Problem 31 - Calculus with Concepts in Calculus

Question

Assume that the equation defines \(z\) implicitly as a function of \(x\) and \(y\), and use "implicit partial differentiation" to find \(\partial z / \partial x\) and \(\partial z / \partial y\). $$ x-y z+\cos x y z=2 $$

Step-by-Step Solution

Answer

\(\frac{\partial z}{\partial x} = \frac{1}{yz + y z^2 \sin(xyz)}\), \(\frac{\partial z}{\partial y} = \frac{xz \sin(xyz) - z}{y + xz^2 \sin(xyz)}\).

Step 1: Differentiate the Equation with Respect to x

To find \( \frac{\partial z}{\partial x} \), implicitly differentiate the equation \( x - yz + \cos(xyz) = 2 \) with respect to x, treating y and z as functions of x. The derivative becomes:\[1 - (y \frac{\partial z}{\partial x}) - yz \sin(xyz)(y \frac{dz}{dx}) + yz \sin(xyz) = 0\]Simplify to obtain:\[1 - yz \frac{\partial z}{\partial x} - y z^2 \sin(xyz) \frac{\partial z}{\partial x} = 0\]This forces an expression for \( \frac{\partial z}{\partial x} \).

Step 2: Solve for \(\frac{\partial z}{\partial x}\)

Rearrange the previous result to explicitly solve for \(\frac{\partial z}{\partial x}\) by factoring it out:\[\frac{\partial z}{\partial x} (yz + y z^2 \sin(xyz)) = 1\]Therefore, we find:\[\frac{\partial z}{\partial x} = \frac{1}{yz + y z^2 \sin(xyz)}\]

Step 3: Differentiate the Equation with Respect to y

To find \( \frac{\partial z}{\partial y} \), implicitly differentiate the equation \( x - yz + \cos(xyz) = 2 \) with respect to y, treating x and z as functions of y. The derivative becomes:\[0 - (z + y\frac{dz}{dy}) - zx\sin(xyz)(x\frac{dz}{dy}) + zx\sin(xyz) = 0\]Simplify to obtain:\[-(z + y \frac{\partial z}{\partial y}) - xz^2 \sin(xyz)\frac{\partial z}{\partial y} + xz \sin(xyz) = 0\]

Step 4: Solve for \(\frac{\partial z}{\partial y}\)

Rearrange the equation to solve for \(\frac{\partial z}{\partial y}\):\[-(z + y \frac{\partial z}{\partial y}) - xz^2 \sin(xyz)\frac{\partial z}{\partial y} + xz \sin(xyz) = 0\]This simplifies to:\[\frac{\partial z}{\partial y} (y + xz^2 \sin(xyz)) = xz\sin(xyz)\]Therefore, we find:\[\frac{\partial z}{\partial y} = \frac{xz \sin(xyz) - z}{y + xz^2 \sin(xyz)}\]

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, especially when dealing with functions of multiple variables. They allow us to understand how a function changes as one of its input variables changes, while keeping the other input variables constant.
For example, if you have a function expressed as \( f(x, y) \), the partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), gives you the rate of change of \( f \) as \( x \) changes and \( y \) remains fixed. Similarly, \( \frac{\partial f}{\partial y} \) gives the rate of change with respect to \( y \), holding \( x \) constant.
We often use partial derivatives in physics and engineering when analyzing systems influenced by multiple factors. Understanding how a system responds to changes in one specific variable can be crucial.

• Partial derivatives help describe gradients and slopes in multivariable contexts.
• They are instrumental in deriving other calculus concepts like tangent planes and optimization problems.

Calculus Problems

Calculus problems involving partial derivatives can often appear complex due to the involvement of multiple variables and their interactions. Problems like the one presented require you to apply differentiation rules in a careful manner to achieve correct solutions.
Steps to solving such problems often include:
- Identifying the given function and its dependent and independent variables.
- Applying differentiation step by step with respect to each variable independently, while treating others as constants temporarily.
- Rearranging equations to solve for the unknown partial derivatives. This might involve simplification through algebraic manipulation.
Practice and familiarity with differentiation techniques are crucial when tackling calculus problems. Such exercises build a stronger grasp of how varying each parameter in a multivariable function impacts the overall result.

Implicit Differentiation

Implicit differentiation is a useful method when dealing with equations where variables are intermixed, and it is difficult to solve for one variable explicitly in terms of others.
Unlike explicit differentiation, where you can directly differentiate a variable, implicit differentiation involves differentiating each part of an equation with respect to a given variable, while keeping in mind that some variables themselves are functions of the variable in question.
To use implicit differentiation effectively:

• Consider all variables as functions when differentiating.
• Apply the chain rule where necessary to account for the dependence of variables.
• Once differentiated, solve the resulting equation for the desired derivative.

This method is especially handy in multivariable contexts where isolating a specific variable directly is challenging, like in our problem involving the function \( z \) with respect to \( x \) and \( y \).

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions of several variables. It introduces new concepts and methods for assessing and solving problems where multiple factors interplay.
The study of multivariable calculus includes learning about gradients, divergence, curl, and using multiple integrals. It becomes a crucial foundation for fields such as physics, engineering, and economics, where systems often depend on more than one variable.
In our exercise, multivariable calculus comes into play as we observe changes in \( z \) with respect to changes in both \( x \) and \( y \). The techniques of partial derivatives and implicit differentiation let us explore these relationships comprehensively.

• Multivariable calculus helps in modeling real-world phenomena that cannot be captured by single-variable functions.
• It creates ways to analyze functions over surfaces and volumes, beyond simple lines or curves.

Understanding these principles through exercises helps grasp how multiple inputs can affect a complex system.