parabola with the vertex at (1,0) and opening to the right. You may be tempted to graph the entire parabola, but remember the restriction on t in the equation y = √t. This is only defined then t is greater than or equal to 0. Therefore, y cannot be negative and we only graph the lower half of the parabola.

Let's try one more example. Suppose a curve is represented by the parametric equations x = 3sin t and y = 5cos t. First get the trigonometric function by itself by dividing the first equation by 3 and the second equation by 5. Doing so, gives us x/3 = sin t and y/5 = cos t. Now square and add these two quantities and you'll see this all fall into place. Squaring and adding gives x2/9 + y2/25 = sin2t + cos2t. Recall from the Pythagorean identities that the right side of this equation equals 1, therefore we have x2/9 + y2/25 = 1.

Those familiar with the equations of conic sections will see immediately that this is an ellipse centered at (0,0) with y being the major axis of length 10 and x being the minor axis of length 6. The ellipse will be elongated vertically 10 units and horizontally 6 units.

Parametric equations are very useful in solving problems involving vertical and horizontal distance. This guide should give students the basics on understanding and using parametric equations.