- #1

94JZA80

- 122

- 2

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)...

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

now consider the case where it is not obvious that the derivatives of two

now there are two exercises that ask me to show that two

x

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:

Show that the following two families of curves are orthogonal:

y = cx

x

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 = x

see above

[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 = cx

however, it quite difficult to find generic points of intersection of the functions y = cx

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

**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:x

^{2}+ y^{2}= r^{2}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 = cx

^{2}, where C is a constantx

^{2}+ 2y^{2}= k, where k is a constantFirst 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 = x

^{2}and x^{2}+ 2y^{2}= 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 = 0.5x^{2}+ 2y^{2}= 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 = -0.5x^{2}+ 2y^{2}= 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 x^{2}+ 2y^{2}= k take on imaginary values. and so a graph of the function x^{2}+ 2y^{2}= 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 = cx^{2}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 = cx^{2}and x^{2}+ 2y^{2}= 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 = cx

^{2}is orthogonal to any member of the family of curves x^{2}+ 2y^{2}= 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 x^{2}+ y^{2}= r^{2}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 x^{2}+ y^{2}= r^{2}and ax + by = 0 are orthogonal.however, it quite difficult to find generic points of intersection of the functions y = cx

^{2}and x^{2}+ 2y^{2}= 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

Last edited by a moderator: