Problem 27 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

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

Step-by-Step Solution

Answer

The first partial derivatives of the function \(f(x, y) = x \sqrt{y} + y^2\) at the point (2, 1) are \(\left. \frac{\partial f}{\partial x}\right|_{(2, 1)} = 1\) and \(\left. \frac{\partial f}{\partial y}\right|_{(2, 1)} = 4\).

Step 1: Find the partial derivative with respect to x

To find the partial derivative of the function with respect to x, we treat y as a constant and differentiate with respect to x: \(\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x \sqrt{y} + y^2) = \sqrt{y} + 0 = \sqrt{y}\)

Step 2: Find the partial derivative with respect to y

To find the partial derivative of the function with respect to y, we treat x as a constant and differentiate with respect to y: \(\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x \sqrt{y} + y^2) = x \frac{1}{2\sqrt{y}} + 2y\)

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

Now, we need to evaluate the partial derivatives at the given point (2, 1). For the partial derivative with respect to x: \(\left. \frac{\partial f}{\partial x}\right|_{(2, 1)} = \sqrt{1} = 1\) For the partial derivative with respect to y: \(\left. \frac{\partial f}{\partial y}\right|_{(2, 1)} = 2 \frac{1}{2\sqrt{1}} + 2(1) = 2+2 = 4\)

Step 4: Write the final answer

The first partial derivatives of the function \(f(x, y) = x \sqrt{y} + y^2\) at the point (2, 1) are: \(\left. \frac{\partial f}{\partial x}\right|_{(2, 1)} = 1\) \(\left. \frac{\partial f}{\partial y}\right|_{(2, 1)} = 4\)

Key Concepts

Multivariable Calculus

Multivariable calculus extends the concepts of single variable calculus to higher dimensions. In the context of multivariable functions, like the one in our example function,
\( f(x, y)=x \sqrt{y}+y^{2} \), we are dealing with functions that have more than one input.

A natural progression from differentiation in single-variable calculus is to partial differentiation in multivariable calculus. Here, we find the rate of change of a function with respect to one variable while keeping others constant. This approach is essential when dealing with multivariate systems in fields as diverse as physics, engineering, economics, and statistics.

In practical applications, understanding how a multivariate function behaves as one variable changes while others are held fixed can offer invaluable insights. For instance, in an economic model, one might want to see how the output of a product might change with respect to the cost of one of its raw materials, assuming all other costs remain the same.

Differentiation

Differentiation is a fundamental tool in calculus that measures how a function changes as its input changes. When it comes to multivariable functions,

such as \( f(x, y) = x \sqrt{y} + y^2 \), we don't just look for a single derivative as we do with one-variable functions, but instead, we find several partial derivatives, each with respect to a different variable.

The partial derivative with respect to x is found by treating y as a constant. Mathematically, this looks like
\( \frac{\partial f}{\partial x} = \sqrt{y} \). Conversely, the partial derivative with respect to y treats x as a constant, resulting in
\( \frac{\partial f}{\partial y} = x \frac{1}{2\sqrt{y}} + 2y \).

These partial derivatives express the slope of the function along the x-axis and y-axis, respectively, and they are crucial when we are looking to find the gradient or to optimize the function.

Mathematical Analysis

Mathematical analysis provides a rigorous foundation for calculus, including differentiation and integration. It deals with limits and continuity, which underpin the operations of calculus, particularly when handling more complex variables.

When applied to multivariable functions, analysis delves into topics such as continuity in higher dimensions, partial differentiation (as discussed in our example exercise), and multiple integrals. Evaluating a partial derivative involves looking at the limit of the function's incremental ratio as one variable changes infinitesimally while others are held constant.

In the exercise provided, we analyzed how small changes in x and y affect our function
\( f(x, y) = x \sqrt{y} + y^2 \). Finding the exact change at a specific point, such as (2, 1), relies on these concepts from mathematical analysis. By evaluating the partial derivatives, we determined the precise impact of altering each variable independently at this point, furthering our understanding of the function's behavior in a specific situation.