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Magazine Chapter 4: Problem 239
Find \(\frac{d z}{d t}\) using the chain rule where \(z=3 x^{2} y^{3}, x=t^{4}\) and \(y=t^{2}\)
Short Answer
Expert verified
The derivative is \(\frac{dz}{dt} = 42t^{13}\).
Step by step solution
01
Identify the functions involved
Given the function for \(z\), we have \(z = 3x^2y^3\). The problem also provides functions for \(x\) and \(y\) as functions of \(t\), specifically \(x = t^4\) and \(y = t^2\). We will use these to find \(\frac{dz}{dt}\).
02
Differentiate \(z\) with respect to \(x\) and \(y\)
Compute \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) using partial differentiation: 1. \(\frac{\partial z}{\partial x} = 6xy^3\).2. \(\frac{\partial z}{\partial y} = 9x^2y^2\).
03
Differentiate \(x\) and \(y\) with respect to \(t\)
Differentiate \(x\) and \(y\) with respect to \(t\): 1. \(\frac{dx}{dt} = 4t^3\).2. \(\frac{dy}{dt} = 2t\).
04
Apply the Chain Rule
Apply the chain rule to find \(\frac{dz}{dt}\), which is given by:\[\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}\]Substituting the values, we get:\[\frac{dz}{dt} = 6xy^3 \cdot 4t^3 + 9x^2y^2 \cdot 2t\]
05
Substitute \(x\) and \(y\) in terms of \(t\)
Replace \(x\) and \(y\) with their expressions in terms of \(t\):1. \(x = t^4\), so \(x^2 = (t^4)^2 = t^8\).2. \(y = t^2\), so \(y^3 = (t^2)^3 = t^6\) and \(y^2 = (t^2)^2 = t^4\).Substitute these in the chain rule expression:\[\frac{dz}{dt} = 6t^4(t^6) \cdot 4t^3 + 9(t^8)(t^4) \cdot 2t\]
06
Simplify the expression
Now, simplify the expression:\[\frac{dz}{dt} = 24t^{13} + 18t^{13} = 42t^{13}\]
07
Write the final answer
The derivative of \(z\) with respect to \(t\) is \(\frac{dz}{dt} = 42t^{13}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Differentiation
Partial differentiation is a crucial concept in calculus, especially when dealing with functions of multiple variables. It involves taking the derivative of a function with respect to one variable while keeping other variables constant. This allows us to understand how changes in one variable affect the function.
In our exercise, the function \( z = 3x^2y^3 \) is influenced by both \( x \) and \( y \). To find how \( z \) changes with respect to \( x \) or \( y \), we computed the partial derivatives \( \frac{\partial z}{\partial x} = 6xy^3 \) and \( \frac{\partial z}{\partial y} = 9x^2y^2 \).
These partial derivatives were essential steps in applying the chain rule, as they describe the sensitivity of \( z \) to changes in its variables \( x \) and \( y \).
In our exercise, the function \( z = 3x^2y^3 \) is influenced by both \( x \) and \( y \). To find how \( z \) changes with respect to \( x \) or \( y \), we computed the partial derivatives \( \frac{\partial z}{\partial x} = 6xy^3 \) and \( \frac{\partial z}{\partial y} = 9x^2y^2 \).
These partial derivatives were essential steps in applying the chain rule, as they describe the sensitivity of \( z \) to changes in its variables \( x \) and \( y \).
Function Composition
Function composition is a process that applies one function to the results of another. In simpler terms, it involves combining two or more functions into a single operation.
In our given problem, we have the functions \( x = t^4 \) and \( y = t^2 \). We substitute these into the function for \( z \), which is \( z = 3x^2y^3 \). This means \( x \) and \( y \) are composed with the variable \( t \), making \( z \) depend on \( t \).
By doing this, we're able to find derivatives with respect to a singular variable \( t \), greatly simplifying the differentiation process with the help of the chain rule. This interlinking of different functions into a single input-output relationship is a powerful tool in calculus.
In our given problem, we have the functions \( x = t^4 \) and \( y = t^2 \). We substitute these into the function for \( z \), which is \( z = 3x^2y^3 \). This means \( x \) and \( y \) are composed with the variable \( t \), making \( z \) depend on \( t \).
By doing this, we're able to find derivatives with respect to a singular variable \( t \), greatly simplifying the differentiation process with the help of the chain rule. This interlinking of different functions into a single input-output relationship is a powerful tool in calculus.
Calculus Derivatives
Derivatives are fundamental in calculus, representing the rate of change of a function with respect to a variable. They are used to understand the behavior of functions and model dynamic systems.
In this exercise, we computed several derivatives to eventually determine \( \frac{dz}{dt} \).
In this exercise, we computed several derivatives to eventually determine \( \frac{dz}{dt} \).
- First, partial derivatives \( \frac{\partial z}{\partial x} = 6xy^3 \) and \( \frac{\partial z}{\partial y} = 9x^2y^2 \) were found to see how \( z \) changes with \( x \) and \( y \).
- Next, derivatives of \( x \) and \( y \) with respect to \( t \) were calculated: \( \frac{dx}{dt} = 4t^3 \) and \( \frac{dy}{dt} = 2t \).
- Finally, using the chain rule, \( \frac{dz}{dt} \) was determined, resulting in \( \frac{dz}{dt} = 42t^{13} \).
