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Magazine Chapter 13: Problem 11
Compute \(\partial z / \partial r\) and \(\partial z / \partial s\). $$ z=\sin 2 u \cos 3 v ; u=(r+s)^{2}, v=(r-s)^{2} $$
Short Answer
Expert verified
\( \frac{\partial z}{\partial r} = 4(r+s) \cos 2u \cos 3v - 6(r-s) \sin 2u \sin 3v \); \( \frac{\partial z}{\partial s} = 4(r+s) \cos 2u \cos 3v + 6(r-s) \sin 2u \sin 3v \).
Step by step solution
01
Identify the composite functions
The function for which we need partial derivatives is given by \( z = \sin 2u \cos 3v \), where \( u = (r+s)^2 \) and \( v = (r-s)^2 \). The task is to compute \( \frac{\partial z}{\partial r} \) and \( \frac{\partial z}{\partial s} \).
02
Apply the Chain Rule
To find \( \frac{\partial z}{\partial r} \), apply the chain rule: \( \frac{\partial z}{\partial r} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial r} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial r} \).Similarly, for \( \frac{\partial z}{\partial s} \), apply the chain rule: \( \frac{\partial z}{\partial s} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial s} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial s} \).
03
Differentiate with respect to u and v
Calculate:\( \frac{\partial z}{\partial u} = 2\cos 2u \cos 3v \)\( \frac{\partial z}{\partial v} = -3\sin 2u \sin 3v \)
04
Differentiate u and v with respect to r and s
1. \( \frac{\partial u}{\partial r} = 2(r+s) \) and \( \frac{\partial u}{\partial s} = 2(r+s) \)2. \( \frac{\partial v}{\partial r} = 2(r-s) \) and \( \frac{\partial v}{\partial s} = -2(r-s) \)
05
Compute \( \partial z / \partial r \)
Substitute the derivatives from Steps 3 and 4 into:\[ \frac{\partial z}{\partial r} = \left(2 \cos 2u \cos 3v \right) \left( 2(r+s) \right) + \left( -3 \sin 2u \sin 3v \right) \left( 2(r-s) \right) \]Simplify to obtain:\[ \frac{\partial z}{\partial r} = 4(r+s) \cos 2u \cos 3v - 6(r-s) \sin 2u \sin 3v \]
06
Compute \( \partial z / \partial s \)
Substitute the derivatives from Steps 3 and 4 into:\[ \frac{\partial z}{\partial s} = \left(2 \cos 2u \cos 3v \right) \left( 2(r+s) \right) + \left( -3 \sin 2u \sin 3v \right) \left( -2(r-s) \right) \]Simplify to obtain:\[ \frac{\partial z}{\partial s} = 4(r+s) \cos 2u \cos 3v + 6(r-s) \sin 2u \sin 3v \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a fundamental concept in calculus used for differentiating composite functions. It allows us to find the derivative of a function that is made up of other functions. When we have a function like \( z = \sin(2u) \cos(3v) \) and the variables \( u \) and \( v \) are themselves functions of \( r \) and \( s \), we need the chain rule to find the partial derivatives \( \frac{\partial z}{\partial r} \) and \( \frac{\partial z}{\partial s} \).
To apply the chain rule in this context:
These steps allow us to see how changes in \( r \) and \( s \) affect \( z \), by perturbing through the intermediary variables \( u \) and \( v \).
To apply the chain rule in this context:
- Compute the derivative of \( z \) with respect to \( u \) and \( v \), which are part of the composite function.
- Compute the derivatives of \( u \) and \( v \) with respect to \( r \) and \( s \).
- Combine these using the formula: \( \frac{\partial z}{\partial r} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial r} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial r} \).
These steps allow us to see how changes in \( r \) and \( s \) affect \( z \), by perturbing through the intermediary variables \( u \) and \( v \).
Composite Functions
In calculus, composite functions play a crucial role when working with real-world models because they allow the modeling of complex systems. A composite function is a function that is formed by combining two or more functions.
For example, in our original exercise:
This composition of functions allows us to model the problem by breaking it down into simpler parts. Evaluating or differentiating a composite function involves applying the rules of calculus鈥攑articularly the chain rule鈥攖o each layer of function composition.
For example, in our original exercise:
- The function \( z = \sin(2u) \cos(3v) \) depends on \( u \) and \( v \) instead of directly depending on \( r \) and \( s \).
- Here, \( u = (r+s)^2 \) and \( v = (r-s)^2 \) are themselves functions of \( r \) and \( s \).
This composition of functions allows us to model the problem by breaking it down into simpler parts. Evaluating or differentiating a composite function involves applying the rules of calculus鈥攑articularly the chain rule鈥攖o each layer of function composition.
Differentiation
Differentiation is the process of finding a derivative, which represents how a function changes as its input changes. In the context of the exercise, we are finding partial derivatives as a form of differentiation.
Partial differentiation allows us to find the rate at which a function changes with respect to one variable while holding the others constant. This is vital in multivariable calculus where functions depend on multiple inputs鈥攍ike \( r \) and \( s \) in our example.
Each of these derivatives provides a piece of the puzzle, showing how each component influences the overall function \( z \), thereby allowing us to compute the needed partial derivatives effectively.
Partial differentiation allows us to find the rate at which a function changes with respect to one variable while holding the others constant. This is vital in multivariable calculus where functions depend on multiple inputs鈥攍ike \( r \) and \( s \) in our example.
- We calculate \( \frac{\partial z}{\partial u} \) and \( \frac{\partial z}{\partial v} \) using standard differentiation techniques.
- The expressions for \( \frac{\partial u}{\partial r} \), \( \frac{\partial u}{\partial s} \), \( \frac{\partial v}{\partial r} \), and \( \frac{\partial v}{\partial s} \) are derived by treating all other variables as constants.
Each of these derivatives provides a piece of the puzzle, showing how each component influences the overall function \( z \), thereby allowing us to compute the needed partial derivatives effectively.
