ࡱ> TVS_'` bjbjLULU 7.?.?K    %%%8<%4p&t ~T&rb+4+++333SSSSSSS$VhXtSE 33333S++e8T};};};3jt+ +S};3S};};6L  N+& pa}%i4*}MR<NT0~TMbY5bY8NbY N33};33333SS;^333~T3333   $@#   @#     Section V: Parametric and Implicit Equations  Chapter 2: Introduction to Implicit Equations When we describe curves on the coordinate plane with algebraic equations, we can define the relationship between the x- and y-coordinates of the curve either explicitly or implicitly. To explain the difference, lets consider the equation  EMBED Equation.DSMT4 . % The equation  EMBED Equation.DSMT4  establishes a relationship between x and y. We say that y is defined explicitly in terms of x since the equation gives us a very clear description of the relationship between x and y: no matter what the x-value is, the corresponding y-value is always  EMBED Equation.DSMT4 . % If we subtract  EMBED Equation.DSMT4  from each side of this equation we obtain the equation  EMBED Equation.DSMT4 . Clearly this equation describes the same relationship between x and y as the equation  EMBED Equation.DSMT4 , but when y isnt isolated, the relationship between the variables isnt explicit. Instead, this relationship is implied by the equation, so we say that it is an implicit equation. An equation involving two variables is considered explicit if one of the variables is isolated on one side of the equation, while and equation is considered implicit if neither of the variables is isolated on one side of the equation. As weve seen, equations like  EMBED Equation.DSMT4  can be written either explicitly or implicitly, but many equations can only be written implicitly. For example, its not possible to solve the implicit equation  EMBED Equation.DSMT4  for either variable. If we try to solve the equation for y, we need to use two explicit equations to communicate the information in the implicit equation:  EMBED Equation.DSMT4   example 1: Recall from Example 4 in the previous chapter that the system of parametric equations below describe a unit circle centered at the point  EMBED Equation.DSMT4 :  EMBED Equation.DSMT4  As we learned in the previous chapter, we use the Pythagorean Theorem to eliminate the parameter t and transform this system into a single implicit equation:  EMBED Equation.DSMT4  Thus, the implicit equation  EMBED Equation.DSMT4  describes a unit circle centered at the point  EMBED Equation.DSMT4 .  Since the parametric system  EMBED Equation.DSMT4  describes a circle of radius 1 unit, we should expect that if we multiply both the x- and y-coordinate by the factor r we will stretch (or compress) the circle so that its radius will change from 1 unit to r units, and if we add constants h and k to the x- and y-coordinates, respectively, we will shift the center of the circle from the point  EMBED Equation.DSMT4  to the point  EMBED Equation.DSMT4 . A generalized parameterization of a circle is given in the box below. If  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  then the system of parametric equations below defines a circle of radius r centered at the point  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4    example 2: Eliminate the parameter t from the system of parametric equations given in the box above to obtain an implicit equation that defines a circle of radius r centered at the point  EMBED Equation.DSMT4 . SOLUTION: We can use the Pythagorean Theorem to eliminate the parameter just as we did in Example 1. The Pythagorean Theorem involves  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  so we need to first we need to isolate  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  in the equations in our system:  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4  Now, we can substitute the expressions  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  for  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  in the Pythagorean Theorem and obtain an implicit equation for the circle:  EMBED Equation.DSMT4  If  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  then the implicit equation  EMBED Equation.DSMT4  defines a circle of radius  EMBED Equation.DSMT4  centered at the point  EMBED Equation.DSMT4 .   example 3: Determine the center and radius of the circle defined by the implicit equation  EMBED Equation.DSMT4 . SOLUTION: Notice that the equation  EMBED Equation.DSMT4  has the form  EMBED Equation.DSMT4  where  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 , and  EMBED Equation.DSMT4 . Thus, the center of the circle is at the point  EMBED Equation.DSMT4  and the radius is 3 units. We could check if we are correct by graphing the circle on our graphing calculator, but our calculators arent able to graph implicit equations. If we transform our implicit equation into a system of parametric equations, we could use our graphing calculator to graph the circle. To transform the implicit equation into a system of parametric equations, we mimic what we did in Example 2 but do everything in the opposite order. First, lets get 1 on the right side of the equation by dividing both sides by  EMBED Equation.DSMT4  (i.e., 9):  EMBED Equation.DSMT4  We now have an equation in which the sum of two squares is equal to 1. Since Pythagorean Theorem has this same form (i.e., since  EMBED Equation.DSMT4 ), we can let  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . (Notice that if we substitute these values for  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  in the Pythagorean Theorem we obtain the equation labeled  EMBED Equation.DSMT4  above.) We can solve these equations for x and y to obtain a parameterization of the given implicit equation:  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4  Thus, the system of parametric equations below describes the same circle as the implicit equation  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4  Notice that from the generalization given at the beginning of Example 2, we can see that this parameterization describes a circle of radius 3 units with center  EMBED Equation.DSMT4  which is exactly what we were looking for.   example 4: The graph of the equation  EMBED Equation.DSMT4  is a circle. Identify the center and radius (by using appropriate algebra). SOLUTION: In order to determine the center and radius of the circle, we need to get our implicit equation into the form  EMBED Equation.DSMT4 . To accomplish this, we can complete the square twice. (If you need to review the procedure for completing the square, see page A40 in the appendix of our textbook as well as page 144 in 2.3 in our textbook.)  EMBED Equation.DSMT4  Thus, the center of the circle is at the point  EMBED Equation.DSMT4  and the radius of the circle is 4 units. 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EMBED Equation.DSMT4  Graph the system on your graphing calculator to check.      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