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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 \( \frac{dy}{dx} \geq 0 \) implies the curve of \( y(x) \) is non-decreasing for all \( t \).

Step by step solution

01

Expressing the coordinates

Given the parametric equations: \( x = \frac{t-2}{t+2} \) and \( y = \frac{2t}{t+1} \).
02

Find dx/dt

Differentiate \( x \) with respect to \( t \): \[ x = \frac{t-2}{t+2} \]\( \frac{dx}{dt} = \frac{(t+2) \cdot 1 - (t-2) \cdot 1}{(t+2)^2} = \frac{t+2 - t + 2}{(t+2)^2} = \frac{4}{(t+2)^2} \).
03

Find dy/dt

Differentiate \( y \) with respect to \( t \): \[ y = \frac{2t}{t+1} \]\( \frac{dy}{dt} = \frac{(t+1) \cdot 2 - 2t \cdot 1}{(t+1)^2} = \frac{2t + 2 - 2t}{(t+1)^2} = \frac{2}{(t+1)^2} \).
04

Calculate dy/dx

Use 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

Show dy/dx is non-negative

Since \( \frac{(t+2)^2}{2(t+1)^2} \) involves squares in the numerator and denominator, it is always non-negative for all real \( t \). Thus, \( \frac{dy}{dx} \geq 0 \) for all \( t \).
06

Use the result to sketch y as a function of x

From the exploration above, \( \frac{dy}{dx} \geq 0 \) means that the function \( y(x) \) is always increasing or remains constant for the range of \( t \). This results in a non-decreasing curve when plotted.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parametric Equations
Parametric equations are a way to define a curve using a set of equations. In the given problem, the coordinates of the curve are described by two separate equations: one for x in terms of t, and another for y in terms of t. This representation allows us to express the relationship between x and y indirectly. For example, with the equations given: \( x = \frac{t-2}{t+2} \) and \( y = \frac{2t}{t+1} \), we can describe a curve that may be more challenging to express with a single y = f(x) function. This method is particularly useful when the curve has complex behavior that is difficult to capture with conventional functions.
Differentiation
Differentiation is the process of finding the rate at which a function changes at any point. In this problem, we need to differentiate x and y with respect to the parameter t, which means we'll find the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). For example, differentiating \( x = \frac{t-2}{t+2} \) with respect to t, involves applying the quotient rule, yielding: \( \frac{dx}{dt} = \frac{4}{(t+2)^2} \). Similarly, differentiating \( y = \frac{2t}{t+1} \) produces: \( \frac{dy}{dt} = \frac{2}{(t+1)^2} \). These derivatives describe how x and y change with respect to t.
Chain Rule
The chain rule in calculus is a formula for computing the derivative of the composition of two or more functions. In our scenario, once we've found \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), we use the chain rule to find \( \frac{dy}{dx} \). The chain rule states that \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). Plugging our expressions for \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) into this formula, we get: \( \frac{dy}{dx} = \frac{\frac{2}{(t+1)^2}}{\frac{4}{(t+2)^2}} = \frac{(t+2)^2}{2(t+1)^2} \). This tells us how y changes with respect to x.
Function Sketching
Function sketching involves drawing the graph of a function based on its key features such as slope, intercepts, and behavior at infinity. Given that \( \frac{dy}{dx} \geq 0 \) indicating the slope is always non-negative, the curve described by the parametric equations is non-decreasing. That means as x increases, y either increases or stays constant, it never decreases. This insight is crucial for accurately sketching the shape and behavior of the curve. To sketch the curve, plot points for various t-values, observe how x and y change, and then connect these points to visualize the overall shape of the function y(x).

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Most popular questions from this chapter

Find the first derivatives of (a) \(x /(a+x)^{2}\), (b) \(x /(1-x)^{1 / 2}\), (c) \(\tan x\), as \(\sin x / \cos x\), (d) \(\left(3 x^{2}+2 x+1\right) /\left(8 x^{2}-4 x+2\right)\)

If \(y=\exp \left(-x^{2}\right)\), show that \(d y / d x=-2 x y\) and hence, by applying Leibnitz' theorem, prove that for \(n \geq 1\) $$ y^{(n+1)}+2 x y^{(n)}+2 n y^{(n-1)}=0 $$

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The curve whose equation is \(x^{2 / 3}+y^{2 / 3}=a^{2 / 3}\) for positive \(x\) and \(y\) and which is completed by its symmetric reflections in both axes is known as an astroid. Sketch it and show that its radius of curvature in the first quadrant is \(3(a x y)^{1 / 3}\).

Use the mean value theorem to establish bounds in the following cases. (a) For \(-\ln (1-y)\), by considering \(\ln x\) in the range \(0<1-y
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