Chapter 34: Problem 1
Given \(x=5 \theta-1\) and \(y=2 \theta(\theta-1)\), determine \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) in terms of \(\theta\).
Short Answer
Expert verified
\( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4\theta - 2}{5} \).
Step by step solution
01
Find \( \frac{\mathrm{d}x}{\mathrm{d}\theta} \)
Differentiate the expression for \( x \) with respect to \( \theta \). Since \( x = 5\theta - 1 \), the derivative is \( \frac{\mathrm{d}x}{\mathrm{d}\theta} = 5 \).
02
Find \( \frac{\mathrm{d}y}{\mathrm{d}\theta} \)
Differentiate the expression for \( y \) with respect to \( \theta \). Given \( y = 2\theta(\theta - 1) \), multiply \( 2\theta \) by \( \theta - 1 \) to expand: \( y = 2\theta^2 - 2\theta \). Now differentiate to get \( \frac{\mathrm{d}y}{\mathrm{d}\theta} = 4\theta - 2 \).
03
Use chain rule to find \( \frac{\mathrm{d}y}{\mathrm{d}x} \)
The chain rule states that \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}} \). From Steps 1 and 2, we have \( \frac{\mathrm{d}x}{\mathrm{d}\theta} = 5 \) and \( \frac{\mathrm{d}y}{\mathrm{d}\theta} = 4\theta - 2 \). Thus, \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4\theta - 2}{5} \).
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.
Differentiation
Differentiation is a fundamental concept in calculus, essential for understanding changes in functions. It helps determine how a function's output value changes relative to changes in its input value. In simple terms, differentiation measures the rate at which something varies. When we differentiate a function, we are essentially finding the slope of its graph, or how steep it is at any given point.
For a function expressed as dependent variable vs. an independent variable, such as \( y = f(x) \), its derivative \( \frac{dy}{dx} \) indicates the rate of change of \( y \) with respect to \( x \). In the given exercise, the expressions for \( x \) and \( y \) are in terms of \( \theta \), a parameter, and we differentiate these expressions with respect to \( \theta \) to find \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \).
Once the individual derivatives are found, they aid in solving for \( \frac{dy}{dx} \) using the parametric equations provided. This is particularly useful for analyzing motion and behavior where coordinates depend on another variable like time.
For a function expressed as dependent variable vs. an independent variable, such as \( y = f(x) \), its derivative \( \frac{dy}{dx} \) indicates the rate of change of \( y \) with respect to \( x \). In the given exercise, the expressions for \( x \) and \( y \) are in terms of \( \theta \), a parameter, and we differentiate these expressions with respect to \( \theta \) to find \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \).
Once the individual derivatives are found, they aid in solving for \( \frac{dy}{dx} \) using the parametric equations provided. This is particularly useful for analyzing motion and behavior where coordinates depend on another variable like time.
Chain Rule
The Chain Rule is a crucial tool in calculus used to differentiate composite functions – functions formed by combining two or more functions. Suppose you have a function \( z \) that depends on \( y \), and \( y \) in turn depends on \( x \). To find how \( z \) changes with \( x \), the Chain Rule states you multiply the rate of change of \( z \) with respect to \( y \) by the rate of change of \( y \) with respect to \( x \).
Mathematically, if \( z = g(y) \) and \( y = f(x) \), then the derivative is given by:\[\frac{dz}{dx} = \frac{dz}{dy} \times \frac{dy}{dx}\]In the parametric context of the exercise provided, the Chain Rule adapts to find \( \frac{dy}{dx} \) in terms of \( \theta \):
Mathematically, if \( z = g(y) \) and \( y = f(x) \), then the derivative is given by:\[\frac{dz}{dx} = \frac{dz}{dy} \times \frac{dy}{dx}\]In the parametric context of the exercise provided, the Chain Rule adapts to find \( \frac{dy}{dx} \) in terms of \( \theta \):
- First, find \( \frac{dy}{d\theta} \), the change in \( y \) per change in \( \theta \).
- Next, find \( \frac{dx}{d\theta} \), the change in \( x \) per change in \( \theta \).
- Finally, use the relationship \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \).
Parametric Equations
Parametric equations define a relationship where both dependent and independent variables are expressed in terms of a third variable, called the parameter. This approach is particularly handy for describing curves that do not pass the vertical line test, and hence cannot be represented by a single function of \( x \) or \( y \).
In the given exercise:
In the given exercise:
- \( x \) and \( y \) are both functions of \( \theta \), a parameter.
- The derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \) must be found initially to determine \( \frac{dy}{dx} \).
- This shows the rate of change of \( y \) with respect to \( x \) when both are linked through \( \theta \).