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Magazine Chapter 2: Problem 7
Find \(d y / d x\) if \(x=(t-2) /(t+2)\) and \(y=2 t /(t+1)\) for
\(-\infty
Short Answer
Expert verified
The derivative \( \frac{dy}{dx} = \frac{(t+2)^2}{2(t+1)^2} \) is always non-negative, meaning the curve of \( y \) with respect to \( x \) is non-decreasing.
Step by step solution
01
Understand the problem
We are given two parametric equations: 1. \( x = \frac{t-2}{t+2} \) 2. \( y = \frac{2t}{t+1} \) We need to find the derivative \( \frac{dy}{dx} \) and show that it is always non-negative.
02
Find \( \frac{dx}{dt} \)
First, differentiate \( x = \frac{t-2}{t+2} \) with respect to \( t \): \[ \frac{dx}{dt} = \frac{(t+2) \cdot 1 - (t-2) \cdot 1}{(t+2)^2} = \frac{4}{(t+2)^2} \]
03
Find \( \frac{dy}{dt} \)
Now differentiate \( y = \frac{2t}{t+1} \) with respect to \( t \): \[ \frac{dy}{dt} = \frac{(t+1) \cdot 2 - 2t \cdot 1}{(t+1)^2} = \frac{2}{(t+1)^2} \]
04
Use chain rule to find \( \frac{dy}{dx} \)
Using the chain rule, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \): \[ \frac{dy}{dx} = \frac{\frac{2}{(t+1)^2}}{\frac{4}{(t+2)^2}} = \frac{2}{(t+1)^2} \cdot \frac{(t+2)^2}{4} = \frac{(t+2)^2}{2(t+1)^2} \]
05
Simplify \( \frac{dy}{dx} \)
Simplify the expression: \[ \frac{(t+2)^2}{2(t+1)^2} \] Since squares of any real number are always non-negative, \( \frac{dy}{dx} \) is always non-negative.
06
Interpret the result
Since \( \frac{dy}{dx} \ge 0 \), the function \( y \) as a function of \( x \) is always non-decreasing. This means the curve of \( y \) with respect to \( x \) either increases or remains constant but never decreases.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
A derivative measures how a function changes as its input changes. In simpler terms, it's the rate at which one quantity changes with respect to another. When we talk about the derivative of a function, we're finding the slope of the function at any given point. For example, if we have a function \( y = f(x) \), then the derivative \( \frac{dy}{dx} \) gives us the rate of change of \( y \) with respect to \( x \). Derivatives are central in calculus and useful across various fields such as physics, engineering, and economics.
- Important Note: In parametric equations, we often need to find the derivative of one variable in terms of another which involves a two-step process of differentiation.
- Why Derivatives Matter: They help us understand the behavior of functions, like finding the maximum and minimum values, or understanding how the functions increase or decrease.
Chain Rule
The chain rule is a fundamental differentiation rule used when dealing with composite functions. It allows us to differentiate a composite function by differentiating the outer function and multiplying it by the derivative of the inner function. For a composite function \( y = f(g(x)) \), the chain rule states that:
\[ \frac{dy}{dx} = \frac{dy}{dg} \cdot \frac{dg}{dx} \]
In the context of parametric equations, the chain rule is especially useful. If we have parametric equations where both \(x\) and \(y\) are defined in terms of a third variable \(t\), like in our problem where:
\[ \frac{dy}{dx} = \frac{dy}{dg} \cdot \frac{dg}{dx} \]
In the context of parametric equations, the chain rule is especially useful. If we have parametric equations where both \(x\) and \(y\) are defined in terms of a third variable \(t\), like in our problem where:
- \(x = \frac{t-2}{t+2} \)
- \(y = \frac{2t}{t+1}\)
We first find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) separately. Then, we use the chain rule to find \( \frac{dy}{dx} \) by dividing \( \frac{dy}{dt} \) by \( \frac{dx}{dt}\). This two-step process allows us to find how \(y\) changes with respect to \(x\):
- Step 1: Differentiate \(x\) and \(y\) with respect to \(t\).
- Step 2: Apply the chain rule using \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \)
The chain rule is a bridge that links the rates of change of different functions.
Parametric Equations
Parametric equations allow us to describe a set of related quantities as functions of an independent variable, often denoted as \( t \). These equations are particularly useful for representing curves and trajectories where both the x and y coordinates depend on \( t \).
For the given problem, we have:
For the given problem, we have:
- \(x = \frac{t-2}{t+2}\)
- \(y = \frac{2t}{t+1}\)
Here, \( t \) is the independent parameter that describes both \( x \) and \( y \). To analyze how \( y \) changes with respect to \( x \), we apply differentiation techniques.
The process involves:
- 1. Understanding Parametric Representation: Recognize that both functions \( x \) and \( y \) depend on \(t\).
- 2. Finding Derivatives: Differentiate \( x \) and \( y \) with respect to \( t\) to get \(\frac{dx}{dt} \) and \(\frac{dy}{dt} \).
- 3. Applying the Chain Rule: Combine the derivatives to find \(\frac{dy}{dx} \).
Understanding parametric equations is key to analyzing complex curves and translating the relationship between different variables into a more manageable form using a common parameter.
