Problem 30 - 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 \(\partial z / \partial y\). $$ x^{2} z^{2}-2 x y z+z^{3} y^{2}=3 $$

Step-by-Step Solution

Answer

\(\frac{\partial z}{\partial x} = \frac{2yz - 2xz^2}{2x^2z - 2xy + 3z^2y^2}\) and \(\frac{\partial z}{\partial y} = \frac{2xz - 2z^3y}{2x^2z - 2xy + 3z^2y^2}\).

Step 1: Implicitly Differentiate with Respect to x

To find \(\frac{\partial z}{\partial x}\), implicitly differentiate the entire equation \(x^2z^2 - 2xyz + z^3y^2 = 3\) with respect to \(x\). Use the product rule and chain rule where necessary. Start with \(x^2z^2\):\[ \frac{d}{dx}(x^2z^2) = 2xz^2 + 2x^2z\frac{\partial z}{\partial x} \]Then, for \(-2xyz\):\[ \frac{d}{dx}(-2xyz) = -2yz - 2x\frac{\partial z}{\partial x}y \]And for \(z^3y^2\):\[ \frac{d}{dx}(z^3y^2) = 3z^2y^2\frac{\partial z}{\partial x} \]Combine these to get the derivative of the whole equation: \[ 2xz^2 + 2x^2z\frac{\partial z}{\partial x} - 2yz - 2xy\frac{\partial z}{\partial x} + 3z^2y^2\frac{\partial z}{\partial x} = 0 \]

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

Rearrange the equation obtained from Step 1 to isolate \(\frac{\partial z}{\partial x}\):\[ (2x^2z - 2xy + 3z^2y^2)\frac{\partial z}{\partial x} = 2yz - 2xz^2 \]Solve for \(\frac{\partial z}{\partial x}\):\[ \frac{\partial z}{\partial x} = \frac{2yz - 2xz^2}{2x^2z - 2xy + 3z^2y^2} \]

Step 3: Implicitly Differentiate with Respect to y

To find \(\frac{\partial z}{\partial y}\), implicitly differentiate the entire equation \(x^2z^2 - 2xyz + z^3y^2 = 3\) with respect to \(y\). Start with \(x^2z^2\):\[ \frac{d}{dy}(x^2z^2) = 2x^2z\frac{\partial z}{\partial y} \]Then, \(-2xyz\):\[ \frac{d}{dy}(-2xyz) = -2xz - 2x\frac{\partial z}{\partial y}y \]And for \(z^3y^2\):\[ \frac{d}{dy}(z^3y^2) = 2z^3y + 3z^2y^2\frac{\partial z}{\partial y} \]Combine these to get the derivative of the whole equation:\[ 2x^2z\frac{\partial z}{\partial y} - 2xz - 2xy\frac{\partial z}{\partial y} + 2z^3y + 3z^2y^2\frac{\partial z}{\partial y} = 0 \]

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

Rearrange the equation obtained from Step 3 to isolate \(\frac{\partial z}{\partial y}\):\[ (2x^2z - 2xy + 3z^2y^2)\frac{\partial z}{\partial y} = 2xz - 2z^3y \]Solve for \(\frac{\partial z}{\partial y}\):\[ \frac{\partial z}{\partial y} = \frac{2xz - 2z^3y}{2x^2z - 2xy + 3z^2y^2} \]

Key Concepts

Partial Differentiation

Partial differentiation is a technique in calculus used to differentiate a function of multiple variables with respect to one variable, keeping the other variables constant. It's particularly useful when dealing with functions in two or three dimensions. In the given exercise, we're finding the partial derivatives of an implicitly defined function, where the equation is expressed in terms of several variables including the outcome variable, often implicitly present. This approach allows us to understand how a change in one variable affects the function, while holding the others constant.
For example, to find \(\frac{\partial z}{\partial x}\), we differentiate each part of the function with respect to \(x\), assuming \(y\) and \(z\) are constants where applicable. Similarly, for \(\frac{\partial z}{\partial y}\), we differentiate each term with respect to \(y\). These computations provide insight into how changes in either \(x\) or \(y\) influence the value of \(z\), giving a slice of the function's behavior.

Chain Rule

The chain rule is a fundamental tool in calculus for understanding how to differentiate composite functions. It is particularly handy in implicit differentiation, where one is dealing with equations involving multiple variables. The chain rule allows us to differentiate a function that is dependent on another function, thereby linking their rates of change.
In our exercise, the chain rule facilitates the differentiation of terms such as \(x^2z^2\) and \(z^3y^2\), which involve \(z\), a function of other variables. For instance, when differentiating \(x^2z^2\) with respect to \(x\), the chain rule helps us account for how changes in \(x\) affect \(z^2\), which is itself a function of \(x\). This process requires multiplying the derivative of \(z^2\) with respect to \(x\) by the derivative of the inner function \(z\) with respect to \(x\).
Using the chain rule in this manner allows for the proper handling of nested functions, ensuring accurate computation of derivatives in more complex equations.

Product Rule

The product rule is another crucial differentiation strategy used to differentiate terms where two functions are multiplied together. It states that the derivative of a product of two functions is the first function times the derivative of the second function, plus the second function times the derivative of the first function: \( (uv)' = u'v + uv' \).In the context of the problem, the product rule is particularly applicable to terms like \(-2xyz\). When differentiating with respect to \(x\) or \(y\), we use the product rule to separately consider the derivatives of \(-2xy\) and \(z\), accounting for how each part of the product contributes to the overall rate of change.
The product rule ensures that every component of these multiplicative terms is incorporated into the differentiation process, yielding the correct partial derivatives of each segment in the complex implicit equation.

Calculus Problem Solving

Calculus problem solving often requires a mix of strategies to tackle different components of a question effectively. In this exercise, solving requires employing implicit differentiation, together with the chain rule and product rule, to navigate the intricacies of a multi-variable function.
Here are some strategies that help in solving such problems:

• Careful Identification: Recognize which variable you are differentiating with respect to and treat other variables appropriately as constants if they do not directly depend on the one you're differentiating.
• Sequential Application: Use the chain rule and product rule where the function composition demands it—particularly when a function is nested or two functions multiply.
• Simplification: After differentiation, collect and simplify terms to isolate the derivative you are solving for, ensuring a correct and clear solution.

By keeping these strategies in mind, complex calculus problems become more approachable and manageable, resulting in accurate solutions and a better understanding of the underlying mathematics.