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Magazine Chapter 14: Problem 8
\(7-12\) Use the Chain Rule to find \(\partial z / \partial s / \partial s\) and \(\partial z / \partial t\) $$z=\arcsin (x-y), \quad x=s^{2}+t^{2}, \quad y=1-2 s t$$
Short Answer
Expert verified
鈭倆/鈭俿 and 鈭倆/鈭倀 involve derivatives of arcsine with respect to u, simplified using x and y.
Step by step solution
01
Find 鈭倆/鈭俿 using the Chain Rule
To find the partial derivative of \( z \) with respect to \( s \), we first need the derivatives of \( x \) and \( y \) with respect to \( s \). First, compute the partial derivatives:\[\frac{\partial x}{\partial s} = 2s\]\[\frac{\partial y}{\partial s} = -2t\]Now, using the Chain Rule, apply:\[\frac{\partial z}{\partial s} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial s}\]where \( u = x - y \), giving:\[\frac{\partial z}{\partial s} = \frac{1}{\sqrt{1-(x-y)^2}} \cdot (2s + 2t)\]
02
Simplify 鈭倆/鈭俿
Substitute \( x = s^2 + t^2 \) and \( y = 1 - 2st \) into \( x - y \):\[x - y = (s^2 + t^2) - (1 - 2st) = s^2 + t^2 + 2st - 1\]Now substitute into the partial derivative:\[\frac{\partial z}{\partial s} = \frac{1}{\sqrt{1 - (s^2 + t^2 + 2st - 1)^2}} \cdot (2s + 2t)\]
03
Find 鈭倆/鈭倀 using the Chain Rule
To find \( \frac{\partial z}{\partial t} \), we need the derivatives of \( x \) and \( y \) with respect to \( t \). Compute the partial derivatives:\[\frac{\partial x}{\partial t} = 2t\]\[\frac{\partial y}{\partial t} = -2s\]Using the Chain Rule:\[\frac{\partial z}{\partial t} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial t}\]where \( u = x - y \), giving:\[\frac{\partial z}{\partial t} = \frac{1}{\sqrt{1-(x-y)^2}} \cdot (2t + 2s)\]
04
Simplify 鈭倆/鈭倀
Again, substitute \( x = s^2 + t^2 \) and \( y = 1 - 2st \) into \( x - y \):\[x - y = s^2 + t^2 + 2st - 1\]Substitute into the partial derivative:\[\frac{\partial z}{\partial t} = \frac{1}{\sqrt{1 - (s^2 + t^2 + 2st - 1)^2}} \cdot (2t + 2s)\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
In multivariable calculus, partial derivatives play a crucial role. They help us understand how a function changes when one of its several variables is altered, while the others are held constant.
Partial derivatives are especially useful when working with functions of several variables.
In our exercise, we explore the partial derivatives of the function \( z = \arcsin(x-y) \) with respect to the variables \( s \) and \( t \). These partial derivatives help determine the rate of change of \( z \) as each variable changes independently.
Partial derivatives are especially useful when working with functions of several variables.
- For example, consider a function \( f(x, y) \) which depends on two variables. The partial derivative \( \frac{\partial f}{\partial x} \) examines how \( f \) changes as \( x \) changes, keeping \( y \) constant.
- Similarly, \( \frac{\partial f}{\partial y} \) focuses on changes with respect to \( y \).
In our exercise, we explore the partial derivatives of the function \( z = \arcsin(x-y) \) with respect to the variables \( s \) and \( t \). These partial derivatives help determine the rate of change of \( z \) as each variable changes independently.
Arcsin Function
The arcsin function, sometimes called the inverse sine function, is the inverse of the sine function. It takes a value and provides the angle whose sine is that particular value.
Arcsin is significant in trigonometry and especially in calculus due to its impact on differentiable functions. When dealing with the arcsin function in calculus:
In our example, understanding the derivative of \( \arcsin \) is vital in applying the chain rule to determine the partial derivatives of \( z = \arcsin(x-y) \).
Arcsin is significant in trigonometry and especially in calculus due to its impact on differentiable functions. When dealing with the arcsin function in calculus:
- Its derivative is \( \frac{1}{\sqrt{1-x^2}} \), which plays a role in the chain rule for finding derivatives.
- Arcsin helps in finding angle measurements from particular sine values, which can be critical in various scientific and engineering applications.
In our example, understanding the derivative of \( \arcsin \) is vital in applying the chain rule to determine the partial derivatives of \( z = \arcsin(x-y) \).
Multivariable Calculus
Multivariable calculus extends calculus to functions with more than one variable. This broadens the range of problems you can tackle, from physics to economics. Essential tools in this area include partial derivatives and the chain rule.
This exercise demonstrates concepts of multivariable calculus through the partial differentiation of \( z = \arcsin(x-y) \), where \( x \) and \( y \) themselves are functions of \( s \) and \( t \). It highlights how understanding individual variable changes is crucial in broader applications.
- Partial derivatives help us understand the behavior of multivariable functions by examining how they change concerning each variable individually.
- The chain rule allows us to compute the derivative of composite functions, which are frequently encountered in multivariable contexts.
This exercise demonstrates concepts of multivariable calculus through the partial differentiation of \( z = \arcsin(x-y) \), where \( x \) and \( y \) themselves are functions of \( s \) and \( t \). It highlights how understanding individual variable changes is crucial in broader applications.
Function Composition
Function composition involves creating new functions by combining existing ones. In calculus, understanding composition is crucial for applying the chain rule effectively.
When functions are composed, the output of one function becomes the input of another, leading us to consider how changing an initial input affects the final output. Using function composition:
In the given exercise, function composition is seen where \( z \), \( x \), and \( y \) are interdependent, requiring careful differentiation of each component to understand the overall change in \( z \) related to the variables \( s \) and \( t \).
When functions are composed, the output of one function becomes the input of another, leading us to consider how changing an initial input affects the final output. Using function composition:
- The chain rule enables us to differentiate composite functions by relating their rates of change.
- It is particularly useful when dealing with nested functions, such as \( z = \arcsin(x-y) \) in our case.
In the given exercise, function composition is seen where \( z \), \( x \), and \( y \) are interdependent, requiring careful differentiation of each component to understand the overall change in \( z \) related to the variables \( s \) and \( t \).
