💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

# Find the orthogonal trajectories of the family of curves. Use a graphing device to draw several members of each family on a common screen.$y = \frac {1}{x + k}$

## $x-\frac{y^{3}}{3}=C$

#### Topics

Differential Equations

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

### Video Transcript

Mhm. So to start finding an orthogonal to the family of orthogonal trajectories. Uh we need to take the derivative of the original career and we do so we differentiate both sides and we get that dy dx uh is equal to well, okay, by a coaching rule that you should remember it um it would we would have one plus K X minus um K X over the denominator squared. Um Of course you can see these, cancel it out. And uh now uh for orthogonal trajectory we're gonna basically have since its orthogonal it's derivative will be the negative reciprocal of that. So um for the orthogonal trajectory, the NYPD X will be negative um one plus K X squared. Before we continue, we want to see if we can make this easier for ourselves by looking back at what K actually is, That's what we usually do for these. Um It's not always basically using that very first curve. Uh you can rewrite this such to say that one plus K X is equal to x over Y. And therefore uh okay, is equal to X Over Y -1. Um Over yeah, over X. And when you divide both those two terms by X, you get K is equal to um won over why -1 over X. So we're going to try and see if it makes things easier for us to solve that into this. K and when we do that we get negative one plus. Okay, okay. Um we're gonna have one over y minus one over X times X. Now let's sum do distributive property, multiplying X by this annex by this and we get um uh plus X over Y. Uh minus. Well let's see. One over X times X would just be one. Look at that. Um This cancels out with this. So a derivative for the orthogonal trajectory is just negative X over Y square which is negative X squared over Y squared. And now we can just solve this by bringing wide to the left side and next and the dx to the other side, the right side. So it's kind of a cross multiplication. We get why squared dy is equal to negative X squared dx integrate both sides and we get them. Well let's see On the left side will have 1/3 of White Cube, Dizzy. And then and then on the right side will have negative 1/3 of x cubed plus an arbitrary constant. We can call see or I'll just uh I'll call that C. One for now. And we wanna um we want to have X and Y on the same side and preferably X. Should come before Y. So I'm just gonna add that one third of uh I'm going to add one third of X cubed to each side and we get one third of x cubed Plus 1 3rd of Why huge is able to see one. And finally Multiply both sides by three. We get execute plus uh like you is equal to uh is equal to three C. 1. But I'm just gonna uh call that a new arbitrary constant call that just see. And this is what our this is our family of orthogonal trajectories. Um The next part is to um show the graphs of several members of each family. So um using gizmos, this is what you get. Um I've plotted uh first for four different members of the uh X over one plus K X. Family. Uh first K equals zero and one to a negative too. Um And then orthogonal trajectories where first sequel, zero ct was negative wine one and 2. And the point is you can look at this these plots and cm that the trajectories um They do in fact intersect um when they do intersect the this family of uh curves, they do. So unfortunately you can see that it appears that this is the right right angle, right angle, right angle, right angle, right angle, right angle, etcetera. So based on this, it does seem that we have found our family of orthogonal trajectories correctly

The University of Western Ontario

#### Topics

Differential Equations

Lectures

Join Bootcamp