# n-parameter families of curves

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1 1 n-parameter families of curves For purposes of this iscussion, a curve will mean any equation involving x, y, an no other variables. Some examples of curves are x 2 + (y 3) 2 = 9 circle with raius 3, centere at (0, 3) x 2 = y parabola x = y 2 another parabola x 3 e y + sin(xy) = 2 ln(x 2 + y 2 + 1) I on t know what this looks like, but it s still a curve Some curves can obviously be viewe as functions. For instance, y = x 2 obviously makes y into a function of x, so y/ makes sense anywhere along that curve. We say that y is expresse explicitly as a function of x; that means we have a formula of the form ( ) some expression y =. not involving y x = y 2 makes x into a function of y, so /y makes sense anywhere along that curve. We say that x is expresse explicitly as a function of y; that means we have a formula of the form ( ) some expression x =. not involving x The circle x 2 + (y 3) 2 = 9 satisfies neither of those conitions it is neither explicit for y in x, nor explicit for x in y. We say that the relation between x an y is represente implicitly. However, even the circle gives us an explicit relation locally, at most points. For instance, the upper half-circle { (x, y) : x 2 + (y 3) 2 = 9, y > 3 } can be rewritten as y = x 2, making y into a function of x, so y/ makes sense. (Remember that the symbol means the nonnegative square root of, so it always takes a value greater than or equal to zero. To make it two-value you have to put ± in front of it.) In a similar fashion, the lower half circle gives y as a function of x, where y/ also makes sense; an the right or left half circle gives x as a function of y, where /y makes sense. Thus both erivatives are efine everywhere on the circle except at four points. That s enough for most purposes of this course, which is primarily concerne with the local behavior of functions i.e., what happens near a point. (Global behavior is consiere in more avance courses.) In a similar fashion, on most curves (with exceptions such as a vertical line or a horizontal line), the erivatives are efine at most points (with exceptions at corners, local maxima or minima, etc.). Rather than thinking of one of x, y as a function of the other, it may be more helpful to think of x an y as two relate quantities: When one changes, then the other changes in a corresponing fashion. The erivatives y/ an /y give the relative rates of those changes. Many curves can also be expresse parametrically i.e., with x an y both given as functions of a thir variable. For instance, the circle x 2 +(y 3) 2 = 9 (given by one equation in two variables) can also be expresse as x(θ) = 3 cos θ, y(θ) = sin θ (0 θ 2π)

2 (two equations in 3 variables). But we won t be using that kin of parametrization in this iscussion (except once near the very en when we consier Clairaut s equation). The iscussion in this ocument will be largely concerne with another, very ifferent use of the wor parameter : A one-parameter family of curves is the collection of curves we get by taking an equation involving x, y, an one other variable for instance, c (though any other letter will o just as well). Plugging in ifferent numbers for c gives us ifferent curves; in many cases those curves are relate in some way that is visually simple. Here are some examples: The equation x 2 + (y c) 2 = 9 represents a one-parameter family of curves. That is, it represents infinitely many ifferent curves. Inee, it represents the collection of all circles that have raius 3 an that are centere at points along the line x = 0. A few of the curves it represents are c = 3 x 2 + (y 3) 2 = 9 circle centere at (0,3) c = 1 x 2 + (y + 1) 2 = 9 circle centere at (0, 1) c = π + 17 x 2 + (y π 17) 2 = 9 circle centere at (0, π + 17) For comparison, note that this family can also be written as x 2 + (y c) 2 9 = 0. More generally, if f(x, y) is some function of two variables, then f(x, y c) = 0 is a oneparameter family of curves. To get the graph of one of these curves from the graph of another, just translate (move) it vertically. Similarly, f(x + b, y) = 0 represents a one-parameter family; its curves can be obtaine from each other by horizontal translation. The equation y = ax 2 represents infinitely many functions of x. For each value of a, we get y equal to one function of x. The curves represente are the parabolas with axis of symmetry along the y-axis an with vertex at (0, 0). They iffer from one another in that they are stretche vertically by a factor of a (or stretche horizontally by a). Similarly, a two-parameter family of curves is the collection of curves we get by taking an equation involving x, y, an two other variables, provie that that family of curves cannot also be represente using just one parameter. That last clause is a bit subtle, an will be illustrate by the thir example below. Some examples: The equation (x a) 2 +(y b) 2 = 9 is a 2-parameter family of curves. It represents the collection of all circles of raius 3. Plugging in some number for a an some number for b gives us one particular circle of raius 3, the one centere at (a, b). Note that the numbers a an b can be chosen inepenently of each other. We can rewrite the equation as (x a) 2 + (y b) 2 9 = 0. More generally, the equation f(x a, y b) = 0 represents the set of all translates (horizontal an vertical combine) of the curve f(x, y) = 0. The equation y = ax + b represents all straight lines. This is a 2-parameter family. At first glance, the equation y = ax + bx might look like a 2-parameter family of curves to beginners. But it is actually a 1-parameter family of curves; it simply has not been written in the most economical form. 2

3 Inee, rewrite it as y = (a + b)x. You ll see that it is a straight line, passing through (0, 0), with slope equal to a + b. The same straight line is represente by y = mx if we simply take m = a + b. Here are some of the curves that belong to this family of curves: a b curve m 3 2 y = 5x y = 8x y = 5x y = 8x 8 Choosing a = 3 an b = 2 yiels the same curve as choosing a = 12 an b = 7; both those curves are represente by y = mx with m = 5. In other wors, the set of curves that we can get from y = ax + bx by plugging in some number for a an some number for b is the same as the set of curves that we can get from y = mx by plugging in some number for m. More generally, for any positive integer n, an n-parameter family of curves is the collection of curves we get by taking an equation involving x, y, an n other variables, provie that that family of curves cannot be represente with fewer parameters. Once you ve unerstoo the efinitions, it shoul be fairly obvious that y = ax 2 + bx + c is a 3-parameter family while y = ax + bx is only a 1-parameter family. But some examples are not so obvious. For instance, 3 ( ) y = a sin 2 x + b cos 2 x + c looks like a 3-parameter family of curves, because the functions sin 2 x an cos 2 x an 1 (the constant function) are ifferent functions. But we can rewrite this equation as y = (a b) sin 2 x + (b + c). The curves we can get by plugging particular numbers in for a, b, c are the same as the curves we can get from y = p sin 2 x + q by plugging in particular numbers for p an q. Thus, ( ) is actually a 2-parameter family of curves. Exercises on families of curves. Represent each of the following as an n-parameter family of curves i.e., represent it by an equation in x an y with n aitional variables, where n is as low as possible. Simplify as much as you can; then circle your final answer. (A) All horizontal lines. (B) All circles in the plane. (Hint: Translations an size.) (C) With y equal to a function of x: All functions that are polynomials of egree 3 or less.

4 (D) All parabolas that have vertical lines for their axes of symmetry. (Hint: Translations an vertical stretching, or think in terms of polynomials.) (E) All ellipses whose axes of symmetry are parallel to the x- an y- coorinate axes. (Hint: Start with a circle; apply translations, horizontal stretching, an vertical stretching.) 4 Making Differential Equations A ifferential equation is an equation involving erivatives. The orer of a ifferential equation is the highest number of erivations that have been applie to any term in the equation. For instance, 3 y + 2 y y 3 3x3 + 2xy ex y = 0 is a thir-orer ifferential equation, because it involves 3 y but oes not involve any higher erivatives. 3 (This shoul not be confuse with the egree of a ifferential equation. That s harer to efine, but here s an example: The equation y 2 y = sin3 x coul be sai to be of thir egree, because the term y 2 (y/) 1 has exponents summing to 3.) To solve a ifferential equation means, roughly, to fin the set of all the functions which satisfy that ifferential equation. It turns out that, for the most part, the general solution of an nth orer ifferential equation is an n-parameter family of curves. (We ll come back to that for the most part business later, in the last section of this ocument.) Most of this course is concerne with solving ifferential equations i.e., Given a ifferential equation, fin the solution. But, just to help us master the concept of solution, we ll start with this reverse problem, which is much easier: Given the solution, fin the corresponing ifferential equation. The proceure is quite simple to escribe (though in some problems the the actual computations get messy): 1. Algebraically solve for one of the parameters. For instance, if one of the parameters is a, rewrite the equation so that it is in the form ( ) some expression a =. not involving a

5 (If there is more than one parameter, it oesn t matter which one you start with; pick whichever one seems most convenient.) Actually, you coul instea solve for something of the form ) ) ( some expression involving a an no other letters = ( some expression not involving a The left sie of that equation might be 1/a, or 2a 2, or 3a 3 +sin(a), or something like that. Any equation of this sort will o just as well, an one of these variants may be more convenient than another i.e., it might allow us a simpler choice on the right sie of the equation, an that will be avantageous in subsequent steps. 2. Differentiate both sies of the equation with respect to x. (Or with respect to some other variable, if that is more convenient.) Keep in min that y is a function of x, so its erivative is y. Use the chain rule wherever neee. The erivative of any constant is 0. Thus, we get 0 = ( some new expression not involving a We have eliminate a from the equation. But we have also raise the orer of the ifferential equation by one. 3. Now pick another parameter say b an repeat the process. Continue until all the parameters are gone. Each repetition reuces the number of parameters by one, an raises the orer of the ifferential equation by one; thus an n-parameter solution yiels an nth-orer ifferential equation. 4. Simplify the result as much as possible. Following are some examples. An example with parabolas. Fin the ifferential equation whose solution is y = ax 2. Answer. Rewrite as a = x 2 y. Differentiate both sies with respect to x, viewing y as a function of x. On the left sie we ll just get 0. On the right sie, we have to use the rule for the erivative of a prouct of two functions: ( x 2 y ) = ) ( ) ( ) x 2 y + x 2 y = 2x 3 y + x 2 y. Thus we get the ifferential equation 0 = 2x 3 y + x 2 y. Simplify, to.. 5 xy = 2y or y = 2y x. Checking the answer. If y = ax 2, then y = 2ax. Do those satisfy the answer we arrive at? I.e., o we have xy =? 2y? Well, that equation simplifies to x 2ax =? 2 ax 2, which is inee true. An example with some bizarre curve. 3x 2 y + x cos(xy) + e x ln y = c. I have no iea what this curve looks like, but its ifferential equation is not har to fin.

6 Answer. It s alreay solve for c; all we have to o is ifferentiate both sies. That s going to confuse some stuents, who nee to review implicit ifferentiation from calculus. I ll break up into little steps for you: 3x 2 y + x cos(xy) + e x ln y = c Differentiate both sies; Use the prouct rule, (uv) = u v + uv. Thus [ ( )] [ ] 3x 2 y + 3x 2 y + That simplifies a little, to ( 3x 2 y ) + (x cos(xy)) + (ex ln y) = 0. 6xy + 3x 2 y + cos(xy) + x [ ] [ ] [ ] [ ] cos(xy) + x cos(xy) + ex ln y + e x ln y = 0. [ ] [ ] cos(xy) + e x ln y + e x ln y = 0. We re going to nee the chain rule. It helps to think of y as a function of x; write it as y(x) if you like. ln y = ln y y y = 1 y y. Similarly, xy is a function of x; it coul be written as x y(x) if you like. Using the chain rule again, an also the prouct rule: cos(xy) = cos(xy) (xy) (xy) = [ sin(xy)] Substituting those results into our ifferential equation yiels [ y + x y ] = sin(xy)(y + xy ). 6xy + 3x 2 y + cos(xy) x sin(xy)(y + xy ) + e x ln y + e x y 1 y = 0. If you want to emphasize the ifferential equation aspect of this, you coul group all the y terms together: 6xy + cos(xy) xy sin(xy) + e x ln y + [3x 2 y x 2 sin(xy) + e x y 1 ] y = 0. Or, moving the ifferentials to opposite sies of the equation, [xy sin(xy) 6xy cos(xy) e x ln y] = [3x 2 y x 2 sin(xy) + e x y 1 ] y. 6 An example: Ellipses or hyperbolas with axes on the coorinate axes. That can be written as (1) ax 2 + by 2 = 1. What is the ifferential equation? Answer. Let s begin by solving for a; we get (1 ) a = x 2 (1 by 2 ).

7 Now comes our first ifferentiation. We want to ifferentiate both sies with respect to x. Here a an b are constants, an we will view y as a function of x. Here are some facts from calculus that we will nee. (You may nee to review your calculus.) (a) = 0 ( ) x 2 = 2x 3 7 (uv) = uv ( ) y 2 = 2y y + v u = 2yy Thus, ifferentiating equation (1 ) yiels ( x 2 y 2) = x 2 ( ) y 2 + y 2 ( ) x 2 = 2x 2 yy 2x 3 y 2 (2) 0 = 2x 3 (1 by 2 ) + x 2 ( 2byy ). To simplify a little, multiply through by x 3 /2; we obtain (2 ) 0 = 1 by 2 + bxyy. Solving for b yiels (2 ) b = 1 y 2 xyy. If we ifferentiate both sies of that, the left sie will get replace by 0, but the right sie will get messy. Here s a trick to save some work: Before ifferentiating, moify both sies so that the ifferentiation will be easier. In this example, that can be accomplishe by taking the reciprocal on both sies. Thus we obtain (2 ) 1 b = y2 xyy. Now ifferentiate both sies. The left sie still gets replace by 0, but we save a lot of work on ifferentiating the right sie. The result is (3) 0 = 2yy yy xy y xyy. Simplify to yy = xy y + xyy. This is a secon-orer ifferential equation, since it involves y. But checking this one is messy. An example: more hyperbolas. Fin the ifferential equation for (x a)(y b) = 4. Answer. We can write y as an explicit function of x: y = b + 4 x a.

8 Differentiating once gets ri of b; we have y = 4(x a) 2. (Note that y is less than 0; the function b + 4 is ecreasing on each of the two intervals where it is efine.) Now solve that equation x a algebraically for a; we get Differentiating both sies of that yiels Simplifying, 0 = 1 ± a = x ± ( 1 4 y ) 1/2. ( ) ( ) 1 1 3/2 ( ) y 4 y. ( ) 1 3/2 4 y = ± 1 8 y Square both sies an simplify; (y ) 2 + (y ) 3 = 0. Checking: y = 4(x a) 2 an y = 8(x a) 3, hence 0 =? (y ) 2 + (y ) 3 = [8(x a) 3 ] 2 + [ 4(x a) 2 ] 3 = (64 64)(x a) 6 = 0. 8 Exercises on making ifferential equations. Fin the ifferential equations for these families of curves. Simplify; then circle your final answer. (You are urge to check your answers, too.) (F) Straight lines with slope 2/3. These have equation y = 2 x + b. You shoul get a first-orer 3 ifferential equation. Optional hint: If you look ahea to the next section, you ll see that this one-parameter family is aitive with a nonzero particular term, so the resulting ifferential equation shoul be linear nonhomgeneous. (G) All straight lines. That s y = mx + b. Eliminate both m an b, to obtain a secon-orer ifferential equation. Optional hint: This is aitive with two homogeneous terms an no particular term, so the ifferential equation shoul be linear homogeneous. (H) The circles tangent to the x-axis at (0, 0) have equation x 2 + (y b) 2 = b 2. Fin the corresponing first-orer ifferential equation. Optional hint: Not aitive; answer is not linear. (I) y = ax 2 + bx 1/3. Optional hint: Linear homogeneous. (J) y = a sin x + b cos x. Hint: Instea of solving for a or b an then ifferentiating, use this alternate metho, which works well for this problem but not for many other problems: Differentiate the given equation to obtain formulas for y an y. Thus we have three equations: y = a sin x + b cos x, y = [what], y = [what], where the right sies involve a, b, x, but not y. Then algebraically eliminate a an b from those three equations, to fin one secon-orer ifferential equation involving some or all of x, y, y, y but not explicitly mentioning a or b. (K) y = a sin x + b cos x. Hint: Similar to previous problem, but there s more computation. Don t forget to use the chain rule.

9 9 Envelope Solutions (It might be best to postpone this topic for a while e.g., until one has covere separation of variables an other elementary methos for solving first orer ifferential equations.) I state earlier that, for the most part, the solution of an nth orer ifferential equation is an n-parameter family of curves. That s exactly right when we work with linear equations our topic for most of the semester but we get exceptions with some nonlinear ifferential equations. The next few pages consier some of the simplest exceptions. (I will postpone until later in the semester a iscussion of what linear means in this context.) Theorem on envelopes. Let G(x, y.c) be a function of three variables. Thus the equation G(x, y, c) = 0 represents a one-parameter family of curves, if we view c as a parameter. Let us enote by E c the curve given by parameter value c. Suppose that that one parameter family of curves satisfies some first-orer ifferential equation. Then there may also be an aitional curve satisfying that ifferential equation, which is tangent to each E c at one point. This curve envelopes the E c s, so it is calle their envelope. (In some books it is also calle a singular solution to the ifferential equation.) If an envelope solution exists, it can be foun by this proceure: ( ) G(x, y, c) = 0 an G (x, y, c) = 0 c is a system of two equations in the three variables x, y, c. By algebraic means, we may eliminate c, thus obtaining one equation in just the variables x an y. That equation is the envelope curve. The proof uses the vector version of the chain rule. We won t go into any of the etails of that, but we will give several examples. Example 1. The nonlinear ifferential equation y = x y 1 can be solve by the metho of separation of variables, iscusse later in this course. The general one-parameter solution turns out to be y = 1+( 1 4 x2 +c) 2, a family of fourth-egree polynomials. We can rewrite that as G(x, y, c) = 0, if we use the function G(x, y, c) = 1 + ( 1 4 x2 + c) 2 y. Now compute G c = 1 2 x2 + 2c. We want to eliminate c from the system of two equations ( ) mentione in the theorem that is, the system of these two equations: 1 + ( 1 4 x2 + c) 2 y = 0 an 1 2 x2 + 2c = 0. Probably the easiest way to o that is to solve the secon equation for c; we get c = 1 4 x2. Now plug that in for c in the first equation. We en up with the horizontal line y = 1 for the envelope curve in this example.

10 10 Example 2. The nonlinear ifferential equation y = xy +(y ) 2 can be shown to have one parameter family of solutions y = cx + c 2, a family of straight lines. (Those c s are the same. If you prefer to see the c only once, write the family instea as y = (c x)2 1 4 x2. But that woul make the computations below more complicate.) We can represent the solution family as G(x, y, c) = 0 if we use the function G(x, y, c) = cx + c 2 y. Now compute G = x + 2c. We want to eliminate c from the system of two equations ( ) c mentione in the theorem that is, the system of these two equations: cx + c 2 y = 0 an x + 2c = 0. The secon of those equations yiels c = 1 x. Plug that into the first equation; thus we obtain 2 ( 1x)x + ( x)2 y = 0. That simplifies to y = x 2 /4, a parabola. Actually, we o have to choose G with some skill. For instance, we coul have represente that one-parameter family of straight line solutions as y = (c+ 1 2 x)2 1 4 x2, or even as y x2 = c+ 1 2 x. Thus, we coul have use the function G(x, y, c) = y x2 c 1 x. But that yiels G/ c = 1, 2 an so the secon equation in the system ( ) is 1 = 0. Thus, there is no point (x, y) in the plane that satisfy satisfies both equations of ( ), since no point satisfies the secon equation. This approach oes not yiel the envelope solution i.e., it yiels the wrong answer. [I haven t yet figure out how to avoi this ifficulty, except to recommen that you sketch a graph when possible to see if there might be an envelope solution. Watch for a later eition of this ocument.]

11 Example 3. The preceing example is a particular instance (obtaine with h(c) = c 2 ) of an entire broa class of examples: Let h be any continuously ifferentiable function of one variable (other than a constant function). Then the equation y = xy + h(y ) is known as Clairaut s equation. It s not har to verify that the straight lines y = cx + h(c) form a one-parameter solution to Clairaut s equation. After a bit of computation, we arrive at the envelope solution, which can be expresse parametrically: 11 x(t) = h (t), y(t) = th (t) + h(t). Example 4. The ifferential equation xy + y = 0 has general solution x 2 + y 2 = c. That s the set of all circles centere at 0 a family of concentric circles. Sketching the graph (not shown here) makes it obvious that there are no envelope solutions. The theorem on envelopes agrees with that conclusion it tells us to look for a simultaneous solution of the two equations x 2 + y 2 = c an 1 = 0. Obviously those have no simultaneous solution i.e., there are no points (x, y) in the plane that satisfy both these equations. In this case, that s correct, as we can see by looking at the graph. Exercises on evelopes. Fin the envelope for each of the following families of curves. (You on t nee to fin the ifferential equation.) (L) y = e x sin(x + c). Hint: The envelope consists of two curves. (M) y = sin( 1 2 x2 + c). Hint: The envelope consists of two straight lines. (N) (x c)(y c) = 1. Hint: The envelope consists of two straight lines. (O) y x = 2(c x)(1 c). This problem has a history for me personally: When I was a chil, one of the toys I playe with was a square frame line with evenly space pegs, over which one coul loop some colore elastic bans. One of my favorite patterns consiste of the arrangement shown in the accompanying figure. The straight line for parameter c passes through the points (c 1, 1 c) an (c, c); that gives us the one-parameter family inicate above. I coul see a curve being forme, but I in t know much geometry back then. The only curves I knew well were circles, so I guesse the envelope curve was a quarter of a circle. (I in t know it was calle an envelope.) Actually, it turns out to be a parabola, not a circle. You just have to fin the parabola s equation.

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