Problem 13 - 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. $$ x=\cos ^{3} \theta, y=\sin ^{3} \theta \quad \theta=\frac{\pi}{4} $$

Step-by-Step Solution

Answer

The slope of the function at \(θ= \pi /4\) is \(-1\) and the concavity (second derivative) at this point is \(0\), interpreting as an inflection point.

Step 1: Find \(dx/dθ\) and \(dy/dθ\)

First, compute the derivatives of \(x=cos^3 (θ)\) and \(y=sin^3 (θ)\) with respect to \(θ\). We will need to use the chain rule here. The derivative of \(x\) with respect to \(θ\) is \(-3 cos^2(θ) sin (θ)\), and the derivative of \(y\) with respect to \(θ\) is \(3 sin^2 (θ) cos (θ)\).

Step 2: Determine \(dy/dx\)

Using the results from step 1, find \(dy/dx\) by dividing \(dy/dθ\) by \(dx/dθ\). This gives \(dy/dx = \frac{3 sin^2 (θ) cos (θ)}{-3 cos^2(θ) sin (θ)}\). This simplifies to \(dy/dx = - tan (θ)\).

Step 3: First Derivative at \(θ=\frac{\pi}{4}\)

Evaluate the first derivative \(dy/dx\) at the given point \(θ=\frac{\pi}{4}\) from the solutions provided in step 2. Substituting \(θ=\frac{\pi}{4}\) into \(dy/dx = - tan (θ)\) gives \(dy/dx = -1\). Hence the slope of the function at \(θ= \pi /4\) is \(-1\).

Step 4: Compute \(d^2y/dx^2\)

To find the concavity, we need to calculate the second derivative of \(y\) with respect to \(x\), \(d^2y/dx^2\). Start by differentiating \(dy/dx = - tan (θ)\) with respect to \(θ\), we get \(d(dy/dx) / dθ = - sec^2 (θ)\). Then, we find \(d^2y/dx^2\) by dividing \(d(dy/dx) / dθ\) by \(dx/dθ\), which gives \(d^2y/dx^2 = \frac{- sec^2 (θ)}{-3 cos^2(θ) sin (θ)}\).

Step 5: Second Derivative at \(θ=\frac{\pi}{4}\)

Evaluate the second derivative \(d^2y/dx^2\) at the given point \(θ=\frac{\pi}{4}\) from the solutions provided in step 4. Substituting \(θ=\frac{\pi}{4}\) into \(- sec^2 (θ) / -3 cos^2(θ) sin (θ)\) gives \(d^2y/dx^2 = 0\). Hence the concavity of the function at \(θ= \pi /4\) is \(0\). Here, the second derivative at the given point being zero represents an inflection point.

Key Concepts

Chain Rule Differentiation

Understanding the chain rule in calculus is essential for differentiating composite functions. When we have a function that is the composition of two functions, such as in our exercise with the parametric equations \(x = \cos^3\theta\) and \(y = \sin^3\theta\), we apply the chain rule to obtain the derivatives with respect to parameter \(\theta\).

The chain rule states: If \(u = g(\theta)\) and \(v = f(u)\), then the derivative of \(v\) with respect to \(\theta\) is \(dv/d\theta = (dv/du)\cdot(du/d\theta)\). For our exercise, we differentiate \(x\) and \(y\) with respect to \(\theta\) by considering \(\cos\theta\) and \(\sin\theta\) as intermediate functions and \(\cos^3\theta\) and \(\sin^3\theta\) as the outer functions.

Applying this rule helps us to move from the initial equations to expressions of \(dx/d\theta\) and \(dy/d\theta\), which are then used to find \(dy/dx\), the slope function of the curve.

Second Derivative Test

The second derivative test is a helpful tool in analyzing the concavity of a function and determining local maxima and minima. If the second derivative of a function is positive at a certain point, the function is concave up there. Conversely, if it is negative, the function is concave down.

In the context of parametric equations, we compute the second derivative of \(y\) with respect to \(x\) (\(d^2y/dx^2\)) to determine the concavity of the curve. In our exercise, after finding the second derivative and evaluating it at \(\theta = \frac{\pi}{4}\), we concluded that the second derivative \(d^2y/dx^2 = 0\). This result implies an inflection point. At an inflection point, the curve changes its concavity, meaning it transitions from concave up to concave down, or vice versa.

Slope and Concavity

The concepts of slope and concavity are central to understanding the behavior of curves in calculus. The slope, described by the first derivative \(dy/dx\), gives us the rate at which \(y\) changes with \(x\), or the steepness of the curve at any given point.

In our exercise, when we calculate \(dy/dx\) and evaluate it at \(\theta = \frac{\pi}{4}\), the negative value indicates that the curve is decreasing at that point. The concavity, described by the second derivative \(d^2y/dx^2\), indicates the direction in which the curve is bending. A positive second derivative means the curve is bending upwards, and a negative one means it is bending downwards. When the second derivative is zero, as in our evaluated point, this indicates a possible inflection point where the concavity might change.

First and Second Derivatives

First and second derivatives are fundamental in the study of calculus as they provide information about the function's rate of change and the curvature of its graph. The first derivative \(dy/dx\) represents the instantaneous rate of change of \(y\) with respect to \(x\), which informs about the motion and direction of a particle traveling along a curve.

The second derivative \(d^2y/dx^2\), on the other hand, gives deeper insight into the behavior of \(y\)'s rate of change itself. In essence, while the first derivative tells us about the velocity, the second one tells us about the acceleration. In the exercise, we used the first derivative to determine the slope at the given value of \(\theta\), and the second derivative to find the concavity of the curve and identify any point of inflection.