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Sketch a typical level surface for the function.$$f(x, y, z)=\ln \left(x^{2}+y^{2}+z^{2}\right)$$

$$\begin{array}{c}\text { If } \ln \left(x^{\wedge} 2+y^{\wedge} 2+z^{\wedge} 2\right)=k \\\text { Then } x^{\wedge} 2+y^{\wedge} 2+z^{\wedge} 2=e^{\wedge} k\end{array}$$

Calculus 3

Chapter 14

Partial Derivatives

Section 1

Functions of Several Variables

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

Lectures

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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Sketch a typical level sur…

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in this problem trying to sketch the typical level surface with a given function At this first, try to find an equation. Total level, surface. Oh, what we can spray is there. Thing is equal to the natural Long X squared one square does the square where seems just, um, a constant If we raise both take a to both sides. What we have here into the speed is equal to x squared was who I squared. Buzzy Square? No. Based on this equation, we can see that the level curve is it's very that's fair centred at at the origin. Okay, now, if we were to pick the value the equals one, we would haven't equation. No, actually. Sequel zero. That's just single zero. Then we have is the equation ex cleared plus y squared. Plus, the square is equal to born. Anything to desire of power is one. So we'll see that there's a spare with radio Equalling one. Once we sketch, this is what we have. King. We'll grab the equation and here you can see that the radius is ended equal to one

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