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Find an equation for and sketch the graph of the level curve of the function $f(x, y)$ that passes through the given point.$$f(x, y)=\sqrt{x+y^{2}-3}$$

Level curve is $x+y^{2}=4$

Calculus 3

Chapter 14

Partial Derivatives

Section 1

Functions of Several Variables

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12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

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for this problem, we are given the function F of x Y. It was the square root of x plus y squared minus three. We are asked to find an equation for and sketch the graph the level curve for the function that passes through the point and actually I just realized I don't have that one second. All right, so the level curve is 3 -1 Or the given point rather 3 -1. So we can write our level curve as C equals root X plus y squared minus three. Which in turn means that c squared equals x plus y squared minus three. So we have that y squared equals c squared minus x minus three. And then why equals plus or minus the square root of c squared minus x minus three. Where we also need to figure out what exactly the value of C is going to be. We'll see we know that we have the point or three negative one. So we'd have the square root of three plus negative one squared. So that's three plus one minus three. So that's going to be just one or C equals one. So for trying to sketch the given um level curve, it would be plus or minus the square root of negative x minus two. Which 1 2nd here? Just realized that that should be a Um a plus three here. So that would be negative or The level curve that we're looking for would actually be 4 -1. Which then if we want to sketch have that as our origin And we want to go up to four, so it would be a parabola opening up this sort of way.

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