Problem 2 - Thomas Calculus

Question

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. $$ x=\sin 2 \pi t, \quad y=\cos 2 \pi t, \quad t=-1 / 6 $$

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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 \((-\sqrt{3}/2, 1/2)\).

Step 3: Differentiate Parametric Equations

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

Step 4: 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 \( \frac{dy}{dx} \). \( \frac{dy}{dx} = -\tan(-\pi/3) = \sqrt{3} \).

Step 5: 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 \( y - 1/2 = \sqrt{3}(x + \sqrt{3}/2) \).

Step 6: Simplify the Tangent Line Equation

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

Step 7: 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 \( t = -1/6 \), \( \frac{d}{dt}(\frac{dy}{dx}) = -2\pi (4/3) = -8\pi/3 \). \( \frac{d^{2}y}{dx^{2}} = \frac{-8\pi/3}{2\pi \cos(2\pi(-1/6))} = \frac{-8/3}{1/2} = -16/3 \).

Step 8: 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.

Key Concepts

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.