Problem 2 - Calculus with Concepts in Calculus

Question

Approximate the value of \(f\) at the given point. $$ f(x, y)=\sqrt{x^{2}+y} ;(3.02,-4.98) $$

Step-by-Step Solution

Answer

The value of \( f(3.02, -4.98) \approx 2.035 \).

Step 1: Express the Function

Identify the given function: \( f(x, y) = \sqrt{x^2 + y} \). We need to approximate the value of \( f \) at the point \( (3.02, -4.98) \).

Step 2: Determine Partial Derivatives

Calculate the partial derivatives of the function. Start with \( f_x \):\[ f_x = \frac{d}{dx}(x^2 + y)^{1/2} = \frac{x}{\sqrt{x^2 + y}} \].Now, find \( f_y \):\[ f_y = \frac{d}{dy}(x^2 + y)^{1/2} = \frac{1}{2}\cdot \frac{1}{(x^2 + y)^{1/2}} = \frac{1}{2\sqrt{x^2 + y}} \].

Step 3: Evaluate at Nearest Simple Point

Evaluate the function and its partial derivatives at a nearby simple point, such as \((3, -5)\), because \(3^2 - 5 = 4\) makes calculations simpler.- \( f(3, -5) = \sqrt{3^2 - 5} = \sqrt{4} = 2 \).- \( f_x(3, -5) = \frac{3}{\sqrt{4}} = \frac{3}{2} \).- \( f_y(3, -5) = \frac{1}{2\sqrt{4}} = \frac{1}{4} \).

Step 4: Use Linear Approximation

The change in \( f \) can be approximated using the linearization formula: \[ df = f_x(x_0, y_0)\Delta x + f_y(x_0, y_0)\Delta y \]. Where \( \Delta x = 3.02 - 3 = 0.02 \) and \( \Delta y = -4.98 - (-5) = 0.02 \).Thus:\[ df = \frac{3}{2}(0.02) + \frac{1}{4}(0.02) = 0.03 + 0.005 = 0.035 \].

Step 5: Approximate the Value at (3.02, -4.98)

Combine the result of the previous steps to approximate the function's value:\[ f(3.02, -4.98) \approx f(3, -5) + df = 2 + 0.035 = 2.035 \].

Key Concepts

Partial Derivatives

In multivariable calculus, partial derivatives are an essential concept, especially when dealing with functions of more than one variable. They measure how a function changes as each input variable changes, while keeping all other variables constant. For example, if we have a function \(f(x, y) = \sqrt{x^2 + y}\), we can find its partial derivatives with respect to \(x\) and \(y\).

• The partial derivative with respect to \(x\), denoted as \(f_x\), is found by differentiating \(f\) treating \(y\) as a constant. It results in \(f_x = \frac{x}{\sqrt{x^2 + y}}\).
• The partial derivative with respect to \(y\), denoted as \(f_y\), is found by differentiating \(f\) treating \(x\) as a constant. It results in \(f_y = \frac{1}{2\sqrt{x^2 + y}}\).

Partial derivatives help us understand the directional rate of change of the function along the \(x\) and \(y\) axes.

Linear Approximation

Linear approximation is a method used to estimate the value of a function at a point by using its tangent plane or line. It's particularly useful when the point of interest is very close to a point where you can easily compute the function and its derivatives, known as the expansion point.
To approximate \(f(x, y)\) at \((3.02, -4.98)\), we choose a nearby simple point, like \((3, -5)\), where calculations are straightforward. At this point:

• \(f(3, -5) = 2\)
• \(f_x(3, -5) = \frac{3}{2}\)
• \(f_y(3, -5) = \frac{1}{4}\)

The linear approximation formula used here is: \[df = f_x(3, -5) \cdot \Delta x + f_y(3, -5) \cdot \Delta y\] Where \(\Delta x = 0.02\) and \(\Delta y = 0.02\), which yields \(df = 0.035\). This tells us how the function value changes as we move from \((3, -5)\) to \((3.02, -4.98)\).

Function Approximation

Function approximation involves estimating a function's value near a given point using known information. In the example of \(f(x, y)\), approximation is executed by identifying an accessible nearby point and calculating the derivative values there. This approach simplifies complex calculations by using simpler values.

• By evaluating \(f(3, -5)\) directly, we find a straightforward number to work with: \(2\).
• The derivatives \(f_x(3, -5) = \frac{3}{2}\) and \(f_y(3, -5) = \frac{1}{4}\) facilitate understanding how the function behaves in different directions.

With this information, we utilize these values to estimate \(f(x, y)\) at \( (3.02, -4.98)\). Such techniques are foundational to numerical analysis and other practical applications of calculus.

Calculus Problem Solving

Problem solving in calculus often requires a structured approach to break down complex tasks into manageable steps. In the exercise, to find \(f(3.02, -4.98)\), we followed a series of logical steps:

• Identify and express the function.
• Compute the partial derivatives \(f_x\) and \(f_y\).
• Evaluate these derivatives at a nearby simple point, \((3, -5)\).
• Use the linear approximation formula to estimate the change \(df\).
• Combine this with the function’s value at \((3, -5)\) to estimate \(f(3.02, -4.98)\).

This method highlights how calculus can solve real-world problems by approximating values that are otherwise difficult to compute precisely.