Find the linearizations L(x, y, z) of the functions at the given points. f(x, y, z)=√(x^2+y^2+z^

Find the linearizations L(x, y, z) of the functions at the...

Key Concepts

Multivariable Differentiation

This concept involves working with functions that have more than one independent variable. In such functions, each variable can be varied independently and the behavior of the function is studied by considering the changes with respect to each variable. It lays the groundwork for understanding and computing derivatives in a multi-dimensional context.

Gradient Vector

The gradient vector of a multivariable function is the vector of its partial derivatives. It points in the direction of the greatest rate of increase of the function and its magnitude represents the rate of increase. The gradient is essential for linear approximations, as it provides all the necessary directional derivative information needed for forming the tangent plane at a given point.

Linearization (Tangent Plane Approximation)

Linearization is the process of approximating a function near a given point using its value and first derivatives. In the context of multivariable functions, this approximation takes the form of the equation of the tangent plane. The linearization formula uses the function value at the point of interest and the dot product of the gradient with the displacement vector, providing a simple linear model that approximates the function locally.

Chain Rule

The chain rule is a fundamental principle for differentiating composite functions. When a function is composed of an outer function and an inner function, the chain rule allows us to compute the derivative of the composite function by taking the derivative of the outer function evaluated at the inner function and multiplying it by the derivative of the inner function. This is particularly useful in functions like the square root of a sum of squares, where the derivative must account for the inner function's structure.

Transcript

00:02 For this problem we have f of x, y, z equals the square root of x squared plus y squared plus z squared.

00:19 And we want to find the linearization of this.

00:23 The linearization comes from evaluating the function at a specific point.

00:30 And then adding on the partial derivative of the function with respect to each variable times the quantity, the variable minus the coordinate of that point.

00:43 The good news is because these are all similar functions for our variables, the partial derivatives are all going to be looking very similar as we do those.

00:54 It is a little time -consuming because having the power of one -half, for this in the place of the square root.

01:03 When we're doing our derivative, we have to use the power rule plus the chain rule for each one, because that's that quantity to the one -half power.

01:16 So the linearization of the function is going to be the square root of x squared plus y squared, plus z squared, plus the partial derivative with respect to x.

01:35 So again, using the power rule, we take the power times anything in front of the function times the x squared plus y squared plus z squared to the one less power.

01:51 So that'd be to the negative one -half power times the derivative of our function with respect to x and y, and z or constants when we're doing that so that'll be times 2x and then that gets multiplied by x minus the x coordinate of our point and that's just the x part and now we have to do the same thing for y and z and as i said because these are identical looking expressions for y and z we're going to end up with the same partial derivative just with the different variable so the partial derivative with respect to y is going to be one -half times the quantity x squared plus y squared plus z squared to the negative one -half power times two y times the quantity y minus y -sub -zero and then plus the partial derivative with respect to z so one -half times the quantity x squared plus y squared plus z squared to the negative one -half power times 2z.

03:11 There's our partial derivative times z minus z -sib -0.

03:17 And then within each one of these partial derivatives, we can do a little reducing the one -half and the twos will reduce.

03:28 And when we are evaluating these, remember that the negative exponent is going to put that square root in the denominator.

03:35 So our first point where we are trying to find this linearization for this problem is the point 1 -0 -0.

03:48 So plugging 1 in for x and 0 -in for y and z, our linearization, ends up being 1 squared plus 0 squared plus 0 squared, so the square root of 1 is 1 plus 1 in the denominator under a square root.

04:11 So that'll just become 1 times x minus x0, the 1.

04:20 And then with y equaling 0, get rid of that upon its own, with y equaling 0, all of this becomes 0 times y minus 0.

04:45 And the same thing, with z equaling 0, this entire expression becomes 0 times y minus 0...

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