Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-16, find all first partial derivatives of each function. \(F(w, z)=w \sin ^{-1}\left(\frac{w}{z}\right)\)

Step-by-Step Solution

Answer

The partial derivatives are: \( \frac{\partial F}{\partial w} = \sin^{-1}\left(\frac{w}{z}\right) + \frac{w}{z \sqrt{1 - \left(\frac{w}{z}\right)^2}} \) and \( \frac{\partial F}{\partial z} = -\frac{w^2}{z^2 \sqrt{1 - \left(\frac{w}{z}\right)^2}} \).

Step 1: Identify the problem

We need to find the first partial derivatives of the function \( F(w, z) = w \sin^{-1}\left(\frac{w}{z}\right) \). In this context, partial derivatives involve differentiating the function with respect to each variable, treating other variables as constants.

Step 2: Compute the partial derivative with respect to w

To find the partial derivative of \( F \) with respect to \( w \), we treat \( z \) as a constant:\[\frac{\partial F}{\partial w} = \frac{\partial}{\partial w}\left(w \sin^{-1}\left(\frac{w}{z}\right)\right)\]Using the product rule, this becomes:\[= \sin^{-1}\left(\frac{w}{z}\right) + w \cdot \frac{1}{\sqrt{1 - \left(\frac{w}{z}\right)^2}} \cdot \frac{d}{dw}\left(\frac{w}{z}\right)\]Since \( \frac{d}{dw}\left(\frac{w}{z}\right) = \frac{1}{z} \), we continue:\[= \sin^{-1}\left(\frac{w}{z}\right) + \frac{w}{z \sqrt{1 - \left(\frac{w}{z}\right)^2}}\]

Step 3: Compute the partial derivative with respect to z

Now, we find the partial derivative of \( F \) with respect to \( z \), treating \( w \) as a constant:\[\frac{\partial F}{\partial z} = \frac{\partial}{\partial z}\left(w \sin^{-1}\left(\frac{w}{z}\right)\right)\]Using the chain rule, this becomes:\[= w \cdot \frac{1}{\sqrt{1 - \left(\frac{w}{z}\right)^2}} \cdot \frac{d}{dz}\left(\frac{w}{z}\right)\]Since \( \frac{d}{dz}\left(\frac{w}{z}\right) = -\frac{w}{z^2} \), we get:\[= -\frac{w^2}{z^2 \sqrt{1 - \left(\frac{w}{z}\right)^2}}\]

Step 4: Conclusion: Summarize the results

The first partial derivatives of the function \( F(w, z) = w \sin^{-1}\left(\frac{w}{z}\right) \) are:\[\frac{\partial F}{\partial w} = \sin^{-1}\left(\frac{w}{z}\right) + \frac{w}{z \sqrt{1 - \left(\frac{w}{z}\right)^2}}\]\[\frac{\partial F}{\partial z} = -\frac{w^2}{z^2 \sqrt{1 - \left(\frac{w}{z}\right)^2}}\]

Key Concepts

Product rule

The product rule is a foundational concept in calculus used to differentiate functions that are the products of two other functions. If you have a function in the form of two functions multiplied together, say \( u(x) \) and \( v(x) \), the derivative of their product is given by:

• \( (uv)' = u'v + uv' \)

This rule is essential when dealing with functions of several variables, as we often encounter scenarios where a variable's function is multiplied by another. In our exercise, where we find the partial derivative of \( F(w, z) = w \sin^{-1}\left(\frac{w}{z}\right) \) with respect to \( w \), the product rule is necessary. Here, \( u(w) = w \) and \( v(w, z) = \sin^{-1}\left(\frac{w}{z}\right) \).
The application of the product rule involves finding the derivative of each component function with respect to \( w \), and then adding the results according to the rule.
By applying the product rule, we systematically differentiate the product of two components and thus arrive at the correct partial derivative.

Chain rule

The chain rule is another crucial tool in calculus, specifically in dealing with composite functions. When you have a function nested within another function, the chain rule helps differentiate the whole expression. The rule states:

• \( \frac{d}{dx} [f(g(x))] = f'(g(x))g'(x) \)

In the given exercise, when evaluating the partial derivative of the function \( F(w, z) \) with respect to \( z \), we notice that \( \sin^{-1} \) is a composite function of \( \frac{w}{z} \).
Here, the outer function is the inverse sine, and the inner is \( \frac{w}{z} \). Using the chain rule, we first differentiate the outer function regarding the inner, followed by differentiating the inner function itself with respect to \( z \).
This technique allows us to navigate through multiple layers of functions effectively, ensuring every aspect of each function is accounted for. Thus, we deconstruct the layered structure of \( F(w, z) \) with precision, yielding accurate partial derivatives.

Inverse trigonometric functions

Inverse trigonometric functions are extensions of basic trigonometric functions that reverse the process of trigonometry. They include arcsin, arccos, and arctan. The focus in this exercise is on the arcsin function, written as \( \sin^{-1}(x) \).

• For inferring derivatives, it is pivotal to remember the derivative formula: \( \frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}} \).

The arcsin function is specifically relevant when dealing with angles and has implications in various real-world applications, from physics to engineering calculations.
When calculating the partial derivatives of \( F(w, z) \), we must apply this derivative rule to handle the \( \sin^{-1}\left(\frac{w}{z}\right) \) part of the formula. With care, we replace the variable \( x \) with the expression \( \frac{w}{z} \) and proceed accordingly. Mastery of these rules allows smooth navigation through otherwise complex trigonometric expressions, making them manageable and straightforward.

Calculus of several variables

Calculus of several variables, often known as multivariable calculus, is an advanced field that extends the principles of single-variable calculus into higher dimensions. In this realm, functions consist of multiple independent variables, which opens the floor to partial derivatives and multiple integrals.

• The essence of partial derivatives is to measure the rate of change of a function with respect to one variable while holding the others constant.
• This concept is critical in fields like physics, economics, and any discipline requiring modeling of complex systems.

In our specific example, we need to handle the function \( F(w, z) = w \sin^{-1}\left(\frac{w}{z}\right) \), which involves two dimensions: \( w \) and \( z \). Each variable has its independent impact on the function's behavior.
Grasping multivariable calculus allows engineers and scientists to predict and manipulate the effects of various factors on a given phenomenon, through carefully computed changes. Understanding how changes in one variable can influence the overall outcome equips us with critical analytical tools for problem-solving across multiple domains.