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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.

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

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 1

Differential Equations Everywhere

Differential Equations

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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'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'll obtain that negative X divide by two y is equal to the derivative off Why or d y DX. So it'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've just found. So this will be to why divide by X. Let'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'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's equal to to times the natural log of X plus a constant Let's call it V to match up with the variable V that we'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'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

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