Chapter 14: Problem 5
Find \begin{equation}\quad \text { a. }\left(\frac{\partial w}{\partial y}\right)_{x} \quad \text { b. }\left(\frac{\partial w}{\partial y}\right)_{z}\end{equation} at the point \((w, x, y, z)=(4,2,1,-1)\) if \begin{equation}w=x^{2} y^{2}+y z-z^{3} \quad \text { and } \quad x^{2}+y^{2}+z^{2}=6.\end{equation}
Short Answer
Step by step solution
Understand the Problem
Differentiate with Respect to y Keeping x Constant
Establish Relationship for z being Constant
Differentiate with Respect to y Keeping z Constant
Combine Results
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Multivariable calculus
In multivariable calculus, partial derivatives are a fundamental concept. A partial derivative represents how a function changes as one of its variables changes, while all other variables remain constant. This is crucial in understanding how multi-variable functions behave in different directions.
Let's take a function \( w = f(x, y, z) \). The partial derivative \( \left( \frac{\partial w}{\partial y} \right)_x \) refers to the rate of change of \( w \) with respect to \( y \), while keeping \( x \) constant. If our function includes relationships like \( w = x^2 y^2 + yz - z^3 \), it provides a window into how these variables interconnect and influence each other in a multi-variable context.
Implicit differentiation
For example, in the provided constraint \( x^2 + y^2 + z^2 = 6 \), we cannot easily solve for one variable. Instead, implicit differentiation allows us to differentiate both sides of an equation with respect to one variable, while treating the other variables as functions.
- First, differentiate the entire equation with respect to \( y \).
- Treat other variables as constant or as functions that depend on \( y \).
- Apply the chain rule where necessary to account for these dependencies.
Constraint equations
For the problem at hand, the constraint is \( x^2 + y^2 + z^2 = 6 \). This spherical constraint limits the combination of \( x, y, \) and \( z \) values that the function can consider.
- These constraints are essential to solving real-world problems since they reflect real limitations or conditions.
- Constraint equations assist in understanding permissible variable configurations.
- They offer critical insight while performing differentiation since they may redefine our view of variable independence.