Problem 1 - Active Calculus Multivariable

Question

Use the chain rule to find \(\frac{d z}{d t}\), where $$z=x^{2} y+x y^{2}, \quad x=1-t^{5}, \quad y=3-t^{2}$$ First the pieces: \(\frac{\partial z}{\partial x}=\) ________. \(\frac{\partial z}{\partial y}=\) ________. \(\frac{d x}{d t}=\) ________. \(\frac{d y}{d t}=\) ________. End result (in terms of just \(t)\) \(\frac{d z}{d t}=\) __________.

Step-by-Step Solution

Answer

First the pieces: $\frac{\partial z}{\partial x}= 2xy + y^2$. $\frac{\partial z}{\partial y}= x^2 + 2xy$. $\frac{d x}{d t}= -5t^{4}$. $\frac{d y}{d t}= -2t$. End result (in terms of just $t)$ $\frac{d z}{d t}= (2(1-t^5)(3-t^2) + (3-t^2)^2)(-5t^4) + ((1-t^5)^2 + 2(1-t^5)(3-t^2))(-2t)$.

Step 1: Find the partial derivatives of z with respect to x and y

Calculate the partial derivatives of \(z = x^2y + xy^2\): $$\frac{\partial z}{\partial x} = 2xy + y^2$$ $$\frac{\partial z}{\partial y} = x^2 + 2xy$$

Step 2: Find the derivatives of x and y with respect to t

Calculate the derivatives of x and y with respect to t: For x, $$x = 1 - t^5 \Rightarrow \frac{dx}{dt} = -5t^{4}$$ For y, $$y = 3 - t^2 \Rightarrow \frac{dy}{dt} = -2t$$

Step 3: Apply the chain rule

Apply the chain rule to find the derivative of z with respect to t: $$\frac{d z}{d t} = \frac{\partial z}{\partial x}\frac{d x}{d t} + \frac{\partial z}{\partial y}\frac{d y}{d t}$$

Step 4: Substitute the values

Substitute the values from Steps 1 and 2 into the chain rule equation: $$\frac{d z}{d t} = (2xy + y^2)(-5t^4) + (x^2 + 2xy)(-2t)$$ Now substitute the given values of x and y to get the expression in terms of t: $$x = 1 - t^5$$ $$y = 3 - t^2$$

Step 5: Simplify the final expression

Simplify the expression for \(\frac{d z}{d t}\): $$\frac{d z}{d t} = (2(1-t^5)(3-t^2) + (3-t^2)^2)(-5t^4) + ((1-t^5)^2 + 2(1-t^5)(3-t^2))(-2t)$$ Now, you can further simplify the expression by expanding and collecting like terms. The result will be the final expression for \(\frac{d z}{d t}\) in terms of t.

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental concept in calculus, especially when dealing with functions of multiple variables. It allows us to find the derivative of a composite function. In simpler terms, it helps determine how one variable changes when it is affected by the change in another variable, which in turn is a function of a third variable.

When using the chain rule for a function like

• \(z = f(x, y)\) where \(x\) and \(y\) themselves depend on a variable \(t\),
• the derivative \(\frac{dz}{dt}\) is found by adding two contributions:
• from the change in \(x\): \(\frac{\partial z}{\partial x} \cdot \frac{dx}{dt}\) and from the change in \(y\): \(\frac{\partial z}{\partial y} \cdot \frac{dy}{dt}\).

This approach is crucial when differentiating functions that are nested within other functions. It simplifies problems into smaller, easier-to-tackle pieces.

Exploring Partial Derivatives

Partial derivatives represent how a multivariable function changes as one variable changes, keeping the others constant. They are a way to extend derivative concepts to functions with more than one independent variable. Let's say we have a function \(z = x^2y + xy^2\).

• To find the partial derivative of \(z\) with respect to \(x\), \(\frac{\partial z}{\partial x}\), treat \(y\) as a constant. This results in \(2xy + y^2\).
• Similarly, for the partial derivative of \(z\) with respect to \(y\), \(\frac{\partial z}{\partial y}\), treat \(x\) as constant, giving us \(x^2 + 2xy\).

These derivatives illustrate how changes in each variable will affect the function separately, providing insights into the function's behavior under various conditions.

Basics of Differentiation

Differentiation is the process of finding a derivative, which represents an object's rate of change. For a variable \(x\) that changes with another variable \(t\), like \(x = 1 - t^5\), differentiation helps in determining \(\frac{dx}{dt}\), expressing how \(x\) changes as \(t\) changes.

• For the function \(x = 1 - t^5\), the differentiation gives us \(\frac{dx}{dt} = -5t^4\).
• Similarly, if \(y = 3 - t^2\), differentiating with respect to \(t\) yields \(\frac{dy}{dt} = -2t\).

This method is a crucial component in calculus, laying the groundwork for understanding and solving complex problems, particularly when changes in variables lead to changes in other dependent variables.

Mastering Calculus Problem Solving

Solving calculus problems involves breaking down complex expressions into manageable parts using the concepts discovered. Let's see how this works with our given problem.

• After finding the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\), and the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), we apply them in the chain rule formula.
• Substitute these values into the chain rule equation: \[\frac{d z}{d t} = (2xy + y^2)(-5t^4) + (x^2 + 2xy)(-2t).\]
• Input the given expressions for \(x\) and \(y\), and simplify to get the final result in terms of \(t\).

Engage in further simplification to arrive at a concise expression. This thorough process reflects the beauty of calculus problem solving, where each step is interlinked, allowing you to handle initially daunting tasks with clarity and precision.