Calc I problem regarding orthogonal curves and families of curves

[94JZA80]

First, let me start by apologizing for the length of this post. i do in fact have a question about a problem that i couldn't solve (at least not the way i wanted to solve it), but first there is a fair amount of foreground information i must lay down...beyond this foreground information, this post does follow the suggested format (you just have to scroll down a bit)...

ORTHOGONAL CURVES AND FAMILIES OF CURVES:

Definition - Two curves or families of curves are orthogonal if at each point of intersection their tangent lines are perpendicular (that is, the slopes of their tangent lines at their point(s) of intersection are negative reciprocals of each other).

Now the example given in my textbook is oversimplified for obvious reasons. nevertheless, it poses two functions/curves and asks the student to show that the two curves are in fact orthogonal, i.e. that the slopes of their tangent lines at their point(s) of intersection are negative reciprocals of each other. right away it is evident via differentiation that the derivative of one function (-x/y) is the negative reciprocal of the derivative of the other (y/x). however, when i got to the actual exercises, it quickly became apparent that it is not always obvious that the derivatives of two functions are negative reciprocals of one another, even when it is known that the functions are orthogonal.

and so in the case where it is not obvious that the derivatives of two orthogonal curves are negative reciprocals of one another, it is necessary to calculate the two curves' point(s) of intersection, and then evaluate the derivatives of each function at those points of intersection to see if in fact they are negative reciprocals of one another.

now consider the case where it is not obvious that the derivatives of two orthogonal families of curves are negative reciprocals of one another. again, it is necessary to calculate points of intersection between the families of curves...only this time one must calculate "generic" points of intersection for entire families of curves, unlike the previous situation in which one only needed to calculate "specific" points of intersection between two specific curves.

now there are two exercises that ask me to show that two families of curves are orthogonal. one such exercise involves the following functions:

x² + y² = r²
ax + by = 0

...now despite the fact that dealing with two families of curves can be slightly more complicated than dealing with just two specific curves (due to the fact that the values of the constants a, b, and r can be just about anything), i was able to prove the general solution straight away by using direct substitution to find "generic" points of intersection, and then evaluating the derivatives of both functions at those points of intersection...see work below:

[PLAIN]http://img266.imageshack.us/img266/5749/s26p37med.jpg



...and now on to the exercise that I'm having trouble with - the "families of orthogonal curves" exercise for which i could not prove the general solution using just the generic constants:

Homework Statement

Show that the following two families of curves are orthogonal:

y = cx², where C is a constant
x² + 2y² = k, where k is a constant


First i simply tried implicit differentiation, only to find that the functions' derivatives are not obvious negative reciprocals of each other. then i tried finding generic points of intersection between the families of curves, but was unsuccessful in doing so. so then i did what i thought was the next best way to prove the curves orthogonal - i chose arbitrary values for the constants c and k to create specific functions, and then showed that those specific functions were orthogonal. in other words, i have been able to show that arbitrary members of each family of curves are orthogonal by choosing arbitrary values of c and k. specifically, I've been able to show that the curves are orthogonal in the case when c and k are both positive and equal (specifically i chose c = k = 1, and the curves then become y = x² and x² + 2y² = 1...see section 3 below to see my work). likewise, I've also been able to show that the curves are orthogonal when c and k are positive but not equal (specifically i chose c = 0.5 and k = 4, and the curves then become y = 0.5x² + 2y² = 4...see section 3 below to see my work). and finally, i was able to show that the curves are orthogonal when c is negative and k is positive (specifically i chose c = -0.5 and k = 4, and the curves then become y = -0.5x² + 2y² = 4...see section 3 below to see my work). note that when k takes on a negative value, the roots (x-values) of the equation x² + 2y² = k take on imaginary values. and so a graph of the function x² + 2y² = k contains imaginary coordinates not in the REAL cartesian plane when k is negative. hence they do not produce points of intersection with the function y = cx² on the real cartesian plane. thus it is not necessary to consider any instances in which k is negative (specifically, it is not necessary to investigate either the instance in which c is positive and k is negative or the instance in which both c and k are negative). hence, i feel that the three instances for which i proved orthogonality between two arbitrary curves covers the range of values that both constants c and k can take on, which in turn proves that entire families of the curves y = cx² and x² + 2y² = k are orthogonal.

Homework Equations

see above

The Attempt at a Solution

[PLAIN]http://img529.imageshack.us/img529/879/s26p39p1.jpg

[PLAIN]http://img529.imageshack.us/img529/6634/s26p39p2.jpg ...now as you can see through the work above, I've proven orthogonality for two arbitrary functions in 3 unique cases above. you may or may not agree with me that this proves that any member of the family of curves y = cx² is orthogonal to any member of the family of curves x² + 2y² = k. but that's neither here nor there. what i'd like to know is how to prove the general solution using the generic constants c and k (as opposed to having to prove specific solutions using specific values of c and k). looking back at the other problem (the one involving the equations x² + y² = r² and ax + by = 0 that i was able to solve the general solution for), it was easy to calculate the generic points of intersection. and b/c it was easy to calculate those points of intersection, it was likewise easy to calculate derivatives at those points and show that those derivatives were in fact negative reciprocals of one another, thus proving that the functions x² + y² = r² and ax + by = 0 are orthogonal.

however, it quite difficult to find generic points of intersection of the functions y = cx² and x² + 2y² = k. i tried "direct substitution," "elimination," and "addition of equations" methods, none of which produce simple generic values for x (or y, assuming i solved one of the equations for x and substituted into the other equation)...and therefore x and y produce no simple generic points of intersection. regardless of the method used, neither x nor y factors out of the resulting equation cleanly. and so i resorted to the quadratic equation to solve for x (or y), which of course results in somewhat complicated, messy x (or y) values that aren't very "differentiation-friendly." nevertheless, i found my generic points of intersection via direct substitution and factoring the resulting equation using the quadratic formula. i then calculated the derivatives of both functions at those intersects using whatever differentiation rules necessary (product rule, quotient rule, chain rule, etc.). but b/c the x (or y) values were complicated and messy, the derivatives as calculated at the intersects were also complicated and messy. on top of it all, it wasn't apparent that the derivatives were negative reciprocals of each other.

so to sum it up, i need help proving the general solution without doing it by proving specific/arbitrary solutions for specific/arbitrary values of c and k.TIA,
Eric

 

[Dick]

Take your second example where you got y'=2cx and y'=-x/(2y). You are allowed (since you are working at an intersection point) to substitute y=cx^2 into the second expression for y'. You shouldn't need to find actual intersection points.

 

[94JZA80]

Take your second example where you got y'=2cx and y'=-x/(2y). You are allowed (since you are working at an intersection point) to substitute y=cx^2 into the second expression for y'. You shouldn't need to find actual intersection points.

thank you very much for the insight. i see that substituting y = cx² into the equation y' = -x/(2y) gives y' = -x (2cx²) = -1/(2cx). and i see that -1/(2cx) is the negative reciprocal of the derivative y' = 2cx of the function y = cx². i don't really know why i overlooked the simple fact that i could just substitute the function y (in terms of x) into y' to find the other derivative entirely in terms of x. for some reason i had it in my head that i must calculate derivatives at the two functions' points of intersection...go figure.

 

[94JZA80]

...and so i guess in my first example (the exercise in which i was able to calculate generic points of intersection and the derivatives at those points straight away), all that work was completely unnecessary.

if i take the given equations x² + y² = r² and ax + by = 0, i can calculate their derivatives to be y' = -(x/y) and y' = -(a/b).

if i re-write the equation ax + by = 0 as y = -(a/b)x, i can substitute this value into y': y' = -(x/y) = -(x/-(a/b)x) = -x * -(b/ax) = b/a, which is the negative reciprocal of y' = -(a/b).

i guess i was making the problems far more complicated than they needed to be. thanks again for the help.

Eric

 

[Dick]

i guess i was making the problems far more complicated than they needed to be. thanks again for the help.

You've got it. The relations between x and y given by the two curves should be enough to show the derivatives are negative reciprocals. You CAN do it by explicit intersection (if possible), that's just generally much harder.

 

[94JZA80]

ok, I've got another exercise regarding pairs of orthogonal curves/functions (whose derivatives aren't obvious negative reciprocals of one another upon initial calculation). but this time I'm having trouble showing that they're negative reciprocals by solving either of the original equations for y (or x for that matter) and substituting back into y'.

Homework Statement

Show that the following two curves are orthogonal:

x² - y² = 5
4x² = 9y² = 72

Homework Equations

see above

The Attempt at a Solution

y' = x/y
y' = -(4x/9y)

As i stated above, i was unable to solve either of the original equations for y (or x), and was therefore unable to substitute such a value into y'. i essentially had to resort to my method of finding the points of intersection (this time by addition of equations and elimination methods) and showing that the two functions' derivatives at those points are negative reciprocals of one another - see work below:

[PLAIN]http://img29.imageshack.us/img29/4945/s26p36.jpg


...as you can see above, i had no problem solving the problem using the time consuming method of finding the points of intersection and then calculating the derivatives at those points. is there a way to solve this problem quicker? am i overlooking something this time? or is it possible that proving that these functions are orthogonal actually requires finding their points of intersection first this time around?

 

[Dick]

Take the first equation times 72 and subtract the second equation times 5. That cancels the constant. Now you have just terms involving x^2 and y^2. Can you show x/y must equal +/-3/2? I did this because I saw your derivatives just involve x/y. So you must be able to derive something about the value of x/y from the curves.

 

[94JZA80]

thanks again for the insight. i see i had the right idea by adding/subtracting equations, but your idea of eliminating the constant (as opposed to my idea of eliminating either the x² or y² term) makes it possible to evaluate y' entirely in terms of a single variable x or y. in this case, i solved for y first, so that y' is then calculated entirely in terms of x:

[PLAIN]http://img192.imageshack.us/img192/8053/s26p36p2.jpg


...but i guess this is more or less the same method i used. as you can see, by instead eliminating the x² (or y²) term, i was able to solve for y (or x). i was then able to substitute y (or x) back into either of the original equations to solve for x (or y). in fact, looking back at my previous post, you can see I've done just that. oddly enough though, i overlooked simply substituting those x and y values into y' = x/y and y' = -(4x/9y) straight away. instead i apparently felt it necessary to take myself through the thought process of identifying those x and y values as coordinates of all the points of intersection, and THEN evaluate y' = x/y and y' = -(4x/9y) at those points...but really its essentially the same thing. anyways, thanks for clearing that up.


Eric

 

[94JZA80]

ok, i have one final exercise referring to orthogonal families of curves/functions, and this time I'm really stuck!

Homework Statement

Show that the following two curves are orthogonal:

x² + y² = ax, where a is a constant
x² + y² = by, where b is a constant

Homework Equations

see above

The Attempt at a Solution

again, it is not obvious initially that the above derivatives are negative reciprocals of one another. so right away we know there's going to be some manipulation involved in getting one derivative to resemble the negative reciprocal of the other. after many failed attempts at manipulating the given equations to isolate a single variable, i finally succeeded in calculating y' of both functions entirely in terms of the variable x using an intuitive thought process as opposed to strict manipulation. note that the numbers a and b appear in y' as well, but they are treated as constants in this problem...see work below:

[PLAIN]http://img840.imageshack.us/img840/6308/s26p38.jpg


...and yet i still cannot manipulate either derivative into a form that resembles the negative reciprocal of the other. i can't even revert back to my time consuming method of finding the points of intersection and differentiating at those points to test for orthogonality because the given equations are too complicated to combine/eliminate/cancel and solve for x or y explicitly. so technically i don't even know if these families of curves are orthogonal to begin with. but i have to assume that the text isn't throwing a trick question at me, that the families of curves are orthogonal, and that I'm overlooking something that's keeping me from showing that the derivatives are in fact negative reciprocals of one another (i would check the answer, but the text only gives answers for the odd-numbered exercises). so what do you think of this one? is there a better direction to take than the one i took in my work above? I'm lost on this one...

TIA,
Eric

 

[Dick]

It's really not so hard to solve for the intersection points. In fact, it's pretty easy. You've got y=ax/b. Put that into the quadratics. But I think the easiest way is to multiply your two derivatives (a-2x)/(2y) and (2x)/(b-2y) and see if you can show it's (-1) just using the original equations. BTW, one of the intersection points is clearly (0,0). You might need to make a special argument there. Sometimes this sort of question does take some cleverness to see an easy method.

 

[94JZA80]

thanks for the clues. I'm pretty busy at work today, so i probably won't get to mull over it until this evening. but i'll post any progress i make as soon as i get a chance.

Eric

 

[94JZA80]

ok...i succeeded in showing that y' x y' = -1:

((a - 2x) / 2y) x (2x / (b - 2y))
= (2ax - 4x²) / (2by - 4y²)
= (2(x² + y²) - 4x²) / (2(x² + y²) -4y²)
= (2x² + 2y² - 4x²) / (2x² + 2y² - 4y²)
= (2y² - 2x²) / (2x² - 2y²)
= (2(y² - x²)) / (2(x² - y²))
= (y² - x²) / (x² - y²)
= -(x² - y²) / (x² - y²)
= -1

...which obviously shows that the derivatives are negative reciprocals of one another, and that solves the problem.



...now, as far as finding the point(s) of intersection, you pointed out above that i got as far as y = ax / b, and in fact i had already plugged it into one of the quadratics. i just wasn't confident in the x and y roots i came up with b/c i wasn't sure if i was factoring the quadratic correctly. nevertheless, i got x = 0 and ab² / (a² + b²), as well as y = 0 and a²b / (a² + b²). not knowing which x-coordinate went with which y-coordinate, i plugged x = 0 back into one of the original quadratic x² + y² = ax to find that y = 0. and so it would seem that the points of intersection are (0,0) and (ab² / (a² + b²) , a²b / (a² + b²)). this is as far as i managed to work it before starting to feel like this route was a dead end:

letting y = ax / b, we have:

x² + y² = ax
x² + (ax / b)² = ax
x² + (a² / b²)x² = ax
b²x² = a²x² = ab²x
(a² + b²)x² - ab²x = 0
x(a² + b²)x - ab²) = 0

so x = 0 and

(a² + b²)x - ab² = 0
(a² + b²)x = ab²

x = ab² / (a² + b²)

likewise, when i let x = (b/a)y and solve x² + y² = by for y, i get:
y = 0 and
y = a²b / (a² + b²)


now at this point i substituted my x and y values into y':

y' = ((a - 2x) / 2y)
= (a - (2ab² / (a² + b²))) / (2a²b / (a² + b²))
= ((a(a² + b²) - 2ab²) / (a² + b²)) x ((a² + b²) / 2a²b)
= (a³ + ab² - 2ab²) / 2a²b
= (a³ - ab²) / 2a²b


similarly, y' = 2x / (b - 2y)
= (2ab² / (a² + b²)) / (b - (2a²b / (a² + b²)))
= (2ab² / (a² + b²)) / ((a² + b²) / (b³ + a²b - 2a²b))
= 2ab² / (b³ - a²b)

when i got to this point, i saw that, again, the derivatives did not look like negative reciprocals. that is where i gave up on solving the problem using points of intersection earlier. but your idea of trying to show that the derivatives, when multiplied together, equal -1 gave me the motivation to try the same technique here as well...and sure enough:

((a³ - ab²) / 2a²b) x (2ab² / (b³ - a²b))
= (2a⁴b² - 2a²b⁴) / (2a²b⁴ - 2a⁴b²)
= -(2a²b⁴ - 2a⁴b²) / (2a²b⁴ - 2a⁴b²)
= -1


...so i guess i got intimidated and lost confidence in finding the points of intersection, when really i was on the right track all along. however if it weren't for your suggestion to try and show that the product of the derivatives is -1, i wouldn't have known that the derivatives i had were in fact negative reciprocals of one another. using the rule that the product of reciprocals of opposite sign is -1 was key in solving this one i think.

thanks again,
Eric

 

[Dick]

You could also have taken y'=(a^3-a*b^2)/(2*a^2*b) and canceled an 'a' from the numerator and denominator and gotten y'=(a^2-b^2)/(2ab). If you do a similar thing with the other one, they will look a lot more like negative reciprocals.

 

[94JZA80]

You could also have taken y'=(a^3-a*b^2)/(2*a^2*b) and canceled an 'a' from the numerator and denominator and gotten y'=(a^2-b^2)/(2ab). If you do a similar thing with the other one, they will look a lot more like negative reciprocals.

ah, i see that now. that really just cuts down on the number of manipulations it takes to show that the products of the derivatives is -1.


you've actually provided me with two very important pieces of information to use when approaching "orthogonal curve problems" (or "orthogonal families of curves problems") for which it is not immediately obvious via differentiation that the derivatives of two curves (or the derivatives of the general forms of two families of curves) are negative reciprocals:

1) you've shown me that it still isn't necessary to take the longer route of finding the exact points of intersection between two curves (or the generic points of intersection between two families of curves) and calculating the derivatives at those points to check for orthogonality. that is, you've shown me that if i can manipulate either of the original equations to express y explicitly in terms of x (or x explicitly in terms of y), i can substitute that value for y (or x) into one y' and most likely show that it is the negative reciprocal of the other y'.

2) you've shown me that my ability to prove orthogonality depends heavily on HOW i manipulate the original equations and eliminate variables and/or constants. in some instances, it may be beneficial to eliminate x-terms (or y-terms) in order to express an equation entirely in terms of y (or x). in other instances, it may be better to eliminate constants in the equation. either way, it allows me to evaluate y' entirely in terms of a single variable x or y, making it much easier to show that one derivative is the negative reciprocal of the other.


thanks again for all the insight...i feel like i know this material 100 times better than i did a few days ago.

Eric