Problem 89 - Calculus Volume 3

Question

Find \(d^{2} y / d x^{2}\). $$ x=\sin (\pi t), \quad y=\cos (\pi t) $$

Step-by-Step Solution

Answer

The second derivative is \(-\sec^3(\pi t)\).

Step 1: Understand the Problem

We are given two parametric equations: \(x = \sin(\pi t)\) and \(y = \cos(\pi t)\). We need to find the second derivative \(\frac{d^2 y}{dx^2}\) with respect to \(x\).

Step 2: Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)

Differentiate \(x = \sin(\pi t)\) with respect to \(t\), resulting in \(\frac{dx}{dt} = \pi \cos(\pi t)\). Similarly, differentiate \(y = \cos(\pi t)\) with respect to \(t\), resulting in \(\frac{dy}{dt} = -\pi \sin(\pi t)\).

Step 3: Find \(\frac{dy}{dx}\)

Use the chain rule for derivatives to find \(\frac{dy}{dx}\). It's given by \(\frac{dy/dt}{dx/dt}\). Thus, we have:\[\frac{dy}{dx} = \frac{-\pi \sin(\pi t)}{\pi \cos(\pi t)} = -\tan(\pi t)\]

Step 4: Find \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\)

Differentiate \(-\tan(\pi t)\) with respect to \(t\). The derivative of \(-\tan(\pi t)\) is \[\frac{d}{dt}\left(-\tan(\pi t)\right) = -\pi \sec^2(\pi t)\]

Step 5: Find \(\frac{d^2 y}{dx^2}\)

Divide the derivative found in Step 4 by \(\frac{dx}{dt}\) to get the second derivative:\[\frac{d^2 y}{dx^2} = \frac{-\pi \sec^2(\pi t)}{\pi \cos(\pi t)} = -\frac{\sec^2(\pi t)}{\cos(\pi t)} = -\frac{1}{\cos(\pi t) \cos^2(\pi t)}\]Thus:\[\frac{d^2 y}{dx^2} = -\sec^3(\pi t)\]

Step 6: Simplify the Expression

Recognize that \(-\sec^3(\pi t)\) can be expressed as \(-\frac{1}{\cos^3(\pi t)}\). Since this is already simplified in terms of trigonometric functions, this is our final expression for the second derivative.

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental concept in calculus that helps us differentiate composite functions. When you have a function that is formed by one function being inside another, the chain rule allows you to take derivatives of such compositions efficiently. In the context of parametric equations, where both dependent and independent variables are expressed in terms of a third parameter (often time), the chain rule becomes an essential tool to compute derivatives.

Here’s how it works:

• Suppose you have a function of the type \(y = f(g(t))\).
• To find \(\frac{dy}{dt}\), the chain rule tells us to multiply the derivative of the outer function \(f\) by the derivative of the inner function \(g\).
• So: \(\frac{dy}{dt} = f'(g(t)) \cdot g'(t)\).

In our exercise, the chain rule was crucial for transforming \(x = \sin(\pi t)\) and \(y = \cos(\pi t)\) into \(\frac{dy}{dx}\). We did this by finding \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\), then dividing these derivatives to apply the chain rule effectively.

Finding the Second Derivative

The second derivative provides insight into the concavity of a function. It helps us determine where a function is concave up or concave down, and where inflection points occur. When working with parametric equations, finding the second derivative \(\frac{d^2y}{dx^2}\) involves a few extra steps due to the relationship between derivatives in different variables.

To find the second derivative in the context of parametric equations:

• First, you need \(\frac{dy}{dx}\). We obtained this by taking \(\frac{dy/dt}{dx/dt}\).
• Then, differentiate \(\frac{dy}{dx}\) with respect to \(t\) to find \(\frac{d}{dt}(\frac{dy}{dx})\).
• Finally, divide \(\frac{d}{dt}(\frac{dy}{dx})\) by \(\frac{dx}{dt}\) to obtain \(\frac{d^2y}{dx^2}\).

This approach allows us to identify how the curve behaves beyond just the slope. For our specific problem, after computing several steps, we've concluded that \(\frac{d^2 y}{dx^2} = -\sec^3(\pi t)\), giving us the trigonometric form of the second derivative.

The Role of Trigonometric Functions

Trigonometric functions such as sine and cosine play a pivotal role in parametric equations, especially when dealing with periodic or circular motion. These functions help describe angles and rotations smoothly, as seen in the parametric equations of our exercise: \(x = \sin(\pi t)\), \(y = \cos(\pi t)\).

Understanding trig functions is important because:

• They provide periodicity, meaning functions repeat after a certain interval. This is essential in modeling cycles or waves.
• Sine and cosine are linked by the derivatives and identities, like \(\cos^2(x) + \sin^2(x) = 1\).
• Derivatives of these functions yield other trigonometric functions, such as differentiating \(\sin(x)\) resulting in \(\cos(x)\), and vice versa with a negative sign for the derivative of \(\cos(x)\).

In our problem, converting the trigonometric result from \(-\sec^3(\pi t)\) to \(-\frac{1}{\cos^3(\pi t)}\) helps illustrate how integral trigonometric identities and transformations are when working with these equations. Knowing these concepts allows us to manipulate and simplify expressions effectively.