The questions are in the file and have to be done in Ms word. Use a word equation editor to enter equations and all work must be shown.
AEP 8 - Implicit Functions and Implicit Differentiation Overview In this AEP, you will explore some interesting curves in the plane and the implicit functions that arise. Youwill use implicit differentiation tofindan expression for the slope of each of these curves. You will learn how to use Geogebra to perform implicit differentiation. This topic is related to Learning Targets 7 and 10, and is based on the material in Section 2.7 of Active Calculus. This AEP is required for students aiming for a course grade of B or higher. Remember, AEPsdonothavefixeddeadlines; simplyworkon this itemuntil youare ready to submit it. But remember the Two Items PerWeekRule. Prerequisites and tech requirements You will be ready to begin this AEP followingModule 6C. Technologyused in thisAEP:Youwill be using theGeogebraCASCalculator to plot implicit curves, to check work that you have done by hand, to compute implicit derivatives that would be extremely tedious to find by hand, to plot tangent lines to implicit curves, and to solve systems of equations. AEPDescription and Tasks What this AEP is about Up until Module 6C, the graphs you have been investigating have been graphs of functions: they satisfy the vertical line test. Such graphs arise frequently in applications. Other kinds of graphs are also important - graphs that do not correspond to the graph of a function. For example, this graph is not the graph of a function: Thisgraph is a contour curve for a surface in threedimensions. Here is aGeogebra visualizationof the surface and some of its contour curves. Adjust the slider to n=1 to see the curve in the image above. The surface on the right can be manipulated with your mouse. (If you see a gap in the curve on the left, try zooming in or out. It is not a ”real” gap, but an artifice of using a computer to construct the display.) Your work in Module 6C shows that it is possible to determine the slope of tangent lines to such graphs, provided the slope is defined, using implicit differentiation. Evaluation and interpretation of 1 https://activecalculus.org/single/sec-2-7-implicit.html https://www.geogebra.org/m/yxga9vuu https://www.geogebra.org/m/yxga9vuu implicit derivatives is the 10th Learning Target for MAT 181. This Learning Target does not appear on any of the Checkpoints. It is considered only in this AEP, although the concept arises in other contexts inMAT181. It is of fundamental importance inhigher levelmath, science, andengineering courses, which is why this AEP is required for students aiming for a B or higher in MAT 181. Tasks for this AEP The following example illustrates how towork through the tasks in this AEP. Youdo not need to include anything from this example in your report. Example Consider the curve called the Folium of Descartes. The equation for this curve is ?3 + ?3 = 3??. In this example, we will do the following: 1. Use implicit differentiation by hand to find ?? ?? , and verify the expression using Geogebra. 2. Find the equation of the tangent line to the Folium of Descartes at the point 32 , 3 2 . 3. Graph the curve, the given point, and the tangent line. 4. Find all points on the Folium of Descartes which have a horizontal tangent line or a vertical tangent line. Now let’s perform these steps. 1. Working by hand, remember the key idea is that we envision that ? is an implied function of ?. ? ?? ? 3 + ?3 = ??? (3??) 3?2 + 3?2???? = 3? + 3? ?? ?? Nowwe need to collect the terms that contain the desired derivative ??/??. 3?2 − 3? ???? = 3? − 3? 2 ?? ?? = 3(? − ?2) 3(?2 − ?) ?? ?? = ? − ?2 ?2 − ? Let’s check ourworkusingGeogebra’sImplicitDerivative() command. Whenusing this command, you will need to first re-write the equation of the implicit curve with everything moved to one side, leaving zero on the other side. In this case, we would rewrite the equation of the graph as ?3 + ?3 − 3?? = 0 2 The left side of this equation is the input to the ImplicitDerivative() command. Log in to your Geogebra account, and then open up the Geogebra CAS calculator. It is very important that you open this specific application. The only visual clue that you have done so cor- rectly is that it will say ”Geogebra CAS Calculator” at the top of the Goegebra page in your browser. Another clue is that there is no ”Tools” tab on the left. CAS stands for Computer Algebra System. This application allows you to work withmathemat- ics symbolically. The usual Geogebra applications requiremost quantities to be assigned nu- merical values, hence the automatic creation of sliders for symbols in any expression that you enter. So, in the Geogebra CAS, enter the command ImplicitDerivative(x^3+y^3-3xy) Note that you are entering an expression as the input, not an equation. Examine Geogebra’s output. Note that the order of the terms on the result is not the same as the expression derived above, but both expressions are equivalent. (Why?) 2. Now we construct the tangent line to the Folium of Descartes at the point 32 , 3 2 . To write down an equation of a line, we need a point (which we have) and the slope. In the previous step, we computed an expression for the slope of the graph at any point (?, ?) on the graph. We need only substitute the values for ? and ?: ?? ?? (?,?)=(3/2,3/2) = 3/2 − (3/2) 2 (3/2)2 − 3/2 A careful examination of the right side reveals that it is equal to -1. So the slope of the tangent line at the given point is -1. So we have a point and a slope. We can immediately write down the tangent line using ? = ?1 +?(? − ?1): ? = 32 − ? − 3 2 3. Let’s produce a graph of the equation together with the tangent line we’ve constructed. In Geogebra, enter the original equation into an input region and press enter . In the next input region, enter the equation of the tangent line (don’t forget to press enter ). Finally, you can enter the point itself in the next input region. If all has gone well, you should see an image like this (the colours might be different): 3 4. Finally, wewill find all of the points on the FoliumofDescarteswhich have either a horizontal or vertical tangent line. A horizontal tangent line has a slope of zero, and a vertical tangent line has an undefined slope. Let’s consider finding the locations where the tangent line is horizontal. First, look at the graph. You can see that there is ahorizontal tangent line at the origin. (There is also a vertical tangent line there.) There is another location with a horizontal tangent line at roughly (1.2, 1.6). Finding the exact locations of the horizontal tangent lines requires two conditions to bemet: • the derivative must be zero at the point • the point must be on the graph Looking at the expression for the derivative, we see that the numerator is zero for any point with coordinates (?, ?2) since if ? = ?2 then ?2 − ?will be zero. This condition alone, how- ever, does not uniquely identify a location in the plane. We need identify the point satisfying both ? = ?2 (to make the derivative zero) and ?3 + ?3 = 3?? (to ensure the point is on the graph). If you plot the graph of ? = ?2 in Geogebra, you’ll see that it intersects the graph of the curve at the point where the tangent line is horizontal. To find that point, we couldwork by hand by substituting ?2 for ? everywhere in the equation of the curve, and then solving for ? in the resulting equation. There’s some algebra to deal with. Let’s not do that, though. Instead, let’s ask Geogebra to do that work for us. In the next available input region, enter the following command: Solve({x^3+y^3-3*x*y=0,(x^2-y)/(x-y^2)=0},{x,y}) We have given two inputs to this command. The first input is a set of two equations in ? and ?. The second input is another set, this one containing the variables for which we are asking Geogebra to solve. (”Hey Geogebra, solve this system of equations for both ? and ?, please.”) 4 Geogebra responds with {? = 0, ? = 0} , ? = 3√2, ? = 3√2 2 This isGeogebra’sway of tellingus the two locationswhere the tangent line is horizontal. One point is at (0, 0), and the other is at 3√2, 3√2 2 . To find the locations where the tangent line is vertical, we have two options. One is a bit sneaky, and is based on symmetry. The other uses the same approach we just used, but with an adjustment to make it possible to use Geogebra’s Solve() command. First, let’s use the method based on symmetry. Note that the equation of the Folium of Descartes contains ? and ? in identical roles. That is, you could switch the places of ? and ? in the equation, and the equation would look exactly the same. This algebraic change (switching the roles of ? and?) is equivalent to reflecting the graph of the equation across the line ? = ?. Looking at the graph, you can see that the graph hasmirror symmetry across the line ? = ?. This is the graphical consequence of the ”identical roles” property we see in the equation. Note also that a horizontal line becomes a vertical line after this reflection. Taken all together, this means that the reflection across the line ? = ? of the horizontal tangent line you found earlier is a vertical line through the point 3√2 2 , 3√2 . While this is a nice argument, it works only because of the symmetry of the graph. A more robust method based on solving a system of equations will work even if the graph does not have this symmetry. So let’s do that now. We immediately run into difficulty when we try to give Geogebra the appropriate system of equations. For a horizontal tangent line, the situation is relatively easy: one equation comes from setting the expression for the derivative to zero, and the other is the equation of the curve itself. Now, however, we are looking for a point where the slope is undefined. What equation do we give as the condition on the slope? The problem is that we are looking at the curve from an inappropriate point of view. The tangent line’s slope is undefined only because we are looking at the graph in such a way that the x-axis is horizontal. If we imagine looking at the graph so that the y-axis is horizontal (turning our heads by ?/2 radians), then