Problem 92 - Thomas Calculus

Question

In Exercises \(87-94\) , 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

Tangent line: \(y = \sqrt{3}x + 1 - \frac{\pi\sqrt{3}}{3}\). Value of \( \frac{d^2y}{dx^2} \) is \(-4\).

Step 1: Find the Derivatives

First, we need to find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). Given:\[ x = t - \sin t \quad \text{and} \quad y = 1 - \cos t \]For \( \frac{dx}{dt} \):\[ \frac{dx}{dt} = 1 - \cos t \]For \( \frac{dy}{dt} \):\[ \frac{dy}{dt} = \sin t \]

Step 2: Calculate the Slope of the Tangent

The slope of the tangent line is given by \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).Substituting in the derivatives we found:\[ \frac{dy}{dx} = \frac{\sin t}{1 - \cos t} \]Now, substitute \( t = \frac{\pi}{3} \) to find the slope at this point:\[ \frac{dy}{dx} \bigg|_{t = \frac{\pi}{3}} = \frac{\sin(\frac{\pi}{3})}{1 - \cos(\frac{\pi}{3})} = \frac{\frac{\sqrt{3}}{2}}{1 - \frac{1}{2}} = \sqrt{3} \]

Step 3: Find the Point on the Curve

Substitute \( t = \frac{\pi}{3} \) into the parametric equations to find the coordinates of the point.For \( x \):\[ x = \frac{\pi}{3} - \sin(\frac{\pi}{3}) = \frac{\pi}{3} - \frac{\sqrt{3}}{2} \]For \( y \):\[ y = 1 - \cos(\frac{\pi}{3}) = 1 - \frac{1}{2} = \frac{1}{2} \]The point on the curve is \( \left( \frac{\pi}{3} - \frac{\sqrt{3}}{2}, \frac{1}{2} \right) \).

Step 4: Write the Tangent Line Equation

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

Step 5: Find the Second Derivative

To find \( \frac{d^2y}{dx^2} \), differentiate \( \frac{dy}{dx} \) with respect to \( t \) and then divide by \( \frac{dx}{dt} \).The derivative \( \frac{dy}{dx} = \frac{\sin t}{1 - \cos t} \) is:\[ \frac{d}{dt}\left(\frac{\sin t}{1 - \cos t}\right) = \frac{(1-\cos t)\cos t - \sin^2 t}{(1 - \cos t)^2}\]Now, divide by \( \frac{dx}{dt} \):\[ \frac{d^2y}{dx^2} = \frac{\frac{(1-\cos t)\cos t - \sin^2 t}{(1 - \cos t)^2}}{1 - \cos t} = \frac{(1-\cos t)\cos t - \sin^2 t}{(1 - \cos t)^3}\]Substitute \( t = \frac{\pi}{3} \):\[ \frac{d^2y}{dx^2} = \frac{(1- \frac{1}{2})\times \frac{1}{2} - \left(\frac{\sqrt{3}}{2}\right)^2}{\left(\frac{1}{2}\right)^3} = -4 \]

Step 6: Conclusion

The equation of the tangent line at \( t = \frac{\pi}{3} \) is \( y = \sqrt{3}x + 1 - \frac{\pi\sqrt{3}}{3} \). The value of \( \frac{d^2y}{dx^2} \) at this point is \(-4\).

Key Concepts

Parametric Equations

Parametric equations are a handy way to describe a curve without needing a single equation relating x and y directly. Instead, we use a separate parameter, often denoted as \( t \), to express both \( x \) and \( y \) as individual functions. For example, in this exercise, we have \( x = t - \sin t \) and \( y = 1 - \cos t \). This not only provides a compact representation of curves but also makes it easier to analyze their motion and properties.

Parametric equations have several great advantages:

• They simplify the representation of complex shapes.
• They enable easy analysis of curves with respect to time or another parameter.
• They allow one to easily trace the direction of the curve.

When using these equations to find a tangent line or the second derivative, like in this exercise, it's crucial to remember that changes in \( t \) influence both \( x \) and \( y \). This interdependence makes differentiation a central tool.

Differentiation

Differentiation is the process of finding the derivative of a function, which gives us the rate at which a quantity changes. When working with parametric equations, we look at how both \( x \) and \( y \) change with respect to the parameter \( t \).

In this exercise, we compute:

• The derivative \( \frac{dx}{dt} \): Which gives us the rate of change of \( x \) with respect to \( t \) and is \( 1 - \cos t \).
• The derivative \( \frac{dy}{dt} \): Which provides the rate of change of \( y \) with respect to \( t \) and is \( \sin t \).
• The slope of the tangent line, \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \): Represents how \( y \) changes with respect to \( x \) at any point \( t \).

Understanding these derivatives helps bridge the relationship between the parameter \( t \) and how the curve behaves in the Cartesian plane. Differentiation helps us find not just tangents but also how steep those tangents are at any point.

Second Derivative

The second derivative, represented as \( \frac{d^2y}{dx^2} \), is an extension of our differentiation concepts, telling us how the rate of change of the slope itself changes. It's crucial for understanding the curvature of a graph, providing insight into its concavity and points of inflection.

To find \( \frac{d^2y}{dx^2} \), we differentiate \( \frac{dy}{dx} \) with respect to \( t \) again and then divide by \( \frac{dx}{dt} \).

In our exercise:

• We first differentiate \( \frac{dy}{dx} = \frac{\sin t}{1 - \cos t} \) with respect to \( t \) to get another expression, showing the rate of change of the slope.
• Then, dividing by \( \frac{dx}{dt} \), which is \( 1 - \cos t \), isolates the second derivative \( \frac{d^2y}{dx^2} \).
• At \( t = \frac{\pi}{3} \), we calculate it to be \(-4\), indicating a particular curve behavior at this point.

The second derivative hence provides crucial insights into the dynamic nature of curves, especially concerning their acceleration and bending behavior in this geometric space.