Chapter 13: Problem 1
Use an appropriate form of the chain rule to find \(d z / d t\) $$ z=3 x^{2} y^{3} ; x=t^{4}, y=t^{2} $$
Short Answer
Expert verified
\( \frac{dz}{dt} = 42t^{13} \)
Step by step solution
01
Identify the Function Relationships
First, understand that the function we are dealing with is given by \( z = 3x^2y^3 \), where \( x = t^4 \) and \( y = t^2 \). Here, \( z \) is expressed in terms of \( x \) and \( y \), but both \( x \) and \( y \) are themselves functions of \( t \).
02
Apply the Chain Rule
We use the chain rule to find \( \frac{dz}{dt} \). The chain rule specifies that \( \frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} \). Let's calculate each of these parts.
03
Calculate Partial Derivatives of z
Compute the partial derivatives: \( \frac{\partial z}{\partial x} = 6xy^3 \)\( \frac{\partial z}{\partial y} = 9x^2y^2 \).
04
Compute Derivatives of x and y with respect to t
Now calculate the derivatives of \( x \) and \( y \) with respect to \( t \): \( \frac{dx}{dt} = 4t^3 \) and \( \frac{dy}{dt} = 2t \).
05
Substitute and Simplify
Substitute the partial derivatives and the derivatives with respect to \( t \) into the chain rule formula: \( \frac{dz}{dt} = (6xy^3)(4t^3) + (9x^2y^2)(2t) \). Substitute \( x = t^4 \) and \( y = t^2 \): \( \frac{dz}{dt} = 6(t^4)(t^2)^3(4t^3) + 9(t^4)^2(t^2)^2(2t) \).
06
Simplify the Expression
Calculate each term: First term: \( 6t^4(t^6)(4t^3) = 24t^{13} \) Second term: \( 9(t^8)(t^4)(2t) = 18t^{13} \). Combine them to find:\( \frac{dz}{dt} = 24t^{13} + 18t^{13} = 42t^{13} \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are a fundamental concept in multivariable calculus. They enable us to understand how a function changes with respect to one variable while keeping the others constant.In the exercise, we dealt with the function \( z = 3x^2y^3 \), where \( z \) depends on two variables, \( x \) and \( y \). A partial derivative of \( z \) with respect to \( x \), denoted as \( \frac{\partial z}{\partial x} \), is calculated by treating \( y \) as a constant. Similarly, \( \frac{\partial z}{\partial y} \) is found by treating \( x \) as constant.
- For \( x \), the partial derivative is \( \frac{\partial z}{\partial x} = 6xy^3 \).
- For \( y \), we get \( \frac{\partial z}{\partial y} = 9x^2y^2 \).
Functions of a Parameter
In this particular problem, we explored functions of a parameter through the variables \( x \) and \( y \) as functions of \( t \).These types of functions let us track how a change in one parameter affects multiple related variables. Here, \( x = t^4 \) and \( y = t^2 \) illustrate how the parameters were expressed in terms of \( t \).
- \( x \) as a function of \( t \) means \( x \)'s rate of change is expressed by \( \frac{dx}{dt} = 4t^3 \).
- \( y \) being \( t \)-dependent has its rate of change expressed by \( \frac{dy}{dt} = 2t \).
Multivariable Calculus
Multivariable calculus extends calculus concepts into higher dimensions, allowing us to work with functions of multiple variables.In this exercise, we used the chain rule, a pivotal method in multivariable calculus. The chain rule helps calculate derivatives of composite functions. By finding \( \frac{dz}{dt} \) through the relation \( \frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} \), we combine the influence of multiple variables (\( x \) and \( y \) through \( t \)).Here's a summary of applying this:
- Identify function dependencies and compute partial derivatives.
- Calculate the rate of change for \( x \) and \( y \) as functions of \( t \).
- Substitute these into the chain rule formula and solve.