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Problem 67 Hard Difficulty

Two curves are orthogonal if their tangent lines are perpendicular at each point of intersection. Show that the given families of curves are orthogonal trajectories of each other; that is, every curve in one family is orthogonal to every curve in the other family. Sketch both families of curves on the same axes.
$ y = cx^2, x^2 + 2y^2 = k $


The two curves are orthogonal

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Video Transcript

we're told to curbs are orthogonal, their tangent lines are perpendicular at each point of intersection. Were given to families of curves and were asked to show that the given families of curbs are orthogonal trajectories of each other. That is that every curve in one family it was orthogonal to every curve and the other family. And then we're asked to sketch both families of curves in the same axes. The first family is Y equals C. X squared. These are parabolas with vertex at the origin and a vertical axis. And the other family is X squared plus Y squared. Or I should say two Y squared. Yeah. Yeah equals K. These are ellipses imagine so which are centered at the origin and with a five horizontal major axis was trade expensive back weaker tone. So in order to verify that these are orthogonal trajectories. is there really a three part method to this first, I'm going to begin with the equation for. Mhm. Well on the second, the equation for our parabolas, Y equals X. Squared. I think the derivative of both sides with respect to X. We get dy dx equals two C. X. Um goddamn it. I just want I just want to see the bridge blow up. So we have dy dx now for this to be find the orthogonal trajectory. We want to substitute, it's negative dx. Dy for dy dx because we want our tangent lines be orthogonal to tangent lines of the original curves. So now we have the equation negative dx Dy equals to see X. Kill that motherfucker cap his ass. Yeah. Just bullshit. Why don't he's helping years on the same side? Just making me So that's how you walk now. This is a separable differential equation which we can easily solve. We have D. X over X equals negative two C. D. Wine and we can integrate both sides taking anti derivatives. He's fucking the whole shit up. It's making me this is the natural log of the absolute value of X equals negative to see times Y plus some constant, which I guess I'll call a right. All right, lose what is Dave just told him to be right. So he was tweeting surgeon dude, trust the government for you. Right. Just either that corporations such parties got from nestle doesn't want people to have this. It doesn't make Yes. Well, news charts, this is one method a better method might be instead. Yeah. Yeah. Let's solve our first family of equations for the constant C. So we have Y over X squared equals are constant C. Now we'll take their derivative of both sides with respect to X. So we have the bottom X squared times the derivative of the top dy dx minus the top. Wide times the derivative of the bottom two. X. All over the bottom. X. to the 4th. On the right hand side, this is zero. Okay, now we replace Rome. Want to solve for dy dx We have x squared dy dx minus two. X Y equals zero. Therefore dy dx equals two. Why Coover X? Now we replace with negative dx dy to get negative dx dy equals two. Y over X. Yeah. And then this is a separable differential equation which we can write as dx times X equals negative two. Y. Dy integrating both sides. I get one half X squared equals negative Y squared plus some constancy. And therefore we have X squared plus two. Y squared equals two times some constant big C. And if we set two times big C equal to the constant K. Oh, it's just then it follows that we get x squared plus two y squared equals. Okay. Yeah. Therefore we have shown at least two equations. These two families. Yeah, discuss of curves. Look if you have rock bringing a cross strait, well these are orthogonal trajectories. It wasn't there like some of each other legislation? Yes. Go to drink immediately afterwards on and said, yeah, gentlemen, allow me to demonstrate. Now we'll sketch families of both curves on the same axes. So the first family. Again, these are just parabolas with vertical axes centered it with vertex at the origin. So they look like this, they just rest. And so I'm a put on my second pair of pants which I brought as they continue. I'm sorry that this is now my 3rd presentation for feeling about explosive diarrhea. I retract both broad legislation. Grilled two sandwiches for everyone. I see I see the other family of curves as I pointed out before. These are ellipses centered at the origin with the horizontal major axis. Mhm. So which is one of my friends screwed up. In fact, the horizontal axis drink by advising Vicky's. Well, it'll look something like this. That This is how the 3rd field gracious home of the american senate ball massive amounts of heat. Yeah. Mhm. Yeah. Did you have actually say about rising ellipses look something like this expert advice? Mm The whole time. Well, on the top I watched that like see like we can't touch the ground. Right, right. Yes. Classic. Did you ever trip on any everything says she called salvia? Yes. From the graph. It seems intuitive if not possible, that these families of curves are in fact orthogonal trajectories.