Problem 1 - Calculus with Concepts in Calculus

Question

Approximate the value of \(f\) at the given point. $$ f(x, y)=\sqrt{x^{2}+y^{2}} ;(3.01,4.03) $$

Step-by-Step Solution

Answer

The approximate value of \( f(3.01, 4.03) \) is 5.03.

Step 1: Identify the Function and Point

The function given is \( f(x, y) = \sqrt{x^2 + y^2} \) and we need to find an approximate value of this function at the point \((3.01, 4.03)\).

Step 2: Find Function Value at Nearby Integer

Calculate the value of \( f(x, y) \) at the nearby integer point \((3, 4)\). This will serve as a base for approximation. So, \( f(3, 4) = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \).

Step 3: Calculate Partial Derivatives

Find the partial derivatives of \( f(x, y) = \sqrt{x^2 + y^2} \):- Partial with respect to \( x \), \( f_x = \frac{x}{\sqrt{x^2 + y^2}} \).- Partial with respect to \( y \), \( f_y = \frac{y}{\sqrt{x^2 + y^2}} \).At \((3, 4)\), we have:\( f_x(3, 4) = \frac{3}{5} = 0.6 \) and \( f_y(3, 4) = \frac{4}{5} = 0.8 \).

Step 4: Approximate Change Using Linearization

Using linear approximation, the change \( \Delta f \) at a point \((3.01, 4.03)\) is given by:\[ \Delta f \approx f_x(3, 4) \cdot (3.01 - 3) + f_y(3, 4) \cdot (4.03 - 4) \]\( = 0.6 \times 0.01 + 0.8 \times 0.03 \).Calculate \( 0.6 \times 0.01 = 0.006 \) and \( 0.8 \times 0.03 = 0.024 \).

Step 5: Calculate Approximation

Now add these changes to the base value from Step 2:\( f(3.01, 4.03) \approx 5 + 0.006 + 0.024 = 5.03 \).

Key Concepts

Partial Derivatives

Partial derivatives are an essential concept in differential calculus, particularly when dealing with functions that have more than one variable. When you have a function like \( f(x, y) = \sqrt{x^2 + y^2} \), which involves both \( x \) and \( y \), partial derivatives help us understand how the function changes with respect to each variable separately.

• Partial Derivative with respect to \( x \) (\( f_x \)): This measures the rate of change of the function as \( x \) changes, while keeping \( y \) constant.
• Partial Derivative with respect to \( y \) (\( f_y \)): Similarly, this measures the rate of change of the function as \( y \) changes, keeping \( x \) constant.

To find these partial derivatives for our function, we differentiate with respect to the variables while treating other variables as constants. Here, \( f_x = \frac{x}{\sqrt{x^2 + y^2}} \) and \( f_y = \frac{y}{\sqrt{x^2 + y^2}} \). By calculating these at specific points, we gain insights into how small changes in \( x \) and \( y \) affect the function's output.

Linear Approximation

Linear approximation is a method used to estimate the value of a function near a given point using tangent lines. It is based on the idea that close to the point of tangency, the function behaves very similarly to its tangent line. By using this method, we can make quick and accurate approximations for the function values.

• Formula: For a function \( f(x, y) \) near the point \((a, b)\), we estimate \( f(x, y) \) by \( f(a, b) + f_x(a, b)\Delta x + f_y(a, b)\Delta y \), where \( \Delta x = x - a \) and \( \Delta y = y - b \).
• In Practice: This means we calculate the base function value at \((a, b)\) and then adjust it by the changes (\(\Delta x\) and \(\Delta y\)) scaled by the rates of change or partial derivatives \( f_x \) and \( f_y \).

In our example, using the function \( f(x, y) = \sqrt{x^2 + y^2} \), we calculated \( f(3, 4) = 5 \) and then adjusted it slightly using the coefficients of \( 0.6 \) for \( x \) and \( 0.8 \) for \( y \) to find an approximation at the new point \((3.01, 4.03)\).

Functions of Two Variables

Functions of two variables, such as \( f(x, y) = \sqrt{x^2 + y^2} \), require us to consider changes along both axes. This type of function takes two input values, \( x \) and \( y \), and maps them to a single output value. Understanding functions with two variables is crucial because

• Real-world problems often involve relationships dependent on more than one quantity.
• These functions can be visualized as surfaces in three-dimensional space, where each point \((x, y)\) corresponds to a height given by \( f(x, y) \).

Such functions are intricate, requiring techniques like partial derivatives and linear approximation to analyze. When we solved our problem, we saw that the value of the function does not just depend on \( x \) or \( y \) separately, but on both in tandem; hence, small variations in either variable impact the outcome.