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Show that the ellipse $ x^2/a^2 + y^2/b^2 = 1 $ a…

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Problem 68 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 = ax^3, x^2 + 3y^2 = b $

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Video Transcript

we're told that two curves are orthogonal if their tangent lines are perpendicular at each point of intersection. Whereas to show that the given families of curves are orthogonal trajectories of each other, That is that every curve in one family's orthogonal to every curve in the other families. Then we have to sketch both families of curves in the same axes. Wow. Well our families of curves are Y equals X cubed. So these are translations or transformations of the cubic function where it's stretched vertically. And the other family of curves is x squared Plus three, Y is squared equals B. So these are ellipses centered at the origin which appear to have a horizontal major axis. I really don jr I'm brian. Well there's three steps here. First you want to differentiate the equation of the family of curves so the constant in the first equation gets eliminated, then we'll obtain a differential equation and finally we'll solve the differential equation we obtain. So you begin with our family of curves, Y equals a X cubed. Whoa, That's a good Richardson. Now I'll solve for our constant so that we get it on one side, I get y over X cubed equals a, wow, that sounds exactly. We want to find the derivative, the slope of these curves, but we want to find an independent of the constant. A. So taking the derivative with respect to X at both sides, we have the bottom X cubed times dy dx minus the top, Y times the derivative of the bottom three X squared all over the bottom X cube squared which is X to the sixth And the right hand side is simply zero. So now I've eliminated a this is good because now our slope is only going to depend on Y and X, solving for the dy dx we have x cubed, dy dx minus three, x squared y equals zero and therefore dy dx equals three Y over X. Now this is the slope of our first family of curves. We know that the slope of our second family of curves. Well, this has to be negative reciprocal. So this is going to be we want to substitute. No. Yeah uh negative dx dy for dy dx. So the tangent lines are perpendicular at the point of intersection. Therefore we get negative dx dy Equals three Y over X. This is a separable differential equation. We can write this as X D X equals negative three Y dy. So now I'll take the anti derivative of both sides. You're so we get X one half X squared equals negative three halves Y squared plus a constant, which I'll call. See it's pretty fair, I don't know what's the cool thing. I get X squared plus three Y squared is equal to to see if I set to C equal to the constant be right. Then the equation becomes X squared plus three Y squared equals B. We recognize these are the same as the family of equations we were given. Oh, therefore it follows that the given families of curves are orthogonal trajectories of each other. So we have verified this fact. Now I'm going to sketch both families of curves and the same axes has said before the first family of curves. These are transformations of the cubic with I guess you could say the vertex of the cubic at the origin and they're just stretched vertically or horizontally. We're really from. So for example if A was one, we'd have something like this and we get steeper. Well, okay, of course we can also get flatter honey. This is 44 minutes. Yeah, we stop bonus 10. We can also if we have a negative a flip it that were like this of good of good shit. Anyway we did my friends right hurry up. You've been in the park and I'm recently a new vehicle. So my body has just have not know what why. Now the second family of curves as I said, these are ellipses with a horizontal major axis centered at the origin. I'm so they probably look something a bit like this. So one. Yeah. Will you suck their dicks? Will you go home? Yes. Yeah. Well we started with gay Star Wars. That was a star was you know, a lot of you are going to say this happen. Mhm. How many times little eating on you want to have sex, right? Not him. Looking at the ellipses and the cubic functions. They do appear to be orthogonal trajectories of each other. That is the tangent lines at the point of intersection. Like here here, here and here do appear to be orthogonal

Chris T.

Ohio State University

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Problem

Show that the ellipse $ x^2/a^2 + y^2/b^2 = 1 $ a…

Problem 68 Hard Difficulty

Answer

SEE PROOF

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Calculus: Early Transcendentals

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