Problem 12 - Calculus Early Transcendentals

Question

\(7-12\) Use the Chain Rule to find \(\partial z / \partial s / \partial s\) and \(\partial z / \partial t\) $$z=\tan (u / v), \quad u=2 s+3 t, \quad v=3 s-2 t$$

Step-by-Step Solution

Answer

\( \frac{\partial z}{\partial s} = \sec^2\left(\frac{u}{v}\right)\left(\frac{2}{v} - \frac{3u}{v^2}\right) \) and \( \frac{\partial z}{\partial t} = \sec^2\left(\frac{u}{v}\right)\left(\frac{3}{v} + \frac{2u}{v^2}\right) \)

Step 1: Identify the given functions

We are given the function \( z = \tan\left(\frac{u}{v}\right) \), with the functions \( u = 2s + 3t \) and \( v = 3s - 2t \). We need to find \( \frac{\partial z}{\partial s} \) and \( \frac{\partial z}{\partial t} \) using the chain rule.

Step 2: Apply the Chain Rule to find \( \frac{\partial z}{\partial u} \) and \( \frac{\partial z}{\partial v} \)

The derivative of \( z \) with respect to its argument \( \frac{u}{v} \) is \( \sec^2\left(\frac{u}{v}\right) \). By the chain rule, we differentiate the argument with respect to \( u \) and \( v \). This gives us:\[ \frac{\partial z}{\partial u} = \frac{1}{v} \sec^2\left(\frac{u}{v}\right), \quad \frac{\partial z}{\partial v} = -\frac{u}{v^2} \sec^2\left(\frac{u}{v}\right) \]

Step 3: Apply the Chain Rule to find \( \frac{\partial z}{\partial s} \)

Since \( z \) depends on \( s \) through both \( u \) and \( v \), we apply the chain rule:\[ \frac{\partial z}{\partial s} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial s} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial s} \]Given \( u = 2s + 3t \) and \( v = 3s - 2t \), we find:\[ \frac{\partial u}{\partial s} = 2, \quad \frac{\partial v}{\partial s} = 3 \]Substitute the derivatives:\[ \frac{\partial z}{\partial s} = \sec^2\left(\frac{u}{v}\right) \left(\frac{2}{v} - \frac{3u}{v^2}\right) \]

Step 4: Apply the Chain Rule to find \( \frac{\partial z}{\partial t} \)

For \( \frac{\partial z}{\partial t} \), similarly:\[ \frac{\partial z}{\partial t} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial t} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial t} \]Given \( \frac{\partial u}{\partial t} = 3 \) and \( \frac{\partial v}{\partial t} = -2 \), we substitute:\[ \frac{\partial z}{\partial t} = \sec^2\left(\frac{u}{v}\right) \left(\frac{3}{v} + \frac{2u}{v^2}\right) \]

Step 5: Write final derivatives

From our calculations:1. \( \frac{\partial z}{\partial s} = \sec^2\left(\frac{u}{v}\right)\left(\frac{2}{v} - \frac{3u}{v^2}\right) \)2. \( \frac{\partial z}{\partial t} = \sec^2\left(\frac{u}{v}\right)\left(\frac{3}{v} + \frac{2u}{v^2}\right) \)

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus, specifically when dealing with functions of several variables. Unlike ordinary derivatives which apply to single-variable functions, partial derivatives focus on how a multivariable function changes with respect to one specific variable, keeping other variables constant. This concept is particularly useful in fields that require optimization and sensitivity analysis, such as economics, engineering, and physics.
To compute a partial derivative of a function, you differentiate it concerning one of its variables while treating all the other variables as constants. For instance, in the exercise you are working with, the function is given by \[ z = \tan\left(\frac{u}{v}\right) \] where both \( u \) and \( v \) are dependent on \( s \) and \( t \):

• \( u = 2s + 3t \)
• \( v = 3s - 2t \)

To find \( \partial z / \partial s \), you would apply the chain rule, calculating the derivative of \( z \) with respect to \( s \) through the intermediary variables \( u \) and \( v \). This involves the derivatives \( \partial u / \partial s \) and \( \partial v / \partial s \), and the derivative of \( z \) with respect to \( u \) and \( v \), as detailed in the step-by-step solution.

Trigonometric Functions Derivatives

When differentiating trigonometric functions, particularly in the context of calculus, there are specific rules to follow. Knowing these rules is crucial for dealing effectively with derivatives involving trigonometric expressions.
One of the key trigonometric functions we encounter in differentiation is the tangent function. Its derivative is given by:\[ \frac{d}{dx} (\tan(x)) = \sec^2(x) \]For the function in your exercise, \( z = \tan\left(\frac{u}{v}\right) \), you need to use this rule while also paying attention to the fact that the argument of the tangent, \( \frac{u}{v} \), is not a simple variable but a ratio depending on \( u \) and \( v \).
Thus, when applying the chain rule:

• First, calculate the derivative of the tangent function itself, which gives \( \sec^2\left(\frac{u}{v}\right) \).
• Then, differentiate the inner function \( \frac{u}{v} \) with respect to \( u \) and \( v \), which is needed for the application of the chain rule.

Through these steps, you arrive at expressions involving \( \frac{\partial z}{\partial u} \) and \( \frac{\partial z}{\partial v} \), which are significant for solving the exercise.

Multivariable Calculus

Multivariable calculus extends the principles of single-variable calculus to functions with more than one independent variable. It's a critical area of study for analyzing systems with multiple factors that influence an outcome.
One of its key techniques is the chain rule for multiple variables, which enables the calculation of derivatives such as \( \partial z / \partial s \) and \( \partial z / \partial t \) when functions are composed of more than one variable. In this exercise, the goal is to differentiate a function in terms of \( s \) and \( t \) even though \( z \) directly depends on \( u \) and \( v \), which in turn depend on \( s \) and \( t \):

• The function \( z = \tan\left(\frac{u}{v}\right) \) intertwines with \( s \) and \( t \).
• The derivatives \( \partial u/\partial s \) and \( \partial v/\partial s \) observe changes in \( u \) and \( v \) with changes in \( s \).
• Likewise, for \( \partial z/\partial t \), it involves \( \partial u/\partial t \) and \( \partial v/\partial t \) observing their changes with respect to \( t \).

Through calculating these partial derivatives, you gain insight into how the function behaves across different variables, which is essential for comprehending the sensitivities within the system the function represents.