Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

For the function \(f(x, y)=\tan \left(\left(x^{2}+y^{2}\right) / 64\right)\), find the second-order Taylor approximation based at \(\left(x_{0}, y_{0}\right)=(0,0)\). Then estimate \(f(0.2,-0.3)\) using (a) the first-order approximation, (b) the second-order approximation, and (c) your calculator directly.

Step-by-Step Solution

Answer

(a) 0, (b) 0.00203125, (c) 0.00203125.

Step 1: Find the function value at the base point

Calculate the value of the function at \((x_0, y_0)=(0,0)\). Substitute \(x=0\) and \(y=0\) into the function: \[ f(0,0) = \tan\left(\frac{0^2+0^2}{64}\right) = \tan(0) = 0. \]

Step 2: Calculate the first-order partial derivatives

Find the first-order partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). \[ \frac{\partial f}{\partial x} = \sec^2\left(\frac{x^2 + y^2}{64}\right) \cdot \frac{2x}{64} = \frac{x \sec^2\left(\frac{x^2 + y^2}{64}\right)}{32}. \]\[ \frac{\partial f}{\partial y} = \sec^2\left(\frac{x^2 + y^2}{64}\right) \cdot \frac{2y}{64} = \frac{y \sec^2\left(\frac{x^2 + y^2}{64}\right)}{32}. \]

Step 3: Evaluate the first-order derivatives at the base point

Substitute \((x_0, y_0)=(0,0)\) into the derivatives to evaluate them:\[ \frac{\partial f}{\partial x}\bigg|_{(0,0)} = \frac{0}{32} = 0, \]\[ \frac{\partial f}{\partial y}\bigg|_{(0,0)} = \frac{0}{32} = 0. \]

Step 4: Calculate the second-order partial derivatives

Find the second-order partial derivatives: \[ \frac{\partial^2 f}{\partial x^2} = \frac{\sec^2\left(\frac{x^2+y^2}{64}\right)}{32} - \frac{x^2 \sec^2\left(\frac{x^2+y^2}{64}\right)\tan\left(\frac{x^2+y^2}{64}\right)}{1024}, \]\[ \frac{\partial^2 f}{\partial y^2} = \frac{\sec^2\left(\frac{x^2+y^2}{64}\right)}{32} - \frac{y^2 \sec^2\left(\frac{x^2+y^2}{64}\right)\tan\left(\frac{x^2+y^2}{64}\right)}{1024}, \] \[ \frac{\partial^2 f}{\partial x \partial y} = \frac{xy \sec^2\left(\frac{x^2+y^2}{64}\right)\tan\left(\frac{x^2+y^2}{64}\right)}{1024}. \]

Step 5: Evaluate the second-order derivatives at the base point

Substitute \((x_0, y_0)=(0,0)\) into the second-order derivatives to evaluate them:\[ \frac{\partial^2 f}{\partial x^2}\bigg|_{(0,0)} = \frac{1}{32}, \]\[ \frac{\partial^2 f}{\partial y^2}\bigg|_{(0,0)} = \frac{1}{32}, \]\[ \frac{\partial^2 f}{\partial x \partial y}\bigg|_{(0,0)} = 0. \]

Step 6: Construct the Taylor approximations

Use the Taylor series expansion to construct the approximations:First-order: \[ f(x, y) \approx f(0,0) + \frac{\partial f}{\partial x}\bigg|_{(0,0)} x + \frac{\partial f}{\partial y}\bigg|_{(0,0)} y = 0. \]Second-order: \[ f(x, y) \approx 0 + \frac{x^2}{64} + \frac{y^2}{64}. \]

Step 7: Estimate using the first-order approximation

Estimate \(f(0.2, -0.3)\) using the first-order approximation. Since the approximation is 0, we have:\[ f(0.2, -0.3) \approx 0. \]

Step 8: Estimate using the second-order approximation

Estimate \(f(0.2, -0.3)\) using the second-order approximation:\[ f(0.2, -0.3) \approx \frac{0.2^2}{64} + \frac{(-0.3)^2}{64} = \frac{0.04}{64} + \frac{0.09}{64} = \frac{0.13}{64} \approx 0.00203125. \]

Step 9: Calculate directly using a calculator

Use a calculator to find \(f(0.2, -0.3)\):First compute the argument of the tangent function:\[ \frac{0.2^2 + (-0.3)^2}{64} = \frac{0.13}{64} \approx 0.00203125. \]Then,\[ f(0.2, -0.3) = \tan(0.00203125) \approx 0.00203125. \]

Step 10: Conclusion

The estimates are:(a) First-order approximation: \(0\)(b) Second-order approximation: \(0.00203125\)(c) Direct calculation: \(0.00203125\)

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus, particularly useful when dealing with functions of more than one variable. In our exercise, we have a function \(f(x, y) = \tan\left(\frac{x^2 + y^2}{64}\right)\). Partial derivatives help us understand how the function changes with respect to each variable individually, keeping the others constant.

To find the partial derivative of \(f\) with respect to \(x\), we hold \(y\) constant. The derivative \(\frac{\partial f}{\partial x}\) measures the rate at which \(f\) changes as \(x\) changes, which is crucial for constructing Taylor series approximations. Similarly, \(\frac{\partial f}{\partial y}\) is found by varying \(y\) and keeping \(x\) constant.

The process involves applying the chain rule. We first took derivatives of the outer function (tangent) with respect to its argument, and then multiplied by the derivative of the inner function \(\frac{x^2 + y^2}{64}\) with respect to \(x\) or \(y\) as needed. Evaluating these derivatives at \((x_0, y_0) = (0,0)\) simplifies calculations and provides an understanding of the function's behavior at this point.

Second-order Approximation

When estimating functions using Taylor series, a second-order approximation allows us to capture the function's curvature, not just its direction at a certain point. This approximation includes not only the linear terms as in the first-order approximation but also quadratic terms.

In our exercise, the second-order approximation for \(f(x, y)\) is given as:

\[ f(x, y) \approx 0 + \frac{x^2}{64} + \frac{y^2}{64}.\]
Here, the coefficients like \(\frac{1}{64}\) are derived from second-order partial derivatives evaluated at the base point \((x_0, y_0) = (0,0)\). These derivatives measure how the rate of change itself changes as you move along a direction.

This approximation captures the small oscillations in \(f\) when \(x\) and \(y\) are near zero. In practical terms, it offers a more precise estimation near the point \((0,0)\), making it a valuable tool for predictions in physics and engineering.

Function Estimation

Function estimation using Taylor series is a powerful tool that simplifies complex calculations. By using approximations, we can predict the value of a function at non-trivial points easily.

In our scenario, we estimated \(f(0.2, -0.3)\) using both first-order and second-order approximations and compared it to a direct computation. While the first-order approximation suggested \(f(0.2, -0.3) \approx 0\), the second-order approximation yielded \(0.00203125\). Similarly, calculating directly confirmed this value, indicating the higher accuracy of the second-order method.

When using Taylor series, it's key to choose the order of approximation based on the required precision. The closer our estimation needs to be to the true value, the more terms we may need to include, be it linear or quadratic terms.