Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

For the function \(f(x, y)=\sqrt{x^{2}+y^{2}}\), find the secondorder Taylor approximation based at \(\left(x_{0}, y_{0}\right)=(3,4)\). Then estimate \(f(3.1,3.9)\) using (a) the first-order approximation, (b) the second-order approximation, and (c) your calculator directly.

Step-by-Step Solution

Answer

First-order approximation of 4.98, second-order approximation of 4.98192, calculator value of 4.982.

Step 1: Calculate Partial Derivatives

To find the second-order Taylor approximation, we first need the partial derivatives of the function. The partial derivative with respect to \(x\) is \(f_x(x,y) = \frac{x}{\sqrt{x^2 + y^2}}\). The partial derivative with respect to \(y\) is \(f_y(x,y) = \frac{y}{\sqrt{x^2 + y^2}}\). The second partial derivatives are \(f_{xx}(x,y) = \frac{y^2}{(x^2 + y^2)^{3/2}}\), \(f_{yy}(x,y) = \frac{x^2}{(x^2 + y^2)^{3/2}}\), and \(f_{xy}(x,y) = -\frac{xy}{(x^2 + y^2)^{3/2}}\).

Step 2: Evaluate Derivatives at Point \((3,4)\)

Calculate the value of the function and its derivatives at \((3, 4)\). The function value is \(f(3,4) = \sqrt{3^2 + 4^2} = 5\). The derivatives are \(f_x(3,4) = \frac{3}{5}\), \(f_y(3,4) = \frac{4}{5}\), \(f_{xx}(3,4) = \frac{16}{125}\), \(f_{yy}(3,4) = \frac{9}{125}\), and \(f_{xy}(3,4) = -\frac{12}{125}\).

Step 3: Construct First Order Taylor Approximation

The first-order Taylor approximation is given by:\[ f(x, y) \approx f(3,4) + f_x(3,4)(x - 3) + f_y(3,4)(y - 4) \]Substitute the known values to get:\[ f(x, y) \approx 5 + \frac{3}{5}(x - 3) + \frac{4}{5}(y - 4) \]

Step 4: Construct Second Order Taylor Approximation

The second-order Taylor approximation is:\[ f(x, y) \approx f(3,4) + f_x(3,4)(x - 3) + f_y(3,4)(y - 4) + \frac{1}{2} f_{xx}(3,4)(x - 3)^2 + \frac{1}{2} f_{yy}(3,4)(y - 4)^2 + f_{xy}(3,4)(x - 3)(y - 4) \]Substitute the known values:\[ f(x, y) \approx 5 + \frac{3}{5}(x - 3) + \frac{4}{5}(y - 4) + \frac{8}{125}(x - 3)^2 + \frac{9}{250}(y - 4)^2 - \frac{12}{125}(x - 3)(y - 4) \]

Step 5: Estimate Using First Order Approximation

Using the first-order approximation formula:\[ f(3.1, 3.9) \approx 5 + \frac{3}{5}(3.1 - 3) + \frac{4}{5}(3.9 - 4) \]Calculate:\[ f(3.1, 3.9) \approx 5 + 0.06 - 0.08 = 4.98 \]

Step 6: Estimate Using Second Order Approximation

Use the second-order approximation:\[ f(3.1, 3.9) \approx 5 + \frac{3}{5}(0.1) + \frac{4}{5}(-0.1) + \frac{8}{125}(0.1)^2 + \frac{9}{250}(-0.1)^2 - \frac{12}{125}(0.1)(-0.1) \]Calculate each term:First order part: 5 + 0.06 - 0.08 = 4.98Second order corrections: \(\frac{8}{125}(0.01) + \frac{9}{250}(0.01) + \frac{12}{125}(0.01) = 0.00192\)Thus, \[ f(3.1, 3.9) \approx 4.98 + 0.00192 = 4.98192 \]

Step 7: Estimate Using a Calculator

With a calculator, directly compute the value of the function at \((3.1, 3.9)\):\[ f(3.1, 3.9) = \sqrt{3.1^2 + 3.9^2} \approx \sqrt{9.61 + 15.21} = \sqrt{24.82} \approx 4.982 \]

Key Concepts

Partial derivatives

Partial derivatives are essential in multivariable calculus and play a vital role in the Taylor approximation process. They are used to measure how a function changes as each input variable is varied independently, while holding the other variables constant. For a function of two variables like our example with function \(f(x, y) = \sqrt{x^2 + y^2} \), we take derivatives with respect to each variable separately. The partial derivative with respect to \(x\), denoted as \(f_x(x, y)\), is \(\frac{x}{\sqrt{x^2 + y^2}}\). This tells us the rate of change of the function along the \(x\)-axis. Similarly, the partial derivative with respect to \(y\), denoted \(f_y(x, y)\), is \(\frac{y}{\sqrt{x^2 + y^2}}\), which tells us how the function changes along the \(y\)-axis. These partial derivatives are the building blocks for the Taylor series expansion because they help us understand how slight changes in \(x\) and \(y\) affect the function's value.

First-order approximation

A first-order approximation, often called a tangent plane approximation for functions of two variables, gives an initial estimate based on the linear components of the Taylor expansion. It approximates the function by the value of the function plus a correction term that depends linearly on the deviation from the point of approximation. With our function \( f(x, y) = \sqrt{x^2 + y^2}\) at the point \((3, 4)\), the first-order approximation is calculated as follows:

• Start with the function value \( f(3, 4) = 5 \).
• Add the terms derived from the partial derivatives evaluated at the point: \( \frac{3}{5}(x - 3) \) and \( \frac{4}{5}(y - 4) \).

Thus, the first-order approximation formula becomes \( f(x, y) \approx 5 + \frac{3}{5}(x - 3) + \frac{4}{5}(y - 4) \). Using this approximation allows us to make quick estimates for different points by extrapolating the function's value based linearly on the nearest known point.

Function estimation

Function estimation using Taylor approximations is a qualitative method to predict function values near a known point. This approach utilizes both the first-order and second-order terms to provide a refined prediction. The first-order approximation primarily contributes to understanding linear shifts, whereas the second-order approximation adds corrections that account for curvature in the graph. With the second-order Taylor approximation for our function at \((3, 4)\), we consider additional terms that involve second-order derivatives:

• Square terms such as \((x - 3)^2\) and \((y - 4)^2\).
• Mixed terms like \((x-3)(y-4)\).

These account for the rate of change of the rates of change, meaning how much the function's rate itself is altering as we move across the plane. For small deviations from the point \((3, 4)\), the second-order approximation offers an enhanced accuracy over the first-order approximation by including these nonlinear effects, resulting in a more precise estimation as illustrated by our example with an estimated value of \(f(3.1, 3.9) \approx 4.98192\). Each level of approximation builds on the last, providing even better estimations for points near our base.