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Numerade Educator



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 = kx^3 $


$2 x^{2}+3 y^{2}=c$


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Video Transcript

for this problem we are asked to find the orthogonal trajectories of the family of curves, Y squared equals K X cubed. And then we're asked you are asked to use graphing device draws several members of each family on a common screen. I will be skipping that step as me flooding that. I won't do much for you. So we have two white or starting out differentiating implicitly will get to Y Y prime equals K X cubed or a three times K X squared, I should say three K X squared on the right hand side. That means then we have that Y prime equals 3/2 times X times K and mr K there 3/2 times K X squared over. Why? Which now means that we have the equation for the slope or the slopes of that family of curves. We can reduced down the way that we're writing this a little bit. So it's three K X squared over two Y. So if we want the orthogonal trajectories. All right. Why with a little perpendicular sign, then the derivative there is going to be negative two, Y over three K X squared. So now we have a differential equation. We want to solve. We'll have that Y prime over why is going to equal negative to over three K X squared? Or we can write that as negative 2/3 K tend to X power of negative two. Now integrating both sides. We'll have that the right hand side. Second here, we still have that negative to over 3K. Then for X power of negative two we'll go to explore of negative two plus one. So that's exposure of negative one over negative two plus one. So over negative one that just can't swaps the sign and needs us with an X. In the denominator. And we want to include our constant integration. Then on the left hand side uh integral of one over Why is going to be one of why? So we have then that's why is going to be equal to some C. Times E. To the power of to over three K. X. To over three K. X. Right there.