Compute the linear approximation of the function at the given point. f(w, x, y, z)=w^2 x y-e^w y

Compute the linear approximation of the function at the...

Key Concepts

Tangent Plane

The tangent plane is the linear surface that best approximates a multivariable function near a given point. It is constructed using the value of the function and its partial derivatives at that point. The equation of the tangent plane is derived from the gradient and serves as the linear approximation to the function in the neighborhood of the point.

Gradient

The gradient is a vector that consists of all the partial derivatives of a multivariable function. It points in the direction of the greatest rate of increase of the function and its magnitude represents the rate of change in that direction. In the context of linear approximation, the gradient is used to form the equation of the tangent hyperplane at a given point.

Partial Derivatives

Partial derivatives measure the rate at which a multivariable function changes with respect to one variable while keeping the others constant. They are essential in constructing the gradient, which in turn is used to determine the linear approximation of the function. Through partial differentiation, we determine the slopes of the function in each coordinate direction.

Linear Approximation

Linear approximation involves approximating a function near a specific point with its tangent plane (or hyperplane in higher dimensions). For multivariable functions, the linear approximation provides a first-order approximation by utilizing the function's values and its partial derivatives at the point of interest. This concept is fundamental in multivariable calculus for simplifying complex functions locally by replacing them with their linear counterparts.

Transcript

00:01 In this problem we're finding the linear approximation of the given function at the given point.

00:09 First we will be finding the partial derivatives for wx, y, and z.

00:18 So first, let's do the partial for w.

00:24 And what we have is 2 wxy minus y times z times e to the w2.

00:40 Y z the partial with respect to x would be equal to w squared times y with respect to y our derivative is w squared times x minus w times z times e to the w y finally we find a partial with respect to c and that gives us negative w y times e to the w y.

01:44 Now for our first point, let's first evaluate the function and then we will evaluate for each of the partial derivatives at that point.

01:55 So here we have f of negative 2 310 that is equal to negative e negative 2.

02:16 Squared times 3 times 1 minus e to 0 it looks out to be 11 okay so evaluating our partial with respect to w gives us 2 times negative 2 times 3 times 1 minus 0 and that gives negative 12 evaluating the partial with respect to x gives us negative 2 squared times 1 which is simply equal to 4 and evaluating the partial with respect to c gives us negative or negative 2 times 1 times e to 0 and that is equal to to here our part our linear approximation so we started with w x y and c would be equal to 11 minus 12 times w minus negative 2 so that will become plus 2 plus 4 times x minus 3 plus okay so i have forgotten my f of y here to evaluate that would be my derivative with respect to y and what we get here is negative 2 squared times 3 minus 0.

05:01 So that gives us a positive 12.

05:06 So continuing, we have 12 times y minus 1, 2 times z minus 0.

05:31 Once we expand everything, we have 11 minus 12 w minus 24.

05:38 Plus 4x minus 12 plus 12 y minus 12 plus 2 z so our linear approximation will be the equation of negative 12 w plus 4x plus 12 y plus 2 z and once we combine all our constant terms we get negative 37 so partly we're given the point 0 .0 .0.

06:34 1, negative 1, and 2.

06:36 We're going to repeat the same process with this point...

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