Problem 25 - Calculus Early Transcendentals

Question

\(21-26\) Use the Chain Rule to find the indicated partial derivatives. $$\begin{array}{ll}{u=x^{2}+y z,} & {x=p r \cos \theta, \quad y=p r \sin \theta, \quad z=p+r} \\ {\frac{\partial u}{\partial p}, \frac{\partial u}{\partial r}, \frac{\partial u}{\partial \theta}} & {\text { when } p=2, r=3, \theta=0}\end{array}$$

Step-by-Step Solution

Answer

\(\frac{\partial u}{\partial p} = 36\), \(\frac{\partial u}{\partial r} = 24\), \(\frac{\partial u}{\partial \theta} = 30\).

Step 1: Substitute Variables into Expression for u

First, substitute the expressions for \(x\), \(y\), and \(z\) into the expression for \(u\): \[u = (pr \cos \theta)^2 + (pr \sin \theta)(p+r)\].

Step 2: Differentiate u with Respect to p

Find \(\frac{\partial u}{\partial p}\) by differentiating the substituted expression with respect to \(p\):- The term \((pr \cos \theta)^2\) becomes \(2pr \cos\theta \cdot r\cos\theta = 2pr^2 \cos^2\theta\).- The term \((pr \sin \theta)(p+r)\) expands to \(pr^2 \sin \theta + p^2r \sin \theta\). Differentiating yields \(r^2 \sin \theta + 2pr\sin \theta\).Combine these partial derivatives:\[\frac{\partial u}{\partial p} = 2pr^2 \cos^2 \theta + r^2 \sin \theta + 2pr \sin \theta\].

Step 3: Evaluate \(\frac{\partial u}{\partial p}\) at Specified Values

Substitute \(p=2\), \(r=3\), and \(\theta=0\) into \(\frac{\partial u}{\partial p}\): - \(\cos 0 = 1\) and \(\sin 0 = 0\), so the expression becomes:\[\frac{\partial u}{\partial p} = 2 \times 2 \times 3^2 \times 1^2 + 3^2 \times 0 + 2 \times 2 \times 3 \times 0 = 36\].

Step 4: Differentiate u with Respect to r

Find \(\frac{\partial u}{\partial r}\) by differentiating the substituted expression with respect to \(r\): - The term \((pr \cos \theta)^2\) becomes \(2pr \cos \theta \cdot p \cos \theta = 2p^2r \cos^2\theta\).- The term \((pr \sin \theta)(p+r)\) expands to \(pr^2 \sin \theta + p^2r \sin \theta\). Differentiating yields \(2pr \sin \theta + p^2 \sin \theta\).Combine these partial derivatives:\[\frac{\partial u}{\partial r} = 2p^2r \cos^2 \theta + 2pr \sin \theta + p^2 \sin \theta + pr^2 \cos \theta \sin \theta\].

Step 5: Evaluate \(\frac{\partial u}{\partial r}\) at Specified Values

Substitute \(p=2\), \(r=3\), and \(\theta=0\) into \(\frac{\partial u}{\partial r}\): - \(\cos 0 = 1\) and \(\sin 0 = 0\), so the expression becomes:\[\frac{\partial u}{\partial r} = 2 \cdot 2^2 \cdot 3 \cdot 1^2 + 2 \cdot 2 \cdot 0 + 2^2 \cdot 0 + 2 \cdot 3^2 \cdot 1 \cdot 0 = 24\].

Step 6: Differentiate u with Respect to theta

Find \(\frac{\partial u}{\partial \theta}\) by differentiating the substituted expression with respect to \(\theta\): - The term \((pr \cos \theta)^2\) becomes \(-2pr \sin \theta \cdot pr \cos \theta = -2p^2r^2 \cos \theta \sin \theta\).- The term \((pr \sin \theta)(p+r)\) differentiates with respect to \(\theta\) as \(pr \cos \theta(p+r)\).Combine these partial derivatives:\[\frac{\partial u}{\partial \theta} = -2p^2r^2 \cos \theta \sin \theta + pr \cos \theta (p+r)\].

Step 7: Evaluate \(\frac{\partial u}{\partial \theta}\) at Specified Values

Substitute \(p=2\), \(r=3\), and \(\theta=0\) into \(\frac{\partial u}{\partial \theta}\): - \(\cos 0 = 1\) and \(\sin 0 = 0\), so the expression becomes:\[\frac{\partial u}{\partial \theta} = -2 \cdot 2^2 \cdot 3^2 \cdot 1 \cdot 0 + 2 \cdot 3 \cdot 1 (2+3) = 30\].

Key Concepts

Partial Derivatives

Partial derivatives are an essential part of understanding functions with multiple variables. When working with a function like \( u = x^{2} + yz \), where \( x \), \( y \), and \( z \) are functions of more than one variable, finding partial derivatives allows us to examine how the function changes as one of the variables changes, while keeping the others constant.

• Notation: Partial derivatives are denoted as \( \frac{\partial u}{\partial x} \), \( \frac{\partial u}{\partial y} \), and so on, indicating differentiation with respect to the specified variable.
• Calculating Partial Derivatives: To find the partial derivative of a function during differentiation, only treat the variable of interest as a variable, and treat all other variables as constants.

In our specific exercise, we computed \( \frac{\partial u}{\partial p} \), \( \frac{\partial u}{\partial r} \), and \( \frac{\partial u}{\partial \theta} \). Each partial derivative calculated shows how the function \( u \) changes as you tweak each respective variable while fixing others. The Chain Rule, a principle which will be discussed next, assists in this process.

Multivariable Calculus

Multivariable calculus extends single-variable calculus concepts to functions with more than one variable. In this context, our variables \( x \), \( y \), and \( z \) are dependent on \( p \), \( r \), and \( \theta \). This shift necessitates the use of the Chain Rule to manage differentiation.
The Chain Rule is a formula for computing the derivative of the composition of two or more functions. It is crucial in multivariable calculus because it allows us to differentiate complex functions where the variable of interest can be expressed through intermediate variables.

• Usage in Our Exercise: In our example, the Chain Rule helped us find partial derivatives by substituting expressions of \( x \), \( y \), and \( z \) in terms of \( p \), \( r \), and \( \theta \), then differentiating these substituted expressions.
• Steps: The first step involved rewriting \( u \) with the new expressions. Following this, the differentiation took place for each specified variable. This approach reveals our function's behavior in a multidimensional domain.

Understanding this higher-dimensional calculus is paramount in fields like physics and engineering, where you deal with multidimensional parameter spaces.

Trigonometric Functions

Trigonometric functions like \( \cos \theta \) and \( \sin \theta \) often appear in problems involving partial derivatives, especially when dealing with variables that function as angles or have cyclic behavior. In our example, \( x \) and \( y \) are defined using polar coordinates through \( \cos \theta \) and \( \sin \theta \).
Trigonometric functions have unique derivatives:

• \( \frac{d}{d\theta}(\cos \theta) = -\sin \theta \)
• \( \frac{d}{d\theta}(\sin \theta) = \cos \theta \)

These derivatives are essential when using the Chain Rule to find the rate of change of a multivariable function with respect to an angular variable, like \( \theta \).
In our solution:

• Application: Trigonometric identities and their derivatives helped simplify and solve differentiations involving the angular component \( \theta \).
• Evaluating at Specific Values: In our problem, substituting \( \theta = 0 \) effectively simplifies the expressions due to \( \cos 0 = 1 \) and \( \sin 0 = 0 \).

Trigonometric functions thus offer a structured approach to resolving changes in processes that have periodic characteristics or involve rotations.