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Find the orthogonal trajectories of the family of…

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Problem 30 Easy Difficulty

Find the orthogonal trajectories of the family of curves.Use a graphing device to draw several members of each family on a common screen.
$$y^{2}=k x^{3}$$

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Calculus 2 / BC

Calculus Early Transcendentals

Chapter 9

Differential Equations

Section 3

Separable Equations

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Video Transcript

Oh yeah. So in order to find the Ortho orthogonal trajectory, we uh we need to get a differential equation so we can use that derivative. Um So what we do is we differentiate both sides with respect to X. And by chain roll. The left side will be uh to y times the derivative. And on the right side it's going to be um uh well they're executed three X squared. So we're gonna have three K X squared. And right away we could use the fact that K is if you re arrange that first equation, K is equal to mhm. Um Why squared over X cubed? So when we get here is a two Y times dy dx is equal to um three uh Y squared X squared over X cubed. Mhm. And we can cancel her. We can divide both sides by Y to uh get this and then um where and then we can um also can we can divide the expert by the X cube to get just over X. This will get you simplify things. And finally uh well we can divide both sides by two. Um to get that are derivative is three Y over two X. And now we're going to look at the orthogonal trajectory and its derivative. Um um for that it's going to be the negative reciprocal of factory it is that we had before. Um So that's gonna be negative two X over three Y mm. Um And and now we can just solve this provincial equation by multiplying by bringing wide to the left side and uh multiply multiply by dx. We basically we get that three Y dy is so essentially you can think of it as a cross multiplying but that is a negative two X. Dx. And now we can different, we can integrate both sides. And these are just done by power rule. You get 3/2. Y squared is equal to um negative X squared plus. Um an arbitrary constant which we can call, We just call that C1 for now and now we can just rearrange this um First by multiplying both sides by two to get three Y squared is equal to uh negative X squared Plus um to see one. And finally we we'd prefer to get have constants on one side. So I'd X squared to both sides. Get X squared. Oh and sorry. Um she also multiplying this by two but yeah we're gonna get them two X squared plus three Y squared is equal to two C one which we can just call see now since we're done. Um So okay so um uh so that is our the equation for our or third orthogonal trajectory care if um and the other part of the question is um showing the showing the graph the graph of this uh with different values of C. So so this graph here from gizmos shows The Orthogonal trajectories with value c values of 1, 2, 3 and four. They're these ellipses that you see here. And I also have applauded the um the original curve, the family of groups with with different K values. And you can you can see that it looks like they are in fact orthogonal are perpendicular to those to the trajectories that we found. Um here for example, it's like a right angle, right angle, right angle, right angle. So that suggests that we have found her or trajectory um a family correctly. Yeah.

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