Chapter 12: Problem 1
In Problems 1-6, find dw/dt by using the Chain Rule. Express your final answer in terms of \(t\). $$ w=x^{2} y^{3} ; x=t^{3}, y=t^{2} $$
Short Answer
Expert verified
\( \frac{dw}{dt} = 12t^{11} \).
Step by step solution
01
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:\[w(t) = (t^3)^2 (t^2)^3\]
02
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.
03
Compute the partial derivative with respect to x
The partial derivative of \( w \) with respect to \( x \) is \[\frac{\partial w}{\partial x} = 2xy^3.\]
04
Compute the partial derivative with respect to y
The partial derivative of \( w \) with respect to \( y \) is \[\frac{\partial w}{\partial y} = 3x^2y^2.\]
05
Derivatives of x and y with respect to t
Compute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):\[\frac{dx}{dt} = 3t^2, \quad \frac{dy}{dt} = 2t.\]
06
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:\[\frac{dw}{dt} = 6t^{11} + 6t^{11} = 12t^{11}.\]
07
Final Answer
The final expression for \( \frac{dw}{dt} \) in terms of \( t \) is \[\frac{dw}{dt} = 12t^{11}.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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 \( 2xy^3 \).
- Take the derivative of \( y^3 \) to get \( 3y^2 \).
- Multiply by the constant \( x^2 \), resulting in \( 3x^2y^2 \).
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:
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 \( t \).
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:
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 \( \frac{dy}{dt} = 2t \).