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- Given a function f (x, y) f (x, y) and a number c c in the range of f f, a level curve of a function of two variables for the value c c is defined to be the set of points satisfying the equation f (x, y) =c f (x, y) = c. Returning to the function g(x, y) = √9−x2 −y2 g (x, y) = 9 − x 2 − y 2, we can determine the level curves of this function.
courses.lumenlearning.com/calculus3/chapter/level-curves/
Feb 28, 2021 · Calculus 3 video that explains level curves of functions of two variables and how to construct a contour map with level curves. We begin by introducing a typical temperature map as an...
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- Houston Math Prep
. THEOREM 15.12. The Gradient and Level Curves. Given a function. f. differentiable at. (a,b) , the line tangent to the level curve of. f. at. (a,b) is orthogonal to the gradient. ∇f(a,b) , provided. ∇f(a,b)≠0. . Proof: Consider the function. z=f(x,y)
Level curvesInstructor: David JordanView the complete course: http://ocw.mit.edu/18-02SCF10License: Creative Commons BY-NC-SAMore information at http://ocw.m...
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Jan 22, 2022 · In order to find a few level curves, I began by calculating the following for a constant c: $e^{\sqrt{x^2-y^2}}=c$, This gives $\sqrt{x^2-y^2}=\ln(c)$ and $c>0$. The first level curve, when $c=1$ : $$g(x,y)=e^{\sqrt{x^2-y^2}}=1\implies x^2-y^2=0\implies y=±x,$$ which I can I can easily draw, but I am having trouble finding more level curves.
Figure 4.8 Level curves of the function g (x, y) = 9 − x 2 − y 2, g (x, y) = 9 − x 2 − y 2, using c = 0, 1, 2, c = 0, 1, 2, and 3 3 (c = 3 (c = 3 corresponds to the origin). A graph of the various level curves of a function is called a contour map .
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The level curves are given by $x^2-y^2=c$. For $c=0$, we have $x^2=y^2$; that is, $y=\pm x$, two straight lines through the origin. For $c=1$, the level curve is $x^2-y^2=1$, which is a hyperbola passing vertically through the $x$-axis at the points $(\pm 1,0)$.
