Problem #1: Consider a y that walks on a table along the path described by the parametric equations x=t+sin(2t), y=cos(t) for OStSG 3) Sketch a graph of the curve. To help visualize the curve, you can graph this eguation using GeoGebra. Set the domain to be zero to six. bl When does the fly change direction? Use calculus to answer this question (see hint below); show all work. Once completed, check your work with the graph from part a, then plot the points, on your graph in part a, corresponding to when the insect changes direction. Label the points on your graph. Hint: a change of direction can be going from an easterly direction to westerly (or vice versa) and from a northernly to southernly direction (or vice versa); it doesn't have to be an exact 180 degree change in the complete opposite direction. Think of it as changing from increasing to decreasing or from moving left to moving right. Problem #2: The Scrambler is a popular carnival ride consisting of two sets of rotational arms. Watch this video for an actual view of the ride. If we view the ride from an aerial position, we will notice a large arm connected to the central support of the ride. This large arm holds a group of buckets/seats. These large arms spin in a clockwise direction. At the end of each large arm is a spoke support holding a group of buckets/seats. The seats are held by a spokes connected to the spoke support. The spokes rotate in a counter-clockwise direction. Watch this desmos animation to see the ride from an aerial View. Bucketheat Spoke Support Central Support Let's suppose the large arm has a length of 2 meters and rotates clockwise. If we set the central support of the entire ride at the origin, we can describe the end of the arm, at time t seconds, using the parametric equations x = 23in (t), y = 2cos (it) Let's suppose the spokes have a length of 1 meter. in relation to the spoke support, we can describe the position of the bucket/seat (as a function oftime) using the parametric equations x = cos (2t) , y = sin (2t) Placing the entire structure at the origin of our x-y plane, the location of the bucket/seat is the sum of both parametric equations (you will learn why when we discuss vectors in calculus 2). In other words, the bucket/seat is in position (x,y) at time t, as described by the parametric equations x = 25in (t) + cos (2t) , y = 2cos (t) + sin (2:) You can graph this equation using GeoGebra. One advantage to using this applet is you can "trace" along the curve as t increases. Set the domain to be zero to 211', this represents one "rotation" around the central spoke. If you watch the desmos animation again, notice that the rider hits a stopping point, then accelerates to the next stopping point, and so on. a) Sketch a graph of the path of a rider using the GeoGebra app. b) Find the first 3 times the rider changes position (comes to a stop); use calculus to answer this question, your answer should not solely come from the graph. Your work: solve an appropriate equation and show all work. All solutions should be in an exact form not a decimal approximation. Hint: From part a, you should notice the rider changes direction at a corner on the graph. Use your knowledge about the derivative at a corner to find these points. c) From the work you did for part b, you should have 4 values of t that correspond to the derivative being undefined, but there are only 3 corners. Why did you have an extra solution? Explain visually by discussing which part of the graph this value oft corresponds. d) Does the rider actually stop at the moment they change direction? In order to answer part d, you need to show the speed is zero at a corner. We can't if use i as velocity because y(t} and x(tl describe the position of the rider; thus the . dy . . . units of 5 IS meters per meter not meters per second. You Will learn In Calculus 2, . . . dx 2 dy 2 and In phySIcs, that speed IS calculated as 5(t) = [E] + [3;] . Show that the speed is zero at the first time you found in part b, you don't have to do all 3 times. Show all work of plugging in the t-value and simplifying, don't just plug in to a calculator. Applications of Derivatives Parametric Eguations Background: Consider the curve given to the ." right. Suppose that it represents the path of (3, 51 a bug crawling on a table. If the bug starts walking from point (0,1), at time t = 0 seconds, it follows the curve in the direction indicated by the arrow. The location of the 10' 11 bug is traced by the curve as time progresses; in other words, the bugs location _ at a point (x,y) is dependent upon time. In this example, the location of the bug for the first 4 seconds is described by the parametric equations x = t2 2t , y = t+ 1 for 0 S t S 4 , where t represents time [in seconds). You should plug in the values t = O, 1, 2, 3, and 4 to see the points plotted on the curve. In general, we can sketch any curve in the x-y plane using a set of parametric equations, where we describe 2: and y as a function of a parameter r. We generally write the equations as x = f(t) , y = g(t) . In the example above, notice that the curve does not represent y as a function of x, since it fails the vertical line test. This is one advantage for using parametric equations, it provides a tool to sketch a curve in the x-y plane despite the fact the curve is not a function of x. This can be useful in many applications as you will learn in Calculus 2,3, and Differential Equations. In most applications the parameter usually represents time, so we use the variable t. Come back to the bug on the table: If we want to find when the bug changes direction we would use a derivative. Why? Think about our work with a function f(x). From our work in Chapter 3 we saw that the graph of a function changes from increasing to decreasing when f'(x) = 0 (a horizontal tangent} as shown in the graph on the right. For parametric curves, we have to "convert" our derivative, since we don't have 3/ = f(x). What we do have is x as a function of t, 2:0"), and y as a function of t, y(t). So, for parametric equations, we have to find the rate of change of y with respect to x using the formula dy dy E y'(t) E=E=xm E In words: find the derivate ofy with respect to t, then divide that by the derivate ofx with respect to t. In Calculus 2 we will derive where this formula comes from, for now let's just use it to do some problem solving