Question

Find \(d w / d t\) by using the Chain Rule. Express your final answer in terms of \(t\). $$ w=x y+y z+x z ; x=t^{2}, y=1-t^{2}, z=1-t $$

Short answer

\( \frac{dw}{dt} = 2t^3 - 1 \)

Step-by-step solution

Identify the Function and Variables

The given function is a composite with three variables: \( w = xy + yz + xz \). The variables \( x, y, \) and \( z \) are expressed in terms of \( t \): \( x = t^2 \), \( y = 1 - t^2 \), and \( z = 1 - t \). We need to find the derivative \( \frac{dw}{dt} \) using the chain rule.

Express dw/dt Using Chain Rule

The Chain Rule tells us that to find \( \frac{dw}{dt} \), we need to take the partial derivatives of \( w \) with respect to each variable, and then multiply by the derivative of that variable with respect to \( t \). So, \( \frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt} \).

Calculate Partial Derivatives

Calculate the partial derivatives of \( w \): - \( \frac{\partial w}{\partial x} = y + z \)- \( \frac{\partial w}{\partial y} = x + z \)- \( \frac{\partial w}{\partial z} = y + x \)

Calculate Derivatives of Each Variable

Find the derivative of each variable with respect to \( t \):- \( \frac{dx}{dt} = \frac{d}{dt}(t^2) = 2t \)- \( \frac{dy}{dt} = \frac{d}{dt}(1 - t^2) = -2t \)- \( \frac{dz}{dt} = \frac{d}{dt}(1 - t) = -1 \)

Substitute and Simplify

Substitute the expressions for partial derivatives and derivatives with respect to \( t \) into the expression for \( \frac{dw}{dt} \): \[\frac{dw}{dt} = (y + z)(2t) + (x + z)(-2t) + (y + x)(-1)\]Substitute \( x = t^2 \), \( y = 1 - t^2 \), \( z = 1 - t \) into the equation and simplify:

Simplify and Solve

After substituting:\[\frac{dw}{dt} = ((1 - t^2) + (1 - t))(2t) + (t^2 + 1 - t)(-2t) + ((1 - t^2) + t^2)(-1)\]Simplify each term:1. \(((1 - t^2) + (1 - t))(2t) = (2 - t^2 - t)(2t)\), expand and simplify to \(4t - 2t^3 - 2t^2\).2. \((t^2 + 1 - t)(-2t) = (-2t^3 - 2t + 2t^2)\).3. \((1 - t^2 + t^2)(-1) = -1\).Combine like terms to get the final expression:\[\frac{dw}{dt} = 2t^3 - 1\].

Key concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus. When you have a function of multiple variables, like our function \( w = xy + yz + xz \), a partial derivative measures how the function changes as one of the variables changes while keeping the others constant.
In the context of our exercise, we're interested in how \( w \) changes with respect to \( x \), \( y \), and \( z \). We compute:

• \( \frac{\partial w}{\partial x} = y + z \)
• \( \frac{\partial w}{\partial y} = x + z \)
• \( \frac{\partial w}{\partial z} = x + y \)

Each of these gives the rate of change of \( w \) with one variable, keeping others constant. This is essential to use the chain rule effectively in our multivariable function.

Composite Function

Composite functions arise when one function is applied to the result of another. In our exercise, the function \( w = xy + yz + xz \) depends on variables \(x, y, \text{ and } z\), which themselves depend on \( t \). Because of this dependency, \( w \) is a composite function.
For a composite function, you evaluate the outer function using the results from the inner function.
Here:

• \( x = t^2 \)
• \( y = 1 - t^2 \)
• \( z = 1 - t \)

These transformations create a chain of functions where \( t \) influences \( w \) through \( x, y, \text{ and } z \). Understanding this concept is crucial when calculating derivatives using the chain rule.

Calculus

Calculus is the branch of mathematics focused on change. It uses concepts like limits, derivatives, and integrals to explore how quantities vary. In this exercise, we focus on derivatives, which help us understand the rate of change. The chain rule, a vital tool in calculus, allows us to differentiate composite functions by connecting changes in dependent variables.
This exercise demonstrates calculus principles by requiring you to calculate the derivative of \( w \) concerning \( t \). Each step shows how calculus provides tools to transition from a complex composite function to tangible rates of change with respect to a single variable. Calculus revels in the subtle and the infinite, yet it provides practical solutions for concrete problems like the one in this exercise.

Derivative with Respect to a Variable

Taking a derivative with respect to a variable allows us to measure how a function changes as that variable changes. In this exercise, we need to find \( \frac{dw}{dt} \), the derivative of \( w \) with respect to \( t \). Here’s how it is done:
We express \( \frac{dw}{dt} \) by using the chain rule, which states you find the sum of each variable's partial derivatives weighted by the derivative of that variable with respect to \( t \).
The chain rule formula is:

• \( \frac{dw}{dt} = \frac{\partial w}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial w}{\partial y} \cdot \frac{dy}{dt} + \frac{\partial w}{\partial z} \cdot \frac{dz}{dt} \).

Finally, after substituting and simplifying, you find \( \frac{dw}{dt} = 2t^3 - 1 \). This expression shows how the function \( w \) changes as \( t \) changes.