Problem 28 - Calculus Early Transcendentals

Question

a. Find the linear approximation for the following functions at the given point. b. Use part (a) to estimate the given function value. $$f(x, y)=\sqrt{x^{2}+y^{2}} ;(3,-4) ; \text { estimate } f(3.06,-3.92)$$

Step-by-Step Solution

Answer

Answer: The estimated value of the function at the point \((3.06, -3.92)\) is approximately \(5.1\).

Step 1: Find the partial derivatives

To find the linear approximation, we need to find the partial derivatives of the function \(f(x, y) = \sqrt{x^{2} + y^{2}}\) with respect to x and y. The partial derivative of f with respect to x is: $$\frac{\partial f}{\partial x}=\frac{1}{2 \sqrt{x^{2}+y^{2}}}(2 x)=\frac{x}{\sqrt{x^{2}+y^{2}}}$$ The partial derivative of f with respect to y is: $$\frac{\partial f}{\partial y}=\frac{1}{2 \sqrt{x^{2}+y^{2}}}(2 y)=\frac{y}{\sqrt{x^{2}+y^{2}}}$$

Step 2: Evaluate partial derivatives at the point

Now, we'll evaluate both partial derivatives at the given point \((3, -4)\). Partial derivative with respect to x at point (3, -4): $$\frac{\partial f}{\partial x}(3, -4)=\frac{3}{\sqrt{3^{2}+(-4)^{2}}}=\frac{3}{5}$$ Partial derivative with respect to y at the given point (3, -4): $$\frac{\partial f}{\partial y}(3, -4)=\frac{-4}{\sqrt{3^{2}+(-4)^{2}}}=-\frac{4}{5}$$

Step 3: Find the linear approximation equation (tangent plane)

Using the linear approximation formula and the calculated values from steps 1 and 2, we can form the equation of the tangent plane: $$f(x, y) \approx f(3, -4) + \frac{3}{5}(x - 3) - \frac{4}{5}(y + 4)$$ Since \(f(3, -4) = \sqrt{(3)^2 + (-4)^2} = 5\), we have: $$f(x, y) \approx 5 + \frac{3}{5}(x - 3) - \frac{4}{5}(y + 4)$$ This is our linear approximation equation.

Step 4: Estimate the function value at the given point

Now, we can use the linear approximation equation we found in step 3 to estimate the function value at point \((3.06, -3.92)\). $$f(3.06, -3.92) \approx 5 + \frac{3}{5}(3.06 - 3) - \frac{4}{5}(-3.92 + 4)$$ After solving the right-hand side, we get: $$f(3.06, -3.92) \approx 5 + \frac{3}{5}(0.06) +\frac{4}{5}(0.08)$$ $$f(3.06, -3.92) \approx 5 + 0.036 + 0.064$$ $$f(3.06, -3.92) \approx 5.1$$ So, the estimated value of the function at the point \((3.06, -3.92)\) is approximately \(5.1\).

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, specifically dealing with functions of multiple variables. To understand them, imagine a function that depends on more than one input, like two variables, say \(x\) and \(y\). For the function \(f(x, y)\), a partial derivative measures how the function changes as one of these variables changes, while keeping the other constant.

• Partial derivative with respect to \(x\): This tells us how \(f\) changes as \(x\) varies, when \(y\) is held steady. It is denoted as \(\frac{\partial f}{\partial x}\).
• Partial derivative with respect to \(y\): This tells us how \(f\) changes as \(y\) varies, while holding \(x\) fixed. It is denoted as \(\frac{\partial f}{\partial y}\).

In the context of the problem, \(f(x, y) = \sqrt{x^2 + y^2}\), the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) measure the rates of change along the \(x\) and \(y\) axes, respectively.

Tangent Plane Equation

The tangent plane equation provides a linear approximation of a surface near a point. For a function of two variables, the tangent plane at a point \((x_0, y_0)\) gives the best linear model for the function around that point.

To construct this plane, we use the function value at the point, plus the gradients from the partial derivatives. For our exercise, the tangent plane was constructed using:

• The value of the function at \((3, -4)\), which is \(5\).
• The partial derivatives \(\frac{3}{5}\) for \(x\) and \(-\frac{4}{5}\) for \(y\) were used to establish the influence of small changes around the point.

Thus, the tangent plane equation is the linear approximation \(f(x, y) \approx 5 + \frac{3}{5}(x - 3) - \frac{4}{5}(y + 4)\). This equation shows how \(f(x, y)\) behaves near \((3, -4)\).

Function Estimation

Function estimation using linear approximation involves predicting a function's value near a known point. This is incredibly useful because complex functions aren't always easy to compute directly.

Using linear approximation, also known as using a tangent plane for multivariable functions, simplifies the estimation process. We approximate the change in the function based on information at a nearby point. In our example:

• The estimated value at \((3.06, -3.92)\) was obtained by substituting into the tangent plane equation.
• Computations were simplified to add corrections based on small deviations from \((3, -4)\).

The result, a predicted function value of approximately \(5.1\), shows how effective and efficient this approach can be.

Multivariable Calculus

Multivariable calculus extends the principles of calculus into functions with more than one variable, enhancing our understanding of how changes in one variable affect another within a system. It encompasses several tools and concepts:

• Partial Derivatives: Highlight the change of multivariable functions with respect to one variable at a time.
• Tangent Planes: Allow for linear approximations in three-dimensional space.
• Function approximations: Enable estimation of function values near known points, crucial for solving real-world problems where exact calculations are infeasible.

Multivariable calculus helps model and solve problems in physics, engineering, and economics, providing a vital toolkit for understanding complex systems and predicting values with precision and insight.