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

## $x^{2}-y^{2}=K$

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Differential Equations

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### Video Transcript

so to find the orthogonal trajectory we need to first find the derivative of this some curve. So we're just gonna differentiate both sides. And we just get that. Do you have a X. Is negative K over X squared? And you look back at our original curve this can this tells us that K. Is equal to X. Y. So um in place of Kay here we can put X. Y. And uh we can simplify that to you have I. D. X. His um negative Y over X. And for our Ortho orthogonal trajectory. Um Hey I'm just going to draw a line to separate that basically that's gonna it's gonna have a derivative that's the negative or super goal of that. Um Because it's orthogonal to it. So that would be X over Y. We just saw this equation by cross multiplying I guess you could think of it as um So you get this. Okay and then you're gonna integrate both sides. Yeah and you get half of y squared. Uh It's able to half of X squared plus C. Um And then you can also play by to get that Y squared is equal to X squared plus to see. And we prefer that we have X before y. And X amount on the same side. So I'm gonna rearrange this to show X squared. Uh So X squared is equal to I squared minus two C. And then uh X explored minus Y squared is equal to negative to see. And we can just renate we can you use a different constant in place of that negative to see. So I'm gonna choose to call that capital K. So that's what our family of orthogonal trajectories is gonna look like. Um in terms of the equation and the other part of the question is to show graphs of different members of the families. Okay. And here is the, is what we get when we graph different um members of the two families here, I've graphed them well, different uh members of kfx um first four K is one than K is negative one case to case negative too. So you can see that would be this curve, this curve, this curve in this curve and then I dressed a few orthogonal trajectories um With different capital K Values 01 -1 and two. You can see um here this, this, this and this and basically the point is you can look and see that every time that an orthogonal trajectory intersects with one of the original with one of the kfx curves. Um you can it appears that it is in fact orthogonal or perpendicular here, it looks like there's a right angle, right angle, right angle, right angle, right angle, right angle etcetera. Their whole bunch that you can look at. So it appears that we have um uh found our orthogonal trajectory correctly

The University of Western Ontario

#### Topics

Differential Equations

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