Problem 14 - Essential Calculus

Question

Find \(d y / d x\) and \(d^{2} y / d x^{2},\) and find the slope and concavity (if possible) at the given value of the parameter. $$ x=\theta-\sin \theta, y=1-\cos \theta \quad \theta=\pi $$

Step-by-Step Solution

Answer

The slope at \( \theta = \pi \) is -1 and at this point the curve is not concave.

Step 1: Differentiate the given functions

Differentiate \( x = \theta - \sin\theta \) and \( y = 1 - \cos\theta \) with respect to \( \theta \). This will yield \( dx/d\theta = 1 - \cos\theta \) and \( dy/d\theta = \sin\theta \).

Step 2: Find the first derivative dy/dx

\[ dy/dx = (dy/d\theta) / (dx/d\theta) \]. Substitute Step 1's results into this equation to get \( dy/dx = \sin\theta / (1 - \cos\theta) \).

Step 3: Find the second derivative \( d^2y/dx^2 \)

Differentiate \( dy/dx \) with respect to \( x \). First, rewrite \( dy/dx \) in terms of \( \theta \) and then differentiate it using chain rule. Calculate \( (d^2y/d\theta^2) / (dx/d\theta) - ((dy/d\theta) / (dx/d\theta)^2 ) * (d^2x/d\theta^2 ) \), which simplifies to \( (1 - \cos\theta) / (1 - \cos\theta)^3 - (\sin\theta * \sin\theta) / (1 - \cos\theta)^2 \).

Step 4: Find the slope and concavity by substituting \( \theta = \pi \)

Finally, substitute \( \theta = \pi \) into the equations for \( dy/dx \) and \( d^2y/dx^2 \) from Steps 2 and 3. For \( dy/dx \), this yields \( -1 \), which is the slope at \( \theta = \pi \). Substituting \( \theta = \pi \) into the equation for \( d^2y/dx^2 \) yields 0, meaning at \( \theta = \pi \), the curve is neither concave up nor concave down.

Key Concepts

Understanding Derivative Calculus in Parametric Equations

Derivative calculus allows us to understand how a function changes as its inputs change. In parametric equations, instead of a single equation y = f(x), we have two separate equations, each in terms of a parameter, usually denoted as \( \theta \). For the given parametric equations \( x = \theta - \sin \theta \) and \( y = 1 - \cos \theta \), finding derivatives involves understanding how x and y change with respect to \( \theta \).

To find \( \frac{dy}{dx} \), we differentiate each equation with respect to \( \theta \). The derivative of x with respect to \( \theta \) is \( \frac{dx}{d\theta} = 1 - \cos \theta \), and for y, it's \( \frac{dy}{d\theta} = \sin \theta \). The first derivative \( \frac{dy}{dx} \) is then calculated as the ratio \( \frac{dy/d\theta}{dx/d\theta} = \frac{\sin \theta}{1 - \cos \theta} \).

Once we have \( \frac{dy}{dx} \), we can find the second derivative \( \frac{d^2y}{dx^2} \) using a more complex method. It involves chain rule and requires differentiating \( \frac{dy}{dx} \) with respect to \( \theta \) and dividing by \( dx/d\theta \), leading to a deeper understanding of how the curve behaves.

Performing Slope Analysis

Slope analysis involves understanding the gradient or steepness of a curve at a specific point. For parametric equations, this involves calculating \( \frac{dy}{dx} \) at a given parameter value. In this example, we looked at the slope when \( \theta = \pi \).

Substituting \( \theta = \pi \) into the equation for \( \frac{dy}{dx} = \frac{\sin \theta}{1 - \cos \theta} \) gives the slope at that specific point. In our case, \( \sin(\pi) = 0 \) and \( \cos(\pi) = -1 \), thus \( \frac{dy}{dx} = \frac{0}{1+1} = 0 \). This indicates that at \( \theta = \pi \), the tangent to the curve is horizontal.

Understanding the slope is crucial as it tells us about the behavior of the curve at certain points, such as whether the curve is increasing, decreasing, or constant.

Exploring Concavity Analysis

Concavity analysis helps us understand the bend or curvature direction of a graph. Utilizing the second derivative \( \frac{d^2y}{dx^2} \), we determine whether a curve is concave up (like a cup) or concave down (like a frown) at a given parameter value.

For the functions detailed, we derived that \( \frac{d^2y}{dx^2} \) becomes necessary as we substitute \( \theta = \pi \). At this point, the previously derived expression for \( \frac{d^2y}{dx^2} \) was noted to be 0. This implies that at \( \theta = \pi \), the curve is neither concave up nor concave down. In simpler terms, there is no bending in the curve at this specific point.

Being able to deduce concavity provides insight into the nature of the function's graph, assisting in sketching the curve and understanding its shape and turning points.