Problem 25 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Evaluate the first partial derivatives of the function at the given point. \(f(x, y)=x^{2} y+x y^{2} ;(1,2)\)

Step-by-Step Solution

Answer

The first partial derivatives of the function at the point (1,2) are \(f_x(1, 2) = 8\) and \(f_y(1, 2) = 5\).

Step 1: Find the first partial derivative with respect to x (#\(f_x\)

To find the first partial derivative with respect to x, we treat y as a constant and differentiate the function with respect to x: \(f(x, y) = x^2 y + x y^2\) Taking the derivative with respect to x: \(f_x(x, y) = \frac{\partial}{\partial x}(x^2 y + x y^2) = 2xy + y^2\)

Step 2: Find the first partial derivative with respect to y (#\(f_y\)

To find the first partial derivative with respect to y, we treat x as a constant and differentiate the function with respect to y: \(f(x, y) = x^2 y + x y^2\) Taking the derivative with respect to y: \(f_y(x, y) = \frac{\partial}{\partial y}(x^2 y + x y^2) = x^2 + 2xy\)

Step 3: Evaluate the partial derivatives at the given point (1,2)

Now, we need to evaluate the partial derivatives, \(f_x\) and \(f_y\), at the given point (1,2): \(f_x(1, 2) = 2(1)(2) + (2)^2 = 4 + 4 = 8\) \(f_y(1, 2) = (1)^2 + 2(1)(2) = 1 + 4 = 5\) So, the first partial derivatives of the function at the point (1,2) are: \(f_x(1, 2) = 8\) \(f_y(1, 2) = 5\)

Key Concepts

Multivariable Calculus

Multivariable Calculus extends the principles of calculus to functions of more than one variable. Instead of exploring changes along a single line, we analyze how a function behaves over multi-dimensional space.
For instance, if we have a function that depends on two variables, like in the example where the function is \( f(x,y) = x^2 y + x y^2 \), we need to see how changes in each variable affect the outcome.
In multivariable calculus, we look at how a small change in one variable while keeping the others constant impacts the output. This is why partial derivatives come into play—they help us measure these specific changes.
When analyzing changes in a three-dimensional space, such as in physics or engineering, multivariable calculus plays a vital role by allowing us to calculate gradients, tangent planes, and more.

Differentiation

Differentiation, in the context of multivariable calculus, involves finding the derivative of a function concerning one of its variables, treating all other variables as constants. This helps us understand how a function behaves in various directions. Differentiation tells us how a function changes when one of its input variables changes.
For example, say we have the function mentioned earlier, \( f(x, y) = x^2 y + x y^2 \). If we differentiate it with respect to \( x \), we treat \( y \) as a constant. Hence, the derivative is \( f_x = 2xy + y^2 \). Similarly, when differentiating with respect to \( y \), treating \( x \) as constant gives \( f_y = x^2 + 2xy \).
This helps in comprehending how small tweaks in one of the variables can alter the function's value, crucial for applications in optimization and finding equilibrium points.

Mathematical Functions

Mathematical functions are rules that assign every input to specific outputs. In multivariable calculus, functions can depend on two or more variables, like in our exercise's function \( f(x, y) = x^2 y + x y^2 \).
Each variable in such functions influences the result differently. In the given function, both \( x \) and \( y \) contribute to the final value, with their contributions changing based on their interactions as depicted in the terms \( x^2 y \) and \( x y^2 \).
Understanding the roles and relationships between the variables within these functions is a cornerstone of calculus, enabling us to perform various analyses, whether it be calculating volumes, optimizing results, or modeling real-world systems.