we-call-a-coordinate-system-u-v-orthogonal-if-its-coordinate-curves-the-two-families-of-curves-u-con

Question

We call a coordinate system $(u, v)$ orthogonal if its coordinate curves (the two families of curves $u=$ constant and $v=$ constant ) are orthogonal trajectories (for example, a Cartesian coordinate system or a polar coordinate system). Let $(u, v)$ be orthogonal coordinates, where $u=x^{2}+2 y^{2},$ and $x$ and $y$ are Cartesian coordinates. Find the Cartesian equation of the $v$ -coordinate curves, and sketch the $(u, v)$ coordinate system.

Michael Jacobsen · Accepted Answer

in this problem were given that a variable U is equal to X squared plus two y squared Merkel to determine what should a variable V B North for UV to be an orthogonal coordinates system To solve this type of problem, Our first step is to differentiate implicitly the original equation with respect to X. Upon differentiating the left hand side will be zero equals two x plus four y times D y DX. Our goal for the next steps of this process is now to solve for D Y DX. Let&#x27;s start by subtracting two extra both sides now negative two X equals For Why Times D. Y. D X. If we divide both sides by four y and cancel the common factors we&#x27;ll obtain that negative X divide by two y is equal to the derivative off Why or d y DX. So it&#x27;s the first step of this process for the next step and work to determine V we formed the derivative de y DX is equal to the negative, a reciprocal of the prior derivative that we&#x27;ve just found. So this will be to why divide by X. Let&#x27;s solve this differential equation by dividing both sides by why to start out so that one over. Why times de y dx will be equal to to divide by X that we can substitute the entire left hand side by the expression. The derivative with respect to X of the natural log off white itself is equal to one over Why times d y the X. So at the next stage, we can integrate both sides with respect to the variable X. So this is the format the differential equation now takes on the left. Inside, the integral sign and the derivative sign are telling us that we&#x27;re looking for the anti derivative of a derivative of the natural log of whites that allows us toe copy natural log of why, after cancellation, express that that&#x27;s equal to to times the natural log of X plus a constant Let&#x27;s call it V to match up with the variable V that we&#x27;re looking for. The next stage, we can make a simplification by converting the coefficient of two into an exponents. This allows us to write that the natural log of why is equal to the natural log of X squared, plus B. Then we can convert the Net. The log rhythmic equation into an exponential equation with base e by writing why is now equal to E to the power of natural log X squared plus B. And now we have that why is equal to each of the power v times e to the power of natural log X squared. As soon as we perform the cancellation between Basie on the natural log, we can write to that. Why is equal to Let&#x27;s go with the V Times X squared, where V is equal to each of the lower case V. So our solution then upon solving for V is the following will say If V is equal to why divide by X squared by solving for V in the last step of the process, then u V will be an orthogonal coordinate system. This is

Mike Gaerlan · Answer

Okay, So to start solving this problem first we need to find B y the X on both of the equations. So we&#x27;ll have the wind. T x on this one is just going to be able to em. Okay, now for this side here to take d y dx, we&#x27;re gonna take DDX a both sides. First, you have DDX X squared. Plus y squared is equal to DDX of a squared. So here. Now, if we take the dy dx of this side, that&#x27;s just gonna be zero on this side. Over here we have two x plus two y and then by chain roll D Y d x. So we can actually also cancel out the twos just dividing by two on both sides of the equation. So we&#x27;ll get that d Y dx. We are going to move over the X. So we get negative X, and then we&#x27;re gonna divide by Why? So with this in a vehicle to negative X over, Why now? We need to find where they intersect with each other. So the intersection is whenever the two equations are equal to each other. So, for example, if I plug in y equals MX. So if I plug that why into this here. So if I take d y d x of this side and I evaluated at or when y is equal to m X, this is equal to negative X over and X which is equal to negative one over m. So, as you can see here, this is the negative reciprocal of this. So therefore they are orthogonal.

Mike Gaerlan · Answer

First we need to find the slopes of the two equations. So to find people I d. X here, we&#x27;re gonna take DDX of both sides of this equation. So we&#x27;ll take DDX of X y is equal to d d x times or DDX of a So on the left hand side here, we&#x27;re going to need to use product will. So we&#x27;re gonna do first times derivative second. So that&#x27;s the Y DX plus second, which is why times derivative first, which is just one. So now we&#x27;re going to take the river the right hand side. That&#x27;s just gonna be with zero. So solving for DVD X, we&#x27;re going to get that D Y d X is gonna vehicle to negative. Why divided by X for this second family of solution or second family of curves? No, Again, we&#x27;re going to take DDX a both sides there, so we get DDX of X squared minus y squared. Is that a vehicle to dy dx of B? So here on this side, we have we&#x27;re going to get two x so derivative of X rays to X minus two. Why d y over d X is going to be equal to and then zero So next, uh, d y d x So solving for that we have the y t x is gonna be equal to we move. The, uh, we can move this over to the other side. So we have to Why do y the x and then just divide by two? Why here? So the twos will cancel out, and we just have X over y. Now, if you notice here this X over y is already the negative reciprocal of this here, so we don&#x27;t really need to do much so d y d x here, and I&#x27;ll say d one d x one here and then I&#x27;ll call this D to here. Um, well, this is already the negative. Reciprocal. Okay, so, uh, we don&#x27;t really need to show much there already with negative receptacles.

Mike Gaerlan · Answer

to find a terra Fogle. First we&#x27;re going to take d y t i or we need to find the idea of both, um, curves. So on the this left one here, do you i d x is just gonna be to see ex, Okay. And then now, on the right hand side to find you i d x, we&#x27;re going to first find DDX of both sides, so that&#x27;s gonna be DDX of this is interviewed two d d x of K. So then now the derivative of this is just gonna be zero. So we have is equal to zero and then derivative of this. Okay, so first, the derivative of X squared term, that&#x27;s one of equal to two X And then we have a plus for Why d y over d X. So then solving for D Y DX. Okay, let&#x27;s divide by two on both of the terms. So if we do that, then this becomes a to So now, uh, solving for the y t x. We&#x27;re gonna move this ex term to the other side. So we get negative X. So we have B. Why the X is gonna be able to then we have negative X divided by what&#x27;s remaining over here is to Why so Divided by two y Next, we&#x27;re going to evaluate our derivative here when, uh, we&#x27;re intersecting this graph here. So that means we&#x27;re going to find d Y d X at X comma C x squared. We&#x27;re gonna plug in C X squared for for X or for wines. We have negative X over two men see x squared. So then this is gonna become negative. Um, well, the one of the X terms will cancel out. So we get negative one over to see ex, which is the negative reciprocal of this here. Right. So this are these two families are orthogonal, right? So this okay is equal to here. Negative one over d Y DX for the other. You know, the other turn. Okay.

Chris Trentman · Answer

we&#x27;re told two curves are orthogonal, their tangent lines are perpendicular at each point of intersection. Were asked to 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. Then we&#x27;re asked to sketch both families of curves in the same axes. The first family of curves is x squared plus y squared equals r squared. These are circles centered at the origin and the second family of curves is A. X plus B. Y equals zero. These are lines through the origin. Intuitively this seems like orthogonal trajectories. Check this. However, when do three things first will differentiate the equation of the family of curves. Then we&#x27;ll paint a differential equation and then we&#x27;ll solve this differential equation. So first well differentiate both sides of our first family with respect to X. So we get two X Plus two y times. Dy DX equals zero. Very common. And therefore solving for dy dx we get dy dx equals negative X over Y. Now, in our second step I&#x27;m gonna replace dy dx Oh yeah with negative dx Dy in this way will generate curves which are orthogonal to our original family, X squared plus y squared equals r squared. So we get negative dx dy equals negative X over Y. Therefore dx dy equals X. Y. This is a separable differential equation which is easy to solve. So we get D X over X equals dy over why. We split up the derivative into differentials, integrating both sides. We have the integral of D X over X equals the integral of Dy over why. And therefore the natural log of the absolute value of X equals the natural log of the absolute value of Y plus some constant C. Hi guys now combining terms and using log properties, we get the natural log of absolute value of X divided by Y equals C. Using inverse functions we have absolute value of X over Y equals E. To the constant C. Right. And therefore we obtained the X over Y equals plus or minus E. To the sea. Yeah. Now we&#x27;ll let plus or minus E to the C. Which is a constant. Be B over a. Sorry, negative B over A. Then it follows that X over Y equals negative B over A. And therefore A X plus B. Y equals zero. Or thanks. We have shown that these two families of curves right, X squared plus Y squared equals R squared and A X plus B. Y equals zero are orthogonal trajectories. I love beef jerky. That was, you know, okay. So I was vegan but I did eat beef jerky.

Frank Lin · Answer

So for this problem again with computed derivative by imprisoning depreciation and as you and we beautify, there are Flo. And we can use these to evacuation to chance things because on their interception on both, the question should be satisfied so we can use those equation to swap variables. And then here we go the Internet before means is to ah, lines are ah is two lies are prevented from each other, thes to tension. And so these are all over them curves.

Problem

Any curve with the property that whenever it inte…

Problem 22 Medium Difficulty

Answer

$\frac{-x}{2 y}$
$y=v x^{2}$

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Integration Review - Intro

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$\frac{-x}{2 y}$$y=v x^{2}$

Differential Equations and Linear Algebra

Catherine R.

Heather Z.

Kristen K.

Samuel H.

Integration Review - Intro

Definite vs Indefinite

Stephen W. Goode, Scott A. AnninDifferential Equations and Linear Algebra

Catherine R.

Heather Z.

Kristen K.

Samuel H.

$\frac{-x}{2 y}$
$y=v x^{2}$

Stephen W. Goode, Scott A. Annin

Differential Equations and Linear Algebra