Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(F(x, y)=\frac{2 x-y}{x y}\), find \(F_{x}(3,-2)\) and \(F_{y}(3,-2)\).

Step-by-Step Solution

Answer

\( F_x(3,-2) = -\frac{1}{18} \), \( F_y(3,-2) = -\frac{1}{2} \).

Step 1: Understand the Partial Derivatives

We need to find the partial derivatives \( F_x(x, y) \) and \( F_y(x, y) \). These partial derivatives represent the rate of change of the function \( F(x, y) \) with respect to \( x \) and \( y \), respectively.

Step 2: Compute \( F_x(x, y) \)

To find the partial derivative \( F_x(x, y) \), we treat \( y \) as a constant and differentiate \( F(x, y) = \frac{2x-y}{xy} \) with respect to \( x \). Using the quotient rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \), where \( u = 2x - y \) and \( v = xy \).First, differentiate \( u \): \( u' = \frac{d}{dx}(2x - y) = 2 \).Differentiate \( v \):\( v' = \frac{d}{dx}(xy) = y \).Apply the quotient rule:\[ F_x = \frac{2 \cdot xy - (2x-y) \cdot y}{(xy)^2} = \frac{2xy - (2xy - y^2)}{x^2y^2} = \frac{y^2}{x^2y^2} = \frac{1}{x^2y} \].

Step 3: Compute \( F_y(x, y) \)

Now, find the partial derivative \( F_y(x, y) \) by treating \( x \) as a constant and differentiating with respect to \( y \). Again using the quotient rule, where \( u = 2x - y \) and \( v = xy \).First, differentiate \( u \):\( u' = \frac{d}{dy}(2x - y) = -1 \).Differentiate \( v \):\( v' = \frac{d}{dy}(xy) = x \).Apply the quotient rule:\[ F_y = \frac{-1 \cdot xy - (2x-y) \cdot x}{(xy)^2} = \frac{-xy - (2x^2 - xy)}{x^2y^2} = \frac{-2x^2}{x^2y^2} = -\frac{2}{y^2} \].

Step 4: Evaluate \( F_x(3,-2) \)

Substitute \( x = 3 \) and \( y = -2 \) into the expression for \( F_x \):\[ F_x(3, -2) = \frac{1}{3^2(-2)} = \frac{1}{9(-2)} = -\frac{1}{18} \].

Step 5: Evaluate \( F_y(3,-2) \)

Substitute \( x = 3 \) and \( y = -2 \) into the expression for \( F_y \):\[ F_y(3, -2) = -\frac{2}{(-2)^2} = -\frac{2}{4} = -\frac{1}{2} \].

Key Concepts

Quotient Rule

In calculus, the Quotient Rule is a method used for differentiating functions that are expressed as a ratio of two differentiable functions. When faced with a function like \( F(x, y) = \frac{2x-y}{xy} \), you need to use the Quotient Rule to find its derivative effectively.
The Quotient Rule says that for a function \( \frac{u}{v} \), the derivative is given by:

• \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \)

Here, \( u \) and \( v \) are functions of \( x \), and \( u' \) and \( v' \) are their respective derivatives. The importance of this rule lies in its ability to help us understand how the rate of change in the numerator affects the rate of change in the denominator.
In the exercise, we apply the Quotient Rule first to solve for the partial derivative \( F_x(x, y) \), treating \( y \) as a constant. By doing so, we end up with \( F_x = \frac{1}{x^2y} \). Similarly, treating \( x \) as a constant for \( F_y(x, y) \) results in \( F_y = -\frac{2}{y^2} \). Understanding and applying this rule is fundamental when dealing with derivatives involving quotients, especially in multivariable functions.

Multivariable Calculus

Multivariable Calculus is an extension of traditional calculus to functions of more than one variable. Where single-variable functions, like \( f(x) \), depend on just one input, multivariable functions, such as \( F(x, y) \), depend on multiple inputs. This adds complexity and richness to the analysis of changes and rates of change.
In this context, finding partial derivatives is crucial. A partial derivative, such as \( F_x(x, y) \) or \( F_y(x, y) \), determines the rate of change of the function with respect to one variable while keeping the other variables constant. This is essentially an expansion of the derivative concept from single-variable calculus to accommodate several dimensions of input.
Each partial derivative provides a piece of the puzzle, illustrating how one dimension of the input influences the overall function. In our exercise, we find these derivatives and substitute specific values in \( F_x \) and \( F_y \) to capture the specific rate of change at the point \((3, -2)\). Mastery of these concepts allows exploration of complex systems and surfaces in the multivariable realm.

Rate of Change

The rate of change in calculus measures how a quantity changes with respect to another. Unlike single-variable functions, multivariable functions, like \( F(x, y) \), can exhibit changes across different axes, making understanding this concept with partial derivatives crucial.Partial derivatives determine how the function value alters with respect to one variable while keeping others constant. This isolated change is central to understanding the broader behavior of the function.

• For instance, \( F_x(3, -2) = -\frac{1}{18} \) describes how \( F(x, y) \) changes as \( x \) increases while \( y \) remains \(-2\).
• Similarly, \( F_y(3, -2) = -\frac{1}{2} \) tells us how the function changes with varying \( y \), keeping \( x \) fixed at 3.

This understanding helps in fields requiring analysis of behavior at particular points, such as engineering or physics, where knowing how slight changes in input impact outcomes is essential.