Module 2: Reflecting on One s Problems

Transcription

1 MATH55 Module : Reflecting on One s Problems Main Math concepts: Translations, Reflections, Graphs of Equations, Symmetry Auxiliary ideas: Working with quadratics, Mobius maps, Calculus, Inverses I. Transformations and Symmetry. Problem a. The graph on the left shows the portion of the circle x + y = 5 which lies in the region x, y 0. The graph on the right shows the same set of points translated 5 units to the right and 1 unit up. Give a definition of the set of points on the right by transforming the definition of the original set of points Solution. The point (,3) on the left graph will be transformed to the point (9,) since moving to the right by 5 adds 5 to the x value, and similarly we add 1 to the y value to move the point up by 1. Thus each point (x,y) in our graph on the left has (5,1) added to its co-ordinates to get the corresponding new point (x+5, y+1) on the second graph. However, this does not imply that we should replace x with x+5 and y with y+1 to transform the definition. Indeed, the equation (x + 5) + (y + 1) = 5 does not include the point (9,) since is definitely not equal to 5! Let us temporarily label the right hand graph with co-ordinates ( x, y ). What we want to say about ( x, y ) is that ( x, y ) lies on the transformed circle, that is, ( x, y )is the result of translating a point on the circle of radius 5 around (0,0) by (5,1) This is equivalent to: When ( x, y ) is transformed by (-5, -1), it gets moved onto the circle around the origin. And so the equation of the transformed circle, in the primed plane, is ( x 5) + ( y 1) = 5. Of course the x-y equation of the circle on the right is (x 5) + (y 1) = 5. Similarly we can define the transformed region by x 5, y 1 0, which can be rewritten as1 x 9, y 1.

2 So our definition is transformed by applying, to the variables in the original definition, the inverse of the transformation we want to apply to the set of points. Summary: Given a definition D(x,y) of a graph G in the plane, let T(G) denote the graph obtained by transforming G by the transformation T. The definition of T(G) is D(T 1 (x, y)). In most cases, D (x,y) is an equation y = f(x) or F(x,y) = 0. Procedure is: i) Solve the T equations for x and y in terms of x and y [i.e find (x, y) = T 1 (x', y') ] ii) Substitute for x and y in D(x,y) with the formulas from (i) [ giving D(T 1 (x', y')) ] iii) Change the prime variables back to the unprimed [giving D(T 1 (x, y)), as desired ]. In practice, we can omit the primes and do (ii) and (iii) together by writing (x, y) T 1 (x, y) as shorthand for the substitution. In the example above the transformations were T (x, y) = (x', y') = (x + 5, y + 1) and so x = x' 5, y = y' 1 ; we write x x 5, y y 1 as an efficient summary of the substitutions to get the new definition. Problem b. Derive the formulas for reflecting a point (x,y) around the following lines: i) the vertical line x = h ii) the 5 degree line x + y = b. Solution. We let (x', y') denote the new point in each case. i) For vertical reflection, y = y and x h = h x, which gives x = h x. In full vector notation, T(x,y) = ( h x, y ). ii) Given the line y = b x and the point (x,y), let (x', y') be its reflection. Let Δx = x' x, Δy = y' y and note that since the line x + y = b has slope -1, then the line through (x,y) and ( x, y ), being perpendicular to it, has slope -1/(-1) =1. So, Δx = Δy. The key idea is that the midpoint of the segment between (x,y) and ( x, y ), which is the point (x + Δx /, y + Δy / ), lies on the reflecting line, so we have (x + Δx / ) + (y + Δy / ) = b, and since Δx = Δy, we have x + Δx + y = b, and then Δx = b x y. Now plug in to get x = x + Δx = x + b x y = b y, y' = y + Δy = y + Δx = b x y+ Δy =b-x h-x h x x x+ Δx =b-y In vector notation, we have T (x, y) = ( b y, b - x ) Solving the system x' = b y, y' = b x for ( x, y ) in terms of ( x, y )verifies that both of these transformations (i) and (ii) is its own inverse, which is obvious geometrically. y y x+y = b (x,y )

3 A graph G is invariant under transformation T if T(G) = G, or equivalently, T 1 (G) = G. We can check a definition D(x,y) of G to see if G is invariant under T by testing whether the definition D ( T 1 (x, y) ) is equivalent to the original definition D(x,y), which is the same as testing whether D ( T ( x,y ) ) is equivalent to D ( x,y ). Invariance under reflection is usually called symmetry about the line of reflection. Problem c. i. Sketch a few graphs which are invariant under vertical translation by k =. A couple of solutions: 1. The set of parabolas {y = x + m m Z}. The vertical column 0 < x < 3 ii. Consider the equation y x + xy = 16. Both of these graphs extends up and down to infinity. 1. Find the equation for the reflection of this curve in the line y = - x. Is the curve symmetric about the line y = - x?. Determine if this curve is symmetric about either axis, the origin or the line y = x. Solutions. 1. Reflection in y = - x amounts to replacing x by y and y by x, which gives ( x) ( y) + ( x)( y) = 16, which simplifies to x y + xy = 16. This equation is NOT equivalent to the original, since e.g. (0,) is on the original graph but is not on the transformed graph. So the graph is not symmetric around y = - x.. We check each reflective symmetry by transforming the defining equation, then comparing to the original. Each reflection is its own inverse. Symmetry T or T 1 F(T 1 (x, y)) = 0 Equivalent to original? x-axis x x, y -y y x xy = 16 No, try (,) y- axis x - x, y y y x xy = 16 No, try (,) origin x - x, y -y y x + xy = 16 Yes y=x x y, y x x y + xy = 16 No, try (,0)

4 II. Graphing Party. Consider the equation y x + xy = 16 (***). This is the same curve we just looked at in Problem c above. Problem. For the equation (**), find the x and y extents, 5 or 90 degree symmetries, critical points and sketch the graph. Then find anything else that looks interesting. First, by Ic our curve (***) is symmetric about the origin but not about the lines x = 0, y = 0, y = ±x. In fact, since it is symmetric around the point (0,0) and is not periodic in the plane (polynomials don t have periodic graphs), it can only be symmetric around lines which go through (0,0). So there are no 5 or 90 degree line symmetries, as we have already checked all these lines through (0,0) in Ic. x extent. The simplest thing to do is to solve for y in terms of x. Can we do that here? We have y + (x)y (x + 16) = 0, so by the quadratic formula y = x ± 16x (1)( 1)(x + 16) which reduces to y = x ± 5x + 16 (***). = x ± 16x + x + 6 = x ± 0x + 6 Which x s will work? The only possible problem is inside the square root, so we must have 5x in order to take the square root. But, for all x, 5x and so we have that the x extent is the set of all real numbers. Also note that our equation does not define a function, but it is equivalent to the union of two functions (**), each of which can be graphed by a calculator. y extent. The straightforward way is to solve for x in terms of y. We have: x (y)x + (16 y ) = 0 and by quadratic formula we get x = y ± 16y (1)(16 y ) = y ± 16y + y 6) = y ± 0y 6) which reduces to x = y ± 5y 16. Again the only issue is that 5y 16 must be non-negative, which amounts to y 16 5 and then gives us y / 5 or y / 5. In interval notation, the y extent is (,c) (c, ) where c denotes

5 Intercepts: The y extent shows that there are no x-intercepts; the y-intercepts satisfy y 0 + (0)y = 16, so the y-intercepts are ±. We also note the intersection with y = x are the points (,) and (-,-). Critical points: Using implicit differentiation, we have y dy dy dy x y x + x + y = 0 which gives us = = dx dx dx x + y = x y x + y. So y = 0 when x y = 0, i.e. x = y. Plugging into the original equation we find horizontal tangents at y (y) + (y)y = 16, which yields y = ±., so the critical 5 points in question are ±( 8 5, 5 ). Vertical tangents would occur when y gets infinite, i.e. x + y = 0, i.e. y = -x. Plugging into the original equation leads to 5y = 16 which has no solutions. So there are no vertical tangents. Finally, we can divide the plane into four regions by using the lines x = y and y = -x. In each of these regions, the curve is monotonic (either increasing or decreasing). A sketch: We could use our calculator to get a sketch. This sketch shows the curve as well as the two lines which separate the increasing and decreasing portions of the curve. x = y y x + xy = y x + xy = 16 y = - x

6 Lastly, we notice that the upper curve is increasing for large positive x but lies below the line y= x/. Could there be a line an asymptote? If there is an asymptote, then the curve is approaching some line for large x. The upper curve is y = x + 5x What line might this be close to for large x? Well, when x is large, y is like -x +x 5, so we might guess at y = ( 5 )x as an asymptote. In order to check this out, we take the limit of the difference of the y coordinates on the curve and the lines. This limit will go to 0 if the line is an asymptote. lim(( x + x 5x + 16) ( 5x x)) = lim( 5x + 16 x 5) = x lim( 5x + 16 x 5) x 5x x 5 5x x 5 = lim x ( 5x x 5x x 5 ) = lim x ( 16 5x x 5 ) which equals 0 since the numerator is constant and the denominator does to infinity. Thus the line y = ( 5 )x is an asymptote for positive x, and by the symmetry around (0,0), it is also an asymptote for negative x. A similar calculation shows that the line y = ( 5 )x is also an asymptote for both positive and negative x. A good sketch, showing the asymptotes as well as the curve, is shown below. By the way, this curve is an example of a hyperbola. 6 y x + xy = 16 y = ( 5 )x y x + xy = y = ( 5 )x Lastly, note that this curve does have symmetry across two lines through (0,0), namely the two lines which bisect the angles between the asymptotes. More on that in Week 9.

7 III. How many solutions? Problem. 1. How many pairs of integers x and y have their product equal to their sum?. How many pairs of integers x and y have their product equal to twice their sum? Generally, how many pairs of integers have product equal to n times their sum? Solutions. 1. [case n = 1 ] We want integer solutions to xy = x + y (*). Just fiddling around, we might see what happens when x = y and we quickly find x = x which gives two solutions (0,0) and (,). We refer to these two solutions as the diagonal solutions. More fiddling around with integers doesn t seem to lead to more solutions, so we begin to suspect that there are no others. There are many ways to explain why this is so, but we will take an approach based on graphing functions and symmetry. Let s solve the equation for y in terms of x, giving us the function y = x x 1. This is an example of a Mobius function, aka linear fractional transformation, which is to say, a fraction of linear functions. Looking at the function, we see that it has a vertical asymptote at x = 1 (and 1 is not in the domain) and is symmetric about y = x (just look at the original equation to see that switching x and y gives an equivalent definition or you can transform the function equation, of course). The symmetry around y = x means that there must be a horizontal asymptote at y = 1 (and 1 is not in the range). One final symmetry is about the line y = x. To check this, the reflection transformation is y -> x, x -> y. Plugging into (*) gives (-y) (-x)= - y + - x, which becomes x - y + xy = x y. Simplifying gives us xy = x + y, as desired. So y = x is a line of symmetry. The derivative of the function is dy (x 1)1 x(1) 1 = = which is negative for all dx (x 1) (x 1) x, and hence (each continuous part of ) the graph is always decreasing. The graph is shown on the next page, along with the two asymptotes and the two lines of symmetry. It is another example of a hyperbola.

8 y y = x 3 1 y = x x x y = - x The symmetries tell us that the curve consists of four congruent parts: a) x < 0. In this part, 0 < y < 1 so there are no integer solutions for x < 0. b) 0 < x < 1. There can t be any integer solutions with 0 < x < 1. c) 1 < x <. There can t be any integer solutions with 1 < x <. d) x >. Then 1 < y < so there are no integer solutions for x >. In fact, we only needed to check one of these four pieces, since an integer solution (i, j) in one piece would have integer reflections ( j, i ), ( i, - j ), (- j, i) in the other three segments. So, the diagonal solutions (0,0,) and (,) are the only integer solutions to xy = x + y [case n = ] We want to find all integer solutions to xy = ( x + y ). Looking on y = x, we find the diagonal solutions (0,0) and (,). Solving for y gives y = x, and similar calculations to the previous example show: x The graph has asymptotes x = and y =, and is symmetric around y = x and y = x. The domain and range consist of all reals other than, and each continuous piece is decreasing. The graph, yet another hyperbola, is shown on the next page.

9 y = x 8 y = x x 6 y = -x Look at the segment with < x < ( the top segment). It is possible that x = 3 gives a new integer solution. Checking it, we see that y = ( x 3 ) / ( 3 ) = 6 which IS an integer, so we DO have a non-diagonal solution (3,6). Using the symmetries, from (3,6) we automatically get three other integer solutions (6,3) ( - 3, -6 ) = (1, -) and (-, 1 ) Notice that there are no other possible integer solutions in the top segment < x < since the only integer co-ordinate in there is 3. It follows that there are no more integer solutions in any segment, besides the ones listed above, since if there were, say (i,j) then one of reflections { (i,j), (j,i), (-i, j), (-j, -i) }, all of which are integer pairs, would be in the top segment. But we already found all the integer solutions in the top segment. Another way of saying the same thing is to note that the symmetries show that each of the four segments must have the same number of non-trivial integer solutions in this case, that number is one. So, for n=, there is only one non-diagonal non-symmetric integer solution, but there are six integer solutions altogether: two are diagonal and the four non-diagonal ones arise together as above. So we count the six integer solutions as 6 = + x 1. The previous case had = + x 0 since there were no non-diagonal solutions in the case n = 1.