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Chapter 11: Problem 2

Find an equation for the line tangent to the curve at the point defined by the given value of \(t\) . Also, find the value of \(d^{2} y / d x^{2}\) at this point. $$ x=\sin 2 \pi t, \quad y=\cos 2 \pi t, \quad t=-1 / 6 $$

Short Answer

Expert verified
The tangent line: \( y = \sqrt{3}x + 2 \). \( \frac{d^2y}{dx^2} = -16/3 \).

Step by step solution

01

Identify the Parametric Equations

The problem provides the parametric equations \( x = \sin(2\pi t) \) and \( y = \cos(2\pi t) \) with the value of \( t = -1/6 \). Our goal is to find the tangent line to the curve at this value of \( t \) and additionally calculate \( \frac{d^{2}y}{dx^{2}} \) at this point.
02

Find the Point on the Curve

Calculate \( x \) and \( y \) using the given \( t \). For \( t = -1/6 \), \( x = \sin(2\pi(-1/6)) = \sin(-\pi/3) = -\sqrt{3}/2 \) and \( y = \cos(2\pi(-1/6)) = \cos(-\pi/3) = 1/2 \). Therefore, the point on the curve is \((-\sqrt{3}/2, 1/2)\).
03

Differentiate Parametric Equations

Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). \( \frac{dx}{dt} = 2\pi \cos(2\pi t) \) and \( \frac{dy}{dt} = -2\pi \sin(2\pi t) \).
04

Calculate the Derivative \( \frac{dy}{dx} \)

The derivative \( \frac{dy}{dx} \) is given by \( \frac{dy}{dt} / \frac{dx}{dt} = \frac{-2\pi \sin(2\pi t)}{2\pi \cos(2\pi t)} = -\tan(2\pi t) \). Plug in \( t = -1/6 \) to find \( \frac{dy}{dx} \). \( \frac{dy}{dx} = -\tan(-\pi/3) = \sqrt{3} \).
05

Equation of Tangent Line

The equation of the tangent line can be formed using the point-slope form: \( y - y_1 = m(x - x_1) \), where \( m = \sqrt{3} \) and \((x_1, y_1) = (-\sqrt{3}/2, 1/2)\). Thus, the equation is \( y - 1/2 = \sqrt{3}(x + \sqrt{3}/2) \).
06

Simplify the Tangent Line Equation

Simplifying gives \( y = \sqrt{3}x + 3/2 + 1/2 \). Thus, \( y = \sqrt{3}x + 2 \).
07

Find Second Derivative \( \frac{d^{2}y}{dx^{2}} \)

We need \( \frac{d^{2}y}{dx^{2}} = \frac{d}{dt}(\frac{dy}{dx}) / \frac{dx}{dt} \). \( \frac{d}{dt}(\frac{dy}{dx}) = \frac{d}{dt}(-\tan(2\pi t)) = -2\pi \sec^{2}(2\pi t) \). At \( t = -1/6 \), \( \frac{d}{dt}(\frac{dy}{dx}) = -2\pi (4/3) = -8\pi/3 \). \( \frac{d^{2}y}{dx^{2}} = \frac{-8\pi/3}{2\pi \cos(2\pi(-1/6))} = \frac{-8/3}{1/2} = -16/3 \).
08

Verify Calculations

Review the computed values for any possible mistakes to ensure accuracy. The tangent line equation and second derivative check out with all calculations.

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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 of defining a curve by expressing its coordinates as functions of a common variable, typically denoted as \( t \). In our example, we have:
  • \( x = \sin(2\pi t) \)
  • \( y = \cos(2\pi t) \)
The parameter \( t \) can be thought of as a time-like variable that allows us to trace the path of the curve in a specified manner. Here, \( t = -1/6 \) helps to locate a specific point on the curve, \((-\sqrt{3}/2, 1/2)\). By adjusting \( t \), we can explore different points on the curve and understand its shape. This method is helpful for describing and manipulating curves in a precise manner.
Derivative Calculation
Calculating the derivative in parametric form involves finding the rates of change of \( x \) and \( y \) with respect to \( t \), denoted by \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) respectively. For our equations:
  • \( \frac{dx}{dt} = 2\pi \cos(2\pi t) \)
  • \( \frac{dy}{dt} = -2\pi \sin(2\pi t) \)
The derivative \( \frac{dy}{dx} \), representing the slope of the tangent to the curve, is calculated as \( \frac{dy}{dt} \div \frac{dx}{dt} \). By evaluating at \( t = -1/6 \):
  • \( \frac{dy}{dx} = -\tan(2\pi t) \)
  • At \( t = -1/6 \), \( \frac{dy}{dx} = \sqrt{3} \)
This slope is crucial to forming the tangent line equation: \( y - y_1 = m(x - x_1) \). Here, \( m \) (the slope) is \( \sqrt{3} \), while the coordinates \((x_1, y_1)\) are \((-\sqrt{3}/2, 1/2)\).
Second Derivative
The second derivative gives us information about the concavity or curvature of the curve at a given point. To find \( \frac{d^2y}{dx^2} \) in parametric form, we differentiate \( \frac{dy}{dx} \) with respect to \( t \), and then divide by \( \frac{dx}{dt} \). Here's a closer look at the process:
  • \( \frac{d}{dt}(-\tan(2\pi t)) = -2\pi \sec^2(2\pi t) \)
At \( t = -1/6 \), compute \( \frac{d}{dt}(\frac{dy}{dx}) = -2\pi \cdot \frac{4}{3} = -\frac{8\pi}{3} \). Finally:
  • \( \frac{d^2y}{dx^2} = \frac{-\frac{8\pi}{3}}{2\pi \cos(2\pi(-1/6))} = \frac{-8/3}{1/2} = -\frac{16}{3} \)
Understanding this calculation is valuable when analyzing the more subtle dynamics of a curve, such as how it bends and twists in the plane.

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Most popular questions from this chapter

The curve with parametric equations $$ x=(1+2 \sin \theta) \cos \theta, \quad y=(1+2 \sin \theta) \sin \theta $$ is called a limacon and is shown in the accompanying figure. Find the points \((x, y)\) and the slopes of the tangent lines at these points for $$ \text { a. }\theta=0 . \quad \text { b. } \theta=\pi / 2, \quad \text { c. } \theta=4 \pi / 3 $$

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