Problem 11 - 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. $$ x=t-\sin t, \quad y=1-\cos t, \quad t=\pi / 3 $$

Step-by-Step Solution

Answer

The tangent line is \( y = \sqrt{3}x - \sqrt{3}(\frac{\pi}{3} - \frac{\sqrt{3}}{2}) + \frac{1}{2} \), and \( \frac{d^2 y}{d x^2} = -2 \).

Step 1: Determine the point on the curve

First, we will find the coordinates of the point on the curve given the parameter value of \(t = \frac{\pi}{3}\). Calculate \(x\) and \(y\) using the parametric equations.\[ x = t - \sin t = \frac{\pi}{3} - \sin \frac{\pi}{3} = \frac{\pi}{3} - \frac{\sqrt{3}}{2} \]\[ y = 1 - \cos t = 1 - \cos \frac{\pi}{3} = 1 - \frac{1}{2} = \frac{1}{2} \]Thus, the point on the curve is \(\left( \frac{\pi}{3} - \frac{\sqrt{3}}{2}, \frac{1}{2} \right)\).

Step 2: Find the slope of the tangent line

To find the slope \( \frac{dy}{dx} \) of the tangent line, determine \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) first, then compute the derivative using the formula \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).\[ \frac{dx}{dt} = 1 - \cos t \]\[ \frac{dy}{dt} = \sin t \]At \( t = \frac{\pi}{3} \),\[ \frac{dx}{dt} = 1 - \cos \frac{\pi}{3} = 1 - \frac{1}{2} = \frac{1}{2} \]\[ \frac{dy}{dt} = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \]Therefore, \[ \frac{dy}{dx} = \frac{\sin t}{1 - \cos t} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \].The slope of the tangent line is \( \sqrt{3} \).

Step 3: Write the equation of the tangent line

With the slope \( m = \sqrt{3} \) and the point \( \left( \frac{\pi}{3} - \frac{\sqrt{3}}{2}, \frac{1}{2} \right) \), use point-slope form to write the equation of the tangent:\[ y - y_1 = m(x - x_1) \]Substitute \( y_1 = \frac{1}{2} \), \( m = \sqrt{3} \), and \( x_1 = \frac{\pi}{3} - \frac{\sqrt{3}}{2} \):\[ y - \frac{1}{2} = \sqrt{3}(x - (\frac{\pi}{3} - \frac{\sqrt{3}}{2})) \]Hence, the equation of the tangent line is:\[ y = \sqrt{3}x - \sqrt{3} \left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} \right) + \frac{1}{2} \].

Step 4: Compute \( \frac{d^2y}{dx^2} \)

For the second derivative, calculate\( \frac{d^2y}{dx^2} = \frac{d}{dt} \left( \frac{dy/dt}{dx/dt} \right) / \frac{dx}{dt} \).We first find \( \frac{d}{dt} \left( \frac{dy/dt}{dx/dt} \right) \) using the derivative chain rule. Using\( u = \sin t \) and\( v = 1 - \cos t \):\( \frac{d}{dt}(\frac{dy/dt}{dx/dt}) = \frac{(v \cos t - u \sin t)}{v^2} \)Now calculate \( \frac{d^2y}{dx^2} \) at \( t = \frac{\pi}{3} \).\( \cos t = \frac{1}{2}, \sin t = \frac{\sqrt{3}}{2} \):\( \frac{(\frac{1}{2} \cdot \cos \frac{\pi}{3} - \frac{\sqrt{3}}{2} \cdot \sin \frac{\pi}{3})}{(\frac{1}{2})^2} \)Computing,\( = -2 \).Thus, \( \frac{d^2y}{dx^2} = -2 \).

Key Concepts

Parametric Equations

Parametric equations are a way to describe a curve in terms of a parameter, typically denoted as \( t \). Instead of expressing \( y \) solely as a function of \( x \), we have both \( x \) and \( y \) defined in terms of \( t \). This method is extremely useful for capturing complex curves that might be difficult to express with a simple function. In our example, the equations provided are \( x = t - \sin t \) and \( y = 1 - \cos t \). These two equations are used to describe every point on the curve as \( t \) varies.

By substituting specific values of \( t \), you calculate corresponding points \((x, y)\) on the parametric curve. This allows for flexibility in representing intricate paths, which is beneficial in fields like physics or computer graphics. Using a parametric form provides insights into the motion or pathway described by changing \( t \), far clearer than just a standard \( y = f(x) \) format.

First Derivative

The first derivative of a parametric equation deals with finding the slope of the tangent to the parametric curve at any given point. This is represented by \( \frac{dy}{dx} \), the rate of change of \( y \) with respect to \( x \). To compute this, you need to find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \), and their ratio gives \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).

In our exercise,\( \frac{dx}{dt} = 1 - \cos t \) and \( \frac{dy}{dt} = \sin t \). When evaluated at \( t = \frac{\pi}{3} \), \( \frac{dy}{dx} = \sqrt{3} \), indicating the slope of the tangent at that point. This slope describes how steep the line is, providing crucial information for describing linear behavior at that curve point.

Understanding this step supports constructing tangent lines, which are pivotal in calculus to approximate values and analyze curves.

Second Derivative

The second derivative \( \frac{d^2y}{dx^2} \) for a parametric curve involves understanding the curvature, or how the slope is changing, which can be important in analyzing concavity or rectifying curves. To find this, you first determine \( \frac{d}{dt} \left( \frac{dy/dt}{dx/dt} \right) \) and divide by \( \frac{dx}{dt} \).

In the provided solution: Define \( u = \sin t \) and \( v = 1 - \cos t \), then use the differentiation rule to find \( \frac{d}{dt} \left( \frac{dy/dt}{dx/dt} \right) = \frac{(v \cos t - u \sin t)}{v^2} \). Evaluating at \( t = \frac{\pi}{3} \) gives \( \frac{d^2y}{dx^2} = -2 \). The negative value signifies that the curve is concave downward at this point.

This derivative builds on the first, providing a deeper insight into the behavior of the curve, revealing acceleration patterns or even potential points of inflection.