ࡱ> :<9fa jbjb,, "`NNLj       6'6'6'8n','6()L)))*. +!+ 5555555,b8R:~5 -+**-+-+5,  )) 6@,,,-+X ) )5,Db,    -+5,,:1,  E2( 76'+H1 E2L6<61R2;,2;E2, $Situation: y = sin (2x) Prepared at Penn State Mid-Atlantic Center for Mathematics Teaching and Learning 050718 Heather Godine Prompt This lesson takes place in a high school classroom. During a lesson on transformations of the sine function a student asks, Why is the graph of y=sin (2x) a horizontal shrink of the graph of y = sin (x) instead of a horizontal stretch? Commentary Mathematical Foci Mathematical Focus 1 The function y=sin(x) is periodic, and the horizontal length of one period of y=sin(x) is  EMBED Equation.3  radians. The function y=sin(2x) is a composition of the functions y=sin(x) and y=2x. Since the composition maps the outputs of y= 2x to the inputs of the sine function, an entire period will be the output of the function, y=sin (2x), for x-values from 0 to ( . Therefore, two periods or waves of y=sin(2x) will have a combined length of  EMBED Equation.3  radians. When a periodic function has a horizontal shrink transformation, the length of each wave of the transformed function will be shorter than length of each wave of the original function. Therefore, since each wave of y=sin(2x) has length  EMBED Equation.3  radians and each wave of y=sin(x) has length  EMBED Equation.3  radians, the graph of y=sin(2x) will be a horizontal shrink of the graph of y=sin(x).  figure 1 xy=sin (x)y=sin (2x)0000.2617991670.2590.5000.5235983330.5000.8660.78539750.7071.0001.0471966670.8660.8661.3089958330.9660.5001.5707951.0000.0001.8325941670.966-0.5002.0943933330.866-0.8662.35619250.707-1.0002.6179916670.500-0.8662.8797908330.259-0.5003.141590.0000.0003.403389167-0.2590.5003.665188333-0.5000.8663.9269875-0.7071.0004.188786667-0.8660.8664.450585833-0.9660.5004.712385-1.0000.0004.974184167-0.966-0.5005.235983333-0.866-0.8665.4977825-0.707-1.0005.759581667-0.500-0.8666.021380833-0.259-0.5006.283180.0000.000 figure 2  figure 3 Mathematical Focus 2 Consider a right triangle with vertices at the origin, on the x-axis, and a point on the unit circle [figure 4]. The hypotenuse of the triangle is a radius of the unit circle. Transform the triangle by moving the vertex on the unit circle around the circle. One period of the function y=sin( EMBED Equation.3 ) is created when the vertex on the unit circle travels a distance of  EMBED Equation.3  radians. The function y=sin(2 EMBED Equation.3 ) is also created by moving the vertex around the unit circle. Since the input of the function is 2 EMBED Equation.3 , one period of y=sin(2 EMBED Equation.3 ) will be created when the vertex on the unit circle travels a distance of  EMBED Equation.3  radians. As a result, the horizontal length of one period of y=sin(2 EMBED Equation.3 ) will be  EMBED Equation.3  radians, producing a horizontal shrink of y=sin( EMBED Equation.3 ).  figure 4 Mathematical Focus 3 The derivative of y=sin(x) is y=cos(x), and the derivative of y=sin(2x) is y=2*cos(2x). Since both cos(x) and cos(2x) are bounded by -1 and 1, the absolute value of 2cos(2x) will always be double the absolute value of cos(2x). Since the first derivative represents the rate of change of a function, the absolute value of the rate of change of y=sin(2x) will be double the rate of change of y=sin(x) at every point where the rate of change is non-zero. Therefore, the graph of y=sin(2x) will be steeper than the graph of y=sin(x) at every point where the rate of change is non-zero. If the graph of y=sin(x) were stretched horizontally, the transformed graph would slope less steeply than the graph of y=sin(x). Since the transformed graph of y=sin(2x) is steeper than the graph of y=sin(x) at every point where the rate of change is non-zero, the graph of y=sin(2x) cannot be a horizontal stretch of the graph of y=sin(x). 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