Ind the orthogonal trajectories of the family of curves

In summary, the author is trying to solve a problem in calculus and is stuck. He gets help from a friend and finds that he is on the right track, but is confused about how to graph the solution. He eventually realizes that he will be able to graph it if he substitutes k for x-1/y in the differential equation and integrates.
  • #1
Bel_Oubli
18
0

Homework Statement



Find the orthogonal trajectories of the family of curves. Use a graphing device to draw several members of each family on a common screen.
y = x/(1+kx)

2. The attempt at a solution

I have been trying this problem for hours, and I get a different answer every time. I have a big exam over this material tomorrow, and I just discovered that someone in this forum may be able to help me. Any help will be GREATLY appreciated!

The following is my most recent attempt. This was after I got help from a friend. I know that it must be wrong, but it's as far as I have gotten.

y = x/(1+kx)
y' = dy/dx = x/[(x+kx)^2]
slope of tangent line = x/[(x+kx)^2]
slope of orthogonal line = -(1+kx)^2
dy = [-(1+kx)^2]dx

After that, I am completely stuck. I do not think I am anywhere close to the answer. Also, I have no idea how to go about graphing it. Please help!
 
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  • #2
I'm ignoring what I think are typos, like the correct y' is 1/(1+kx)^2. Your final line looks good though. Now just integrate both sides. Use a substitution u=1+kx. Or just square it out if you have to.
 
  • #3
Do you really think I am on the right track? I thought I was supposed to take K out early on, but I do not know when, or what to replace it with. I just think that my final answer is not supposed to have K in it.
 
  • #4
Bel_Oubli said:
Do you really think I am on the right track? I thought I was supposed to take K out early on, but I do not know when, or what to replace it with. I just think that my final answer is not supposed to have K in it.

Yes, I think you are on the right track. And I don't see why you think the final answer shouldn't have a k in it.
 
  • #5
My final answer is y=-[(1+KX)^3]/3K + C. I think that should be correct. To graph it, will I just graph the original equation y = x/(1+kx) and then add y=-[(1+KX)^3]/3K + C for different values of K and C?

In my math book, it says that K is an arbitrary constant and that you have to do it for all values of K, so you find K as follows:
y = x/(1+kx)
1+kx = x/y
k=(x-1)/y

I just do not know when I would add it into the problem. Perhaps this is correct:
y = x/(1+kx)
y' = dy/dx = x/[(x+kx)^2]
And replacing k with (x-1)/y...
slope of tangent line = x/[(x+[(x-1)/y]x)^2] = (xy^2)/(x^4)

Does that look right? I'm so confused...
 
  • #6
For someone who has done all the right stuff, you are mighty confused. Let f(x)=x/(1+kx) and g(x)=c+(1+kx)^3/(-3k). You have shown that f'(x)=-1/g'(x). So if you put in any values of c and k, then if f(x) and g(x) intersect, they will intersect at a right angle. They are orthogonal.
 
  • #7
LOL, I think I'm just getting stressed out because it's almost the end of the semester, and I want to be prepared for Calculus III next semester. Thank you so much for helping me; I will just go to bed and hope I pass the exam tomorrow.

Thanks again!
(:
 
  • #8
Good strategy, you probably will if you get some sleep. You are doing a lot of stuff right.
 
  • #9
Dick, the differential equation defining the family y= x/(1+kx) should NOT have a "k" in it. The k determines which member of the family you have. Once you have determined y'= f(x,y), write y'= -1/f(x,y) and integrate that to find the orthogonal trajectories. You will get a constant of integration in that, of course.

I would be inclined to do it this way: write y= x/(1+kx) as y(1+ kx)= x. That allows you to do two things. First, differentiating both sides, y'(1+ kx)+ ky= 1. Also y+ kxy= x so kxy= x- y and you have both kx= (x-y)/y and ky= (x-y)/x. y'(1+ (x-y)/y)+ (x-y)/x= 1.
1+ (x-y)/y= x/y and 1- (x-y)/x= y/x. The equation of the original family is (x/y)y'= (y/x) or y'= y2/x2.
 
  • #10
Thanks, Halls. Or just write (1+kx)=x/y and put that into the derivative the OP already has y'=1/(1+kx)=y^2/x^2. I had a feeling something was going awry...
 
  • #11
Wow, my answer was really wrong. I guess it does not matter since my exam has already passed, but thanks for the help, guys.
 

What is the concept of orthogonal trajectories?

The concept of orthogonal trajectories refers to a family of curves that intersect at right angles. In other words, the tangent lines of these curves are perpendicular to each other at their points of intersection.

How do you find the orthogonal trajectories of a family of curves?

To find the orthogonal trajectories of a family of curves, you can use the method of differentiation. First, find the differential equation of the original family of curves. Then, use this equation to find the differential equation of the orthogonal trajectories by swapping the variables and changing the sign of the constant term.

What is the significance of orthogonal trajectories in mathematics?

Orthogonal trajectories have several applications in mathematics, physics, and engineering. They can be used to solve problems related to electric and magnetic fields, fluid dynamics, and heat transfer. They also have applications in optimization and curve fitting.

Is it possible for two families of curves to have more than one orthogonal trajectory?

Yes, it is possible for two families of curves to have more than one orthogonal trajectory. In fact, some families of curves have an infinite number of orthogonal trajectories. This depends on the nature of the original family of curves and the differential equation used to find the orthogonal trajectories.

Can orthogonal trajectories be used to find the solution of a differential equation?

No, orthogonal trajectories do not directly provide the solution to a differential equation. However, they can be used to simplify the process of finding a solution by converting a difficult problem into a simpler one. They can also be used to verify the correctness of a solution obtained through other methods.

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