/*! 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 44 The curve with parametric equati... [FREE SOLUTION] | 91Ó°ÊÓ

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The curve with parametric equations $$x=t, \quad y=1-\cos t, \quad 0 \leq t \leq 2 \pi$$ is called a sinusoid and is shown in the accompanying figure. Find the point \((x, y)\) where the slope of the tangent line is a. largest. \(\quad\) b. smallest.

Short Answer

Expert verified
Largest slope at \( \left( \frac{\pi}{2}, 1 \right) \); smallest slope at \( \left( \frac{3\pi}{2}, 1 \right) \).

Step by step solution

01

Differentiate Parametric Equations

To find the slope of the tangent line for the parametric equations, we first differentiate both equations with respect to the parameter \( t \). We have \( x = t \), so \( \frac{dx}{dt} = 1 \). For \( y = 1 - \cos t \), the derivative is \( \frac{dy}{dt} = \sin t \).
02

Find Slope Expression

The slope of the tangent line for parametric curves is given by \( \frac{dy/dt}{dx/dt} \). Thus, the slope \( m \) is \( m = \frac{\sin t}{1} = \sin t \).
03

Determine Maximum and Minimum Slope

Since the slope \( \sin t \) varies between -1 and 1 (the range of the sine function), the slope of the tangent line is also maximized and minimized at these values.
04

Identify Points for Maximum Slope

The maximum slope occurs at \( \sin t = 1 \), which happens at \( t = \frac{\pi}{2} \). Substitute \( t \) into the parametric equations: \( x = \frac{\pi}{2} \) and \( y = 1 - \cos\left(\frac{\pi}{2}\right) = 1 - 0 = 1 \). Therefore, the point \( (x, y) \) is \( \left( \frac{\pi}{2}, 1 \right) \).
05

Identify Points for Minimum Slope

The minimum slope occurs at \( \sin t = -1 \), which happens at \( t = \frac{3\pi}{2} \). Substitute this \( t \) into the parametric equations: \( x = \frac{3\pi}{2} \) and \( y = 1 - \cos\left(\frac{3\pi}{2}\right) = 1 - 0 = 1 \). Therefore, the point \( (x, y) \) is \( \left( \frac{3\pi}{2}, 1 \right) \).

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Key Concepts

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

Tangent Line
Understanding the concept of a tangent line is essential when analyzing curves. A tangent line touches a curve at exactly one point, moving in the same direction as the curve at that particular point. Its slope represents the instantaneous rate of change of the curve. In the exercise, the tangent line's slope is given by the expression \( \frac{dy/dt}{dx/dt} \). For parametric equations like the sinusoid, which uses \( x = t \) and \( y = 1 - \cos t \), the tangent line's slope simplifies to \( \sin t \). This is the derivative of \( y \) concerning \( x \) in terms of the parameter \( t \). The tangent line at any point on a curve is important as it provides information about the behavior of the curve and points out where certain events like peaks or troughs occur. When the slope is zero, the tangent line is horizontal, indicating a local maximum or minimum in terms of \( y \).
Parametric Differentiation
Parametric differentiation allows us to explore and describe curves defined by parametric equations. It involves differentiating each parametric equation separately concerning the parameter—here, \( t \). For the curve given in the exercise, we differentiate \( x = t \) to get \( \frac{dx}{dt} = 1 \). The corresponding \( y \) equation, \( 1 - \cos t \), differentiates to \( \sin t \). The resulting derivative \( \frac{dy}{dt} \) describes how the \( y \)-value changes as \( t \) changes. By dividing these derivatives, \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \), we obtain the slope of the tangent line, which is the rate at which \( y \) changes with respect to \( x \). This slope is crucial for analyzing the curve's behavior and for solving the exercise's main question on where the slope is maximal and minimal.
Parametric Curve Analysis
Parametric curve analysis is a powerful approach to examine the properties of a given curve. Analyzing parametric equations helps us find key features, such as where the slope of the tangent line is the largest or smallest, as seen in the exercise with the sinusoid curve defined by \( x = t \) and \( y = 1 - \cos t \). This procedure involves evaluating the sine function, which controls the slope of our curve. Since \( \sin t \) varies from \(-1\) to \(1\), the points of maximum and minimum slope occur at \( t = \frac{\pi}{2} \) (maximum) and \( t = \frac{3\pi}{2} \) (minimum). By substituting these values of \( t \) back into the parametric equations, we identify the specific points on the curve \( \left( \frac{\pi}{2}, 1 \right) \) and \( \left( \frac{3\pi}{2}, 1 \right) \). Analyzing these points provides insights into the nature and shape of the curve formed by the sinusoid equations.

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