Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-6, find dw/dt by using the Chain Rule. Express your final answer in terms of \(t\). $$ w=x^{2} y^{3} ; x=t^{3}, y=t^{2} $$

Step-by-Step Solution

Answer

\( \frac{dw}{dt} = 12t^{11} \).

Step 1: Rewrite the expression for clarity

The function given is \( w = x^2 y^3 \) where \( x = t^3 \) and \( y = t^2 \). Rewriting, we have:\[w(t) = (t^3)^2 (t^2)^3\]

Step 2: Differentiate using the Chain Rule

To find \( \frac{dw}{dt} \), use the Chain Rule. The Chain Rule says that if \( w = f(x, y) \), then \[\frac{dw}{dt} = \frac{\partial w}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial w}{\partial y} \cdot \frac{dy}{dt}.\]Now compute the necessary partial derivatives.

Step 3: Compute the partial derivative with respect to x

The partial derivative of \( w \) with respect to \( x \) is \[\frac{\partial w}{\partial x} = 2xy^3.\]

Step 4: Compute the partial derivative with respect to y

The partial derivative of \( w \) with respect to \( y \) is \[\frac{\partial w}{\partial y} = 3x^2y^2.\]

Step 5: Derivatives of x and y with respect to t

Compute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):\[\frac{dx}{dt} = 3t^2, \quad \frac{dy}{dt} = 2t.\]

Step 6: Substitute and simplify

Substitute all derivatives into the chain rule expression:\[\frac{dw}{dt} = \left(2xy^3\right) \cdot 3t^2 + \left(3x^2y^2\right) \cdot 2t.\]Substitute \( x = t^3 \) and \( y = t^2 \) into the equation:\[\frac{dw}{dt} = 2(t^3)(t^2)^3 \cdot 3t^2 + 3(t^3)^2(t^2)^2 \cdot 2t.\]Simplify the expression:\[\frac{dw}{dt} = 6t^{11} + 6t^{11} = 12t^{11}.\]

Step 7: Final Answer

The final expression for \( \frac{dw}{dt} \) in terms of \( t \) is \[\frac{dw}{dt} = 12t^{11}.\]

Key Concepts

Partial Derivatives

Partial derivatives are a core concept in multivariable calculus, especially when dealing with functions of multiple variables. In simple terms, a partial derivative measures how a multivariable function changes as one of its input variables changes, while all other variables are held constant.

For example, in the function given in the exercise, \[ w = x^2 y^3, \] we have two variables, \( x \) and \( y \). To find the partial derivative with respect to \( x \), denoted \( \frac{\partial w}{\partial x} \), we treat \( y \) as a constant. Thus, the partial derivative with respect to \( x \) becomes:

• Take the derivative of \( x^2 \) to get \( 2x \).
• Multiply by the constant \( y^3 \) resulting in \( 2xy^3 \).

Similarly, for the partial derivative with respect to \( y \), denoted \( \frac{\partial w}{\partial y} \), we treat \( x \) as constant. We calculate it as:

• Take the derivative of \( y^3 \) to get \( 3y^2 \).
• Multiply by the constant \( x^2 \), resulting in \( 3x^2y^2 \).

Understanding partial derivatives lays the groundwork for the chain rule in multivariable functions, which we explore in the next sections.

Calculus Differentiation

Calculus differentiation involves finding the rate at which a function changes at any given point. It's a fundamental concept used to determine things like velocity, acceleration, and changes in systems.

In this exercise, we use derivatives to compute the rate of change of \( w \) with respect to time \( t \). This involves differentiating functions of one variable such as \( x = t^3 \) and \( y = t^2 \). Here's how to differentiate these functions:

• For \( x = t^3 \), the derivative with respect to \( t \) is \( \frac{dx}{dt} = 3t^2 \), which tells us how \( x \) changes as \( t \) changes.
• For \( y = t^2 \), the derivative with respect to \( t \) is \( \frac{dy}{dt} = 2t \), indicating the rate of change of \( y \) with respect to \( t \).

These derivatives are important because they provide the necessary components to apply the chain rule for finding \( \frac{dw}{dt} \), the derivative of the given composite function with respect to time.

Function Composition

Function composition involves combining several functions into one, such that the output of one function becomes the input to another. This is a vital concept when dealing with more complex derivatives, like those involving the chain rule.

In the exercise, \( w = x^2 y^3 \) is composed of smaller functions of \( t \) through \( x = t^3 \) and \( y = t^2 \). Thus, \( w \) is expressed as a composite function of \( t \), wrapped within \( x \) and \( y \). When using these representations, each component function contributes to the overall rate of change:

• \( x(t) = t^3 \) contributes through its derivative \( \frac{dx}{dt} = 3t^2 \).
• \( y(t) = t^2 \) contributes via its derivative \( \frac{dy}{dt} = 2t \).

By substituting these into their respective places in the chain rule, they enable the efficient computation of \( \frac{dw}{dt} \), capturing the full interaction of changes as they are transmitted through composed functions. Function composition emphasizes how changes in simple functions propagate through to more complex relationships.