Problem 28 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Evaluate the first partial derivatives of the function at the given point. \(g(x, y)=\sqrt{x^{2}+y^{2}} ;(3,4)\)

Step-by-Step Solution

Answer

The first partial derivatives of the function \(g(x, y) = \sqrt{x^2 + y^2}\) at the point (3, 4) are: \(\frac{\partial g}{\partial x}(3, 4) = \frac{3}{5}\) and \(\frac{\partial g}{\partial y}(3, 4) = \frac{4}{5}\).

Step 1: Compute the partial derivative of g(x, y) with respect to x

To find the partial derivative of g(x, y) with respect to x, we need to differentiate g(x, y) with respect to x while treating y as a constant. The given function is: \(g(x, y) = \sqrt{x^2 + y^2}\) To differentiate, we can rewrite g(x, y) in the form: \(g(x,y) = (x^2 + y^2)^{1/2}\) Now we can differentiate with respect to x using the chain rule: \(\frac{\partial g}{\partial x} = \frac{1}{2} (x^2+y^2)^{-1/2}(2x)\) Simplify: \(\frac{\partial g}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}}\)

Step 2: Compute the partial derivative of g(x, y) with respect to y

To find the partial derivative of g(x, y) with respect to y, we need to differentiate g(x, y) with respect to y while treating x as a constant. Using the rewritten form: \(g(x,y) = (x^2 + y^2)^{1/2}\) Now we can differentiate with respect to y using the chain rule: \(\frac{\partial g}{\partial y} = \frac{1}{2} (x^2+y^2)^{-1/2}(2y)\) Simplify: \(\frac{\partial g}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}}\)

Step 3: Evaluate the first partial derivatives at the given point (3, 4)

We have found the partial derivatives of g(x, y) with respect to x and y: \(\frac{\partial g}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}}\) \(\frac{\partial g}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}}\) Now we need to evaluate these derivatives at the given point (3, 4): \(\frac{\partial g}{\partial x}(3, 4) = \frac{3}{\sqrt{3^2 + 4^2}} = \frac{3}{5}\) \(\frac{\partial g}{\partial y}(3, 4) = \frac{4}{\sqrt{3^2 + 4^2}} = \frac{4}{5}\) Therefore, the first partial derivatives of the function g(x, y) at the point (3,4) are as follows: \(\frac{\partial g}{\partial x}(3, 4) = \frac{3}{5}\) \(\frac{\partial g}{\partial y}(3, 4) = \frac{4}{5}\)

Key Concepts

Chain Rule Differentiation

Understanding chain rule differentiation is key when dealing with functions that are compositions of other functions, which is a common scenario in multivariable calculus. When a function, say h(z), is defined as another function of z, where z itself is a function of one or more variables, we employ the chain rule to compute derivatives with respect to these variables.

The chain rule tells us that the derivative of h with respect to a variable, for example x, is the derivative of h with respect to z multiplied by the derivative of z with respect to x. In the language of partial derivatives, if we have a function g(x, y) = h(z) and z = f(x, y), the partial derivatives are given by:

• For x: \( \frac{\partial g}{\partial x} = \frac{\partial h}{\partial z} \cdot \frac{\partial z}{\partial x} \)
• For y: \( \frac{\partial g}{\partial y} = \frac{\partial h}{\partial z} \cdot \frac{\partial z}{\partial y} \)

Applying the chain rule makes it simpler to tackle the differentiation of complex multivariable functions by breaking them down into simpler parts that are easier to differentiate individually.

First Partial Derivatives Evaluation

When we discuss the first partial derivatives evaluation, we're referring to calculating the derivatives of a function with respect to each of its variables while holding the other variables constant. This process provides us with the rates at which the function changes along axes aligned with those variables.

To evaluate the first partial derivatives at a specific point, you follow these steps:

• Find the general expression for the partial derivative with respect to each variable.
• Simplify the expression if necessary.
• Substitute the coordinates of the point into the simplified derivative expressions.

This process has been clearly illustrated in the exercise with the function \( g(x, y) = \sqrt{x^2 + y^2} \) evaluated at the point \( (3, 4) \) where we obtained the simplified partial derivatives and then substituted the point's coordinates to get the specific rate of change at that location, demonstrated as \( \frac{3}{5} \) and \( \frac{4}{5} \) for the x and y directions, respectively.

Multivariable Calculus

The realm of multivariable calculus extends the concepts of single-variable calculus into higher dimensions. In this discipline, we deal with functions that have more than one input and possibly more than one output. To analyze how these functions behave, we use partial derivatives to understand the sensitivities of functions with respect to changes in each of their input variables.

In contrast to single-variable calculus, where the derivative at a point gives a complete picture of a function's behavior at that point, in multivariable calculus, we need to consider all the partial derivatives to understand the function's local behavior. These partial derivatives, when evaluated at a point, can be visualized as forming a vector called the gradient, which points in the direction of the steepest ascent of the function.

Functions of multiple variables can model complex systems in physics, engineering, economics, and many other fields, making multivariable calculus a fundamental tool for understanding the dynamics of systems depending on several variables simultaneously.