Linear approximations and tangent planes. Linear approximation: multivariable version
The linear approximation for the function f of (x, y) is supposed to give a good approximation of how f behaves if we change x a little bit or we change y a little bit or both.
When we start at a point (×0, yO) and then imagine adding a small amount of change to x or y, we want to approximate the original value of fat (x0, yO). This is done by adding the changes that come from increasing delta × or decreasing delta y. When considering the derivatives of functions in multiple variables, we must remember that there are two different things we could change. We could change the values of x, or we could change the values of y. Thus, there are two different derivatives that describe the effect of these two different changes: one that describes how delta x affects × and another that describes how delta y affects y. The derivative of x with respect to × describes how a function changes if we increase × by a small amount. The derivative of y with respect to y describes how the function changes if we change y by a small amount. Let’s look at an example to see how it works. Linear approx for F(x, y) f(x_0+Δx,y_0+Δy)~~f(x_0,y_0)+f_x(x_0,y_0)Δx+f_y(x_0,y_0)Δy Ex. f(x,y)=x^2+y^2 f_x(x,y)=2x f_y(x,y)=2y (x_0,y_0)=(-1,1) f_x(-1,1)=-2 f_y(-1,1)=2 f(-1+Δx,1+Δy)~~2-2Δx+2Δy In our example, we will use the first function that you all discussed in recitation yesterday: f(x, y) = x2 + y2. In recitation, you computed its derivatives. So the x derivative. When taking a derivative of a function that includes both variables, it is important to think of one variable as a constant. For instance, if you were taking the derivative of *2 + 7x, the resulting function would have a constant value of 2x. And the derivative of that would be 0. Let (×0, yO) be (negative 1, 1). Then f(x0, y0) = 2. If (x0, yO) is substituted into f(x), then f(-1) = -2. We can now write the linear approximation to the function by plugging in negative 1 plus delta x, 1 plus delta y, and finding that the linear approximation is approximately 2. The first term to evaluate is the squared derivative of f(x) with respect to x, times delta x. This equals 2. Next, evaluate the derivative of f(x) with respect to y, times delta y; this is equal to negative 2 times delta y.
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