/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 Find \(d y / d x\) and \(d^{2} y... [FREE SOLUTION] | 91Ó°ÊÓ

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Find \(d y / d x\) and \(d^{2} y / d x^{2}\) at the given point without eliminating the parameter. $$ x=\theta+\cos \theta, \quad y=1+\sin \theta ; \theta=\pi / 6 $$

Short Answer

Expert verified
\( \frac{dy}{dx} = \sqrt{3} \), \( \frac{d^2y}{dx^2} = 1 \) at \( \theta = \pi/6 \).

Step by step solution

01

Derive Expressions for dx/dθ and dy/dθ

To find the derivatives \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) without eliminating the parameter, we first find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \). Given \( x = \theta + \cos \theta \), differentiate to get \( \frac{dx}{d\theta} = 1 - \sin \theta \). Similarly, for \( y = 1 + \sin \theta \), we have \( \frac{dy}{d\theta} = \cos \theta \).
02

Calculate dy/dx Using Chain Rule

Apply the chain rule to find \( \frac{dy}{dx} \), which is \( \frac{dy/d\theta}{dx/d\theta} \). This leads to \( \frac{dy}{dx} = \frac{\cos \theta}{1 - \sin \theta} \).
03

Substitute θ = π/6 into dy/dx

Substitute \( \theta = \frac{\pi}{6} \) into the expression for \( \frac{dy}{dx} \). With \( \sin(\frac{\pi}{6}) = \frac{1}{2} \) and \( \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \), this gives \( \frac{dy}{dx} = \frac{\frac{\sqrt{3}}{2}}{1 - \frac{1}{2}} = \frac{\sqrt{3}}{1} = \sqrt{3} \).
04

Differentiate dy/dθ Again to Find d²y/dx²

First, differentiate \( \frac{dy}{d\theta} = \cos \theta \) to get \( \frac{d^2y}{d\theta^2} = -\sin \theta \). Then, use the quotient rule to differentiate \( \frac{dy}{dx} = \frac{\cos \theta}{1 - \sin \theta} \) with respect to \( \theta \): \( \frac{d}{d\theta}(\frac{dy}{dx}) = \frac{(-\sin \theta)(1 - \sin \theta) + \cos^2 \theta}{(1 - \sin \theta)^2} \).
05

Evaluate d²y/dx² at θ = π/6

Substitute \( \theta = \frac{\pi}{6} \) into the second derivative using \( \sin(\frac{\pi}{6}) = \frac{1}{2} \) and \( \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \). Simplifying gives \( \frac{d^2y}{dx^2} = \frac{-\frac{1}{2}(\frac{1}{2}) + \left(\frac{\sqrt{3}}{2}\right)^2}{(\frac{1}{2})^2} = 1 \).

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Key Concepts

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

Chain Rule
The chain rule is a fundamental technique in calculus used for differentiating composite functions. A composite function is simply a function that has another function inside it, like a nested process.
When you use the chain rule, you are essentially unpacking and differentiating these layers.
To understand the chain rule, consider two functions: \( f(u) \) and \( u(x) \). The chain rule states that the derivative of the composite function \( f(u(x)) \) with respect to \( x \) is the derivative of \( f \) with respect to \( u \) multiplied by the derivative of \( u \) with respect to \( x \). In mathematical terms:
  • \( \frac{d}{dx}[f(u(x))] = \frac{df}{du} \cdot \frac{du}{dx} \)
In the context of parametric differentiation, like in our problem, we use the chain rule to express \( \frac{dy}{dx} \) in terms of \( \theta \) by dividing \( \frac{dy}{d\theta} \) by \( \frac{dx}{d\theta} \). This allows us to differentiate a parametric equation without needing to explicitly solve for \( y \) as a function of \( x \).
Understanding the chain rule helps in managing more complex functions that vary with parameters such as \( \theta \), making it an indispensable tool in calculus.
Quotient Rule
The quotient rule is used when one function is divided by another and we need to find the derivative of their ratio.
It's especially helpful when dealing with parametric equations that yield derivatives expressed as quotients.
To apply the quotient rule, if you have two functions \( u(\theta) \) and \( v(\theta) \) such that \( z(\theta) = \frac{u(\theta)}{v(\theta)} \), the derivative \( \frac{dz}{d\theta} \) is given by:
  • \( \frac{d}{d\theta} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{d\theta} - u \cdot \frac{dv}{d\theta}}{v^2} \)
In our example, for finding the second derivative \( \frac{d^2y}{dx^2} \), the quotient rule helps us differentiate \( \frac{dy/d\theta}{dx/d\theta} \), which involves some complex interactions between trigonometric functions.
This process not only requires differentiating each top and bottom function but then recombining them according to the rule.
This makes the quotient rule a vital step in solving for derivatives that can't be isolated to a simple function form.
Understanding each step where the minus sign and the multiplication affect the terms helps prevent common mistakes.
Second Derivative
The second derivative, \( \frac{d^2y}{dx^2} \), gives us insight into the concavity of a function and can indicate points of inflection.
It’s a progression from the first derivative, \( \frac{dy}{dx} \), showing how the rate of change itself changes from one moment to the next.
In essence, it's the derivative of the derivative.When handling parametric equations, finding the second derivative requires using both the chain rule and the quotient rule expertly.
First, differentiate the first derivative again concerning the parameter \( \theta \).
Then adjust by the derivative of \( \frac{dx}{d\theta} \) squared, as noted in our problem.For example, starting from \( \frac{dy}{dx} = \frac{cos \theta}{1 - sin \theta} \), you differentiate again, applying rules carefully to ensure each term is accounted for its role in constructing \( \frac{d^2y}{dx^2} \). This resulting expression captures how momentum in \( y \) with respect to \( x \) evolves, making it a powerful analytical tool to describe motion and acceleration characteristics in calculus.
Thus, the second derivative enriches our understanding beyond rate of change, digging into deeper facets of behavior of curves.

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