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