Problem

Fix a real number $m$. Let $S_{1}$ denote the fam…

Problem 31 Hard Difficulty

Determine the orthogonal trajectories to the given family of curves. Sketch some curves from each family.
$$x^{2}+y^{2}=2 c y$$

Answer

$y=(x-c) c_{1}$

Related Courses

Calculus 2 / BC

Differential Equations and Linear Algebra

Chapter 1

First-Order Differential Equations

Section 8

Change of Variables

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Video Transcript

So for this problem, we've been given this trajectory of curves, and we've been asked to find the trajectory of people for Ganahl curves and then sketch a few of those. So first thing to do is we need to work out the trajectory off the off the purpose that we've been given. And we do that by differentiating this equation with respect to X. When we do this term by term, so different shading with respect to X first term becomes two x second term becomes too. Why then multiplied by d Y, the X and the right hand side becomes to see times d y the X. Now we can rearrange us to get a differential equation which governs the trajectory off the curves that we have been given on this differential equation. Rearranges to give us that he wanted the X It's equal two x over C minus y. Now, in this current form, we can't say that this explicitly governs extract Aries because we still have this constant time. See, we want this entire equation to be in terms of X and y only. So what we need to do is need to go back to our original equation here. Rewrite this for C. So if we do this here, we find that C is equal to x squared plus y squared over to why? Where we've just divided for by two. Why for all equation This thing gives us that the one of the X equals X over exc lead plus y squared over two. Why minus one And then if we most play through on this side by two Wine This is equal to two x y over x squared plus y squared minus 21 squad and then we have some cancellation on the bottom. So final equation that governs of shek trees for given curves is two x y over x squared minus y squared. Yeah, now for you or for colonel curves. So if we say that our family of curves that we have given by coordinates x one and wild one your for colonel curves must satisfy But they're equation d y to the X two. When multiplied with dia de y want the X one is equal to minus one and this is satisfied in all case by the curves. If b y the x that eagles two minus X squared minus y squared over two x Y. And what we've done here is we've just flipped the solution we found earlier and then put in minus sign in front. So this is the differential equation, which we want to solve now. Looking at it, it looks like a homogeneous equation, and indeed it is because we can rewrite it as follows. So do you want the X equals minus X squared times one minus y squared over X squared well over X squared times two y over X, Be common factor of X squared on the top and bottom cancels, and we are left with a function that's completely dependent on why over Rex, this means we can make the variable substitution off B equals Y over X, which is again equivalently given by y equals v Times X. This means that D Y DX is equal to V plus X DVD X by product rule. So if we transform this differential equation into the V variables, we have that the plus X DVD X is equal. Tu minus one minus v squad of the TV Andi, if we just bring the minus sign into a fraction this is B squared, minus one over to be. Now we can subtract the from both sides, but it's minus fee here and then bring this all over. One fraction on this becomes free V squared minus one over to V. So the first head in the last 10 are the separable differential equation for the or for colonel trajectories. Now, to solve this, we separate the V variables to the left hand side on the X variables to the right hand side. So we have the two V over free V squared bonnets. One TV is equal to one over x the x To solve this movie, we need to integrate. Now the integral of two V over free V squared minus one, could be solved using with substitution. U equals free v sled minus one. Andi, this transforms Devi as the U equals six v Even putting this into the integral, we find that the interval of two V of a free V squared minus one TV is equal to eventual afraid over you. You on this simply has solution free times, natural log of you which back in all the variables is free house natural log the B squared minus. Sorry if natural log of free V squared minus one city over Integral theatre. One of Rex by simply the natural log of X. So our entire solution reads free. That's natural. Log free V squared minus one is equal to the natural log of X times. Plus the integration constant, which we will write is the natural log is okay so we can combine the log laws here on the left hand side. So this is now equal to Ellen XK, and then we can just remove the natural log. So because we have it on both sides Andi, when we get on the left hand side, we need to take the free that's in front of it. Power service gives us the natural log free V squared, minus one, well cubed. And then we remains the natural locks. So the solution for V is given by free V squared minus one. Cubed is equal to X K. Where K is the integration constant. The solution for why we simply need to substitute back in that why sled is equal. The square V is equal to y correct. Sorry on. And this gives is that free. Weiss word one sex squared cubes is equal to X multiplied, the extra four multiplied. Okay. And what we've done there is We've substitute multiplied fruit by X cubed different get in any form. So this is our solution, the vehicle funnel trajectories. We've also been asked to sketch and I've sketched some earlier up here. So what we have is the affordable trajectories, but given by he's curves have four branches, one in each quadrant. Andi, I have sketched two calves. The red curve has an integration constant, which is lower than the integration constant of green curve, and we see that it therefore has a lower Grady int.

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