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Graphing parametric equations and rectangular equations on the coordinate system

Graph the parametric equations x = t + 1 and y = t , t 0 , and the rectangular equivalent y = x 1 on the same coordinate system.

Construct a table of values for the parametric equations, as we did in the previous example, and graph y = t , t 0 on the same grid, as in [link] .

Overlayed graph of the two versions of the given function, showing that they are the same whether they are given in parametric or rectangular coordinates.
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Sketch the graph of the parametric equations x = 2 cos θ and y = 4 sin θ , along with the rectangular equation on the same grid.

The graph of the parametric equations is in red and the graph of the rectangular equation is drawn in blue dots on top of the parametric equations.

Overlayed graph of the two versions of the ellipse, showing that they are the same whether they are given in parametric or rectangular coordinates.
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Applications of parametric equations

Many of the advantages of parametric equations become obvious when applied to solving real-world problems. Although rectangular equations in x and y give an overall picture of an object's path, they do not reveal the position of an object at a specific time. Parametric equations, however, illustrate how the values of x and y change depending on t , as the location of a moving object at a particular time.

A common application of parametric equations is solving problems involving projectile motion. In this type of motion, an object is propelled forward in an upward direction forming an angle of θ to the horizontal, with an initial speed of v 0 , and at a height h above the horizontal.

The path of an object propelled at an inclination of θ to the horizontal, with initial speed v 0 , and at a height h above the horizontal, is given by

x = ( v 0 cos θ ) t    y = 1 2 g t 2 + ( v 0 sin θ ) t + h

where g accounts for the effects of gravity and h is the initial height of the object. Depending on the units involved in the problem, use g = 32 ft / s 2 or g = 9.8 m / s 2 . The equation for x gives horizontal distance, and the equation for y gives the vertical distance.

Given a projectile motion problem, use parametric equations to solve.

  1. The horizontal distance is given by x = ( v 0 cos θ ) t . Substitute the initial speed of the object for v 0 .
  2. The expression cos θ indicates the angle at which the object is propelled. Substitute that angle in degrees for cos θ .
  3. The vertical distance is given by the formula y = 1 2 g t 2 + ( v 0 sin θ ) t + h . The term 1 2 g t 2 represents the effect of gravity. Depending on units involved, use g = 32 ft/s 2 or g = 9.8 m/s 2 . Again, substitute the initial speed for v 0 , and the height at which the object was propelled for h .
  4. Proceed by calculating each term to solve for t .

Finding the parametric equations to describe the motion of a baseball

Solve the problem presented at the beginning of this section. Does the batter hit the game-winning home run? Assume that the ball is hit with an initial velocity of 140 feet per second at an angle of 45° to the horizontal, making contact 3 feet above the ground.

  1. Find the parametric equations to model the path of the baseball.
  2. Where is the ball after 2 seconds?
  3. How long is the ball in the air?
  4. Is it a home run?
  1. Use the formulas to set up the equations. The horizontal position is found using the parametric equation for x . Thus,

    x = ( v 0 cos θ ) t x = ( 140 cos ( 45° ) ) t

    The vertical position is found using the parametric equation for y . Thus,

    y = 16 t 2 + ( v 0 sin θ ) t + h y = 16 t 2 + ( 140 sin ( 45° ) ) t + 3
  2. Substitute 2 into the equations to find the horizontal and vertical positions of the ball.

    x = ( 140 cos ( 45° ) ) ( 2 ) x = 198  feet y = 16 ( 2 ) 2 + ( 140 sin ( 45° ) ) ( 2 ) + 3 y = 137  feet

    After 2 seconds, the ball is 198 feet away from the batter’s box and 137 feet above the ground.

  3. To calculate how long the ball is in the air, we have to find out when it will hit ground, or when y = 0. Thus,

    y = 16 t 2 + ( 140 sin ( 45 ) ) t + 3 y = 0 Set  y ( t ) = 0  and solve the quadratic . t = 6.2173

    When t = 6.2173 seconds, the ball has hit the ground. (The quadratic equation can be solved in various ways, but this problem was solved using a computer math program.)

  4. We cannot confirm that the hit was a home run without considering the size of the outfield, which varies from field to field. However, for simplicity’s sake, let’s assume that the outfield wall is 400 feet from home plate in the deepest part of the park. Let’s also assume that the wall is 10 feet high. In order to determine whether the ball clears the wall, we need to calculate how high the ball is when x = 400 feet. So we will set x = 400, solve for t , and input t into y .

         x = ( 140 cos ( 45° ) ) t 400 = ( 140 cos ( 45° ) ) t       t = 4.04      y = 16 ( 4.04 ) 2 + ( 140 sin ( 45° ) ) ( 4.04 ) + 3      y = 141.8

    The ball is 141.8 feet in the air when it soars out of the ballpark. It was indeed a home run. See [link] .

Plotted trajectory of a hit ball, showing the position of the batter at the origin, the ball's path in the shape of a wide downward facing parabola, and the outfield wall as a vertical line segment rising to 10 ft under the ball's path.
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Questions & Answers

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