Problem 41 - Calculus with Concepts in Calculus

Question

Find \(\sqrt{f_{x}^{2}+f_{y}^{2}+1}\) $$ f(x, y)=1-x $$

Step-by-Step Solution

Answer

The square root expression evaluates to \( \sqrt{2} \).

Step 1: Compute Partial Derivative with Respect to x

The function given is \( f(x,y) = 1-x \). First, we need to find the partial derivative of \( f \) with respect to \( x \), denoted \( f_x \). Since the function only involves \( x \) as \( 1-x \) and no \( y \) terms, the derivative is simply:\[ f_x = \frac{\partial}{\partial x}(1-x) = -1 \]

Step 2: Compute Partial Derivative with Respect to y

Next, we need to determine \( f_y \), the partial derivative of \( f \) with respect to \( y \). The given function \( f(x,y) = 1-x \) doesn't have any \( y \) term. Therefore, the derivative is:\[ f_y = \frac{\partial}{\partial y}(1-x) = 0 \]

Step 3: Calculate the Expression Inside the Square Root

Now we need to substitute the partial derivatives into the expression \( f_x^2 + f_y^2 + 1 \) and compute it:- Substitute \( f_x = -1 \) and \( f_y = 0 \) into the expression:\[ f_x^2 = (-1)^2 = 1 \]\[ f_y^2 = 0^2 = 0 \]So, the expression becomes:\[ f_x^2 + f_y^2 + 1 = 1 + 0 + 1 = 2 \]

Step 4: Find the Square Root of the Expression

Finally, we find the square root of the result from Step 3:\[ \sqrt{f_x^2 + f_y^2 + 1} = \sqrt{2} \]This is the final answer.

Key Concepts

Multivariable Calculus

Multivariable calculus is a fascinating extension of calculus that considers functions of multiple variables, like functions with both x and y components. This branch of calculus allows us to explore more complex functions and how they change in multi-dimensional spaces. In the given problem, the function \( f(x, y) = 1 - x \) is quite simple. It only varies with respect to the \( x \)-axis, as there is no \( y \) term present. This showcases one of the key concepts in multivariable calculus: understanding how each variable influences the function in different directions.

Deepening our understanding of multivariable calculus helps us model and comprehend complicated systems in fields such as physics, engineering, and economics. With partial derivatives, we are able to see how a small change in one variable affects the whole function while keeping the other variables constant. It's like slicing a 3D object in different directions to see how the slices change, thereby giving us deeper insights into the function's behavior.

Calculus Problem Solving

Solving calculus problems often involves breaking them down into smaller, more manageable steps. In this particular problem, the goal was to find an expression involving partial derivatives related to the function \( f(x, y) = 1 - x \).

A structured approach is critical:

• First, identify the variables involved—in this case, \( x \) and \( y \).
• Next, compute the partial derivatives with respect to these variables.
• After that, plug these derivatives into the given expression.
• Finally, simplify to reach the solution.

This sequence ensures clarity and accuracy in problem solving. Each step builds upon the previous one, which is a fundamental strategy in calculus to tackle complex problems systematically. By following such processes, we can break down daunting equations into easily solvable parts, leading to the correct solution.

Gradient Function

The gradient is a crucial concept in calculus, particularly when dealing with multivariable functions. It represents the vector of partial derivatives for a given function, giving us a direction in which the function increases fastest. For the function \( f(x, y) = 1 - x \), the gradient is computed using the partial derivatives \( f_x \) and \( f_y \).

In our problem:

• \( f_x = -1 \), showing the function decreases in the \( x \)-direction.
• \( f_y = 0 \), indicating no change along the \( y \)-axis, implying the function is constant in this direction.

The gradient helps us appreciate how a function behaves in multi-dimensional space. It's a foundation in fields such as optimization, where we seek to find maxima or minima by following the gradient's direction. The magnitude of the gradient gives a sense of the steepness, and the direction tells you where the steepest ascent is. Understanding the gradient and how it relates to real-world problems is invaluable in making more informed predictions and optimizations.