Problem 194 - Calculus 3

Question

Find the total differential of the function \(z=\frac{x y}{y+x}\) where \(x\) changes from 10 to 10.5 and \(y\) changes from 15 to 13 .

Step-by-Step Solution

Answer

The total differential of the function indicates a decrease of approximately 0.14 in \( z \).

Step 1: Understand the Problem

We are asked to find the total differential of the function \( z = \frac{xy}{y+x} \) as \( x \) changes from 10 to 10.5 and \( y \) changes from 15 to 13. The goal is to find the approximate change in \( z \).

Step 2: Identify the Variables and Changes

The variables are \( x \) and \( y \) with changes \( \Delta x = 10.5 - 10 = 0.5 \) and \( \Delta y = 13 - 15 = -2 \).

Step 3: Compute the Partial Derivatives

Find the partial derivatives, \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \), using the quotient rule: - \( \frac{\partial z}{\partial x} = \frac{y(y+x) - xy}{(y+x)^2} = \frac{y^2}{(y+x)^2} \)- \( \frac{\partial z}{\partial y} = \frac{x(y+x) - xy}{(y+x)^2} = \frac{x^2}{(y+x)^2} \)

Step 4: Evaluate the Partial Derivatives at Initial Values

Substitute \( x = 10 \) and \( y = 15 \) into the expressions for the partial derivatives:- \( \frac{\partial z}{\partial x} = \frac{15^2}{(15+10)^2} = \frac{225}{625} = 0.36 \)- \( \frac{\partial z}{\partial y} = \frac{10^2}{(15+10)^2} = \frac{100}{625} = 0.16 \)

Step 5: Calculate the Total Differential

The total differential \( dz \) is given by:\[ dz = \frac{\partial z}{\partial x} \Delta x + \frac{\partial z}{\partial y} \Delta y \]Substitute the values:\[ dz = 0.36 \times 0.5 + 0.16 \times (-2) = 0.18 - 0.32 = -0.14 \]

Step 6: Conclusion

The total differential approximates the change in \( z \). The function \( z \) decreases by approximately 0.14 as \( x \) increases from 10 to 10.5 and \( y \) decreases from 15 to 13.

Key Concepts

Partial Derivatives

Our exploration begins with the concept of partial derivatives, a fundamental tool in multivariable calculus. When dealing with functions of multiple variables, partial derivatives allow us to investigate how the function changes as we vary just one of these variables, keeping the others constant. This is crucial for understanding how individual changes affect the overall behavior of a function. In the given problem, we are examining the function \( z = \frac{x y}{y+x} \). Here, \( z \) is dependent on both \( x \) and \( y \). Using partial derivatives, we can find how \( z \) changes with respect to \( x \) and with respect to \( y \):

• Partial derivative with respect to \( x \), \( \frac{\partial z}{\partial x} \), tells us how \( z \) changes when we tweak \( x \) slightly, holding \( y \) constant.
• Partial derivative with respect to \( y \), \( \frac{\partial z}{\partial y} \), indicates how \( z \) alters when \( y \) is slightly varied, with \( x \) held fixed.

This concept is especially useful for finding the total differential and helps us predict the function's behavior under small changes in its input variables.

Quotient Rule

In the process of finding partial derivatives, the quotient rule becomes vital when dealing with functions expressed as a ratio of two differentiable functions. If we have a function \( f(x, y) = \frac{u(x, y)}{v(x, y)} \), the quotient rule helps us differentiate it. To apply the quotient rule, you should remember the formula:\[\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\]In the context of the exercise, when computing \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \), you utilize the quotient rule:

• For \( \frac{\partial z}{\partial x} = \frac{v \frac{\partial u}{\partial x} - u \frac{\partial v}{\partial x}}{v^2} \), where \( u = xy \) and \( v = y + x \).
• Similarly, for \( \frac{\partial z}{\partial y} \), replace the derivatives with respect to \( y \).

This rule is indispensable for handling such expressions, making the differentiation process manageable and systematic.

Change in Variables

Understanding changes in variables is essential when calculating the total differential. When variables \( x \) and \( y \) change -- as they do in our problem, with \( x \) going from 10 to 10.5 and \( y \) from 15 to 13 -- this influences the dependent variable \( z \). The total differential, \( dz \), represents how much \( z \) changes as we make these tiny adjustments to \( x \) and \( y \). The formula for the total differential is: \[dz = \frac{\partial z}{\partial x} \Delta x + \frac{\partial z}{\partial y} \Delta y\]Key points to keep in mind:

• \( \Delta x \) and \( \Delta y \) denote the respective changes in \( x \) and \( y \).
• The partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) are multiplied by these changes to approximate the change in \( z \).

By integrating these changes, we gain a clearer insight into how adjustments in the inputs manifest in the output, offering a powerful way to predict the behavior of multivariable functions.