Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(21-30\), find \(d y / d x\) and \(d^{2} y / d x^{2}\) without eliminating the parameter. $$ x=\frac{2}{1+t^{2}}, y=\frac{2}{t\left(1+t^{2}\right)} ; t \neq 0 $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{1+3t^2}{2t} \), \( \frac{d^2y}{dx^2} = \frac{3t^2-1}{2t^3} \).

Step 1: Differentiate x with respect to t

Given the parametric equations: \[ x = \frac{2}{1+t^2} \] We differentiate \( x \) with respect to \( t \) using the chain rule:\[ \frac{d x}{d t} = \frac{d}{d t}\left( \frac{2}{1+t^2} \right) = \frac{0(1+t^2) - 2(2t)}{(1+t^2)^2} = \frac{-4t}{(1+t^2)^2} \]

Step 2: Differentiate y with respect to t

Given the parametric equation: \[ y = \frac{2}{t(1+t^2)} \] We differentiate \( y \) with respect to \( t \) using the product and chain rule:\[ \frac{d y}{d t} = \frac{d}{d t} \left( \frac{2}{t(1+t^2)} \right) = \frac{(0)t(1+t^2) - 2(1 + t^2) - 2t(2t)(t)}{t^2(1 + t^2)^2} = \frac{-2(1+t^2) - 4t^2}{t^2(1+t^2)^2} \] Simplify:\[ \frac{d y}{d t} = \frac{-2 - 2t^2 - 4t^2}{t^2(1+t^2)^2} = \frac{-2 - 6t^2}{t^2(1+t^2)^2} \]

Step 3: Find dy/dx using dy/dt and dx/dt

We have already computed:\[ \frac{d x}{d t} = \frac{-4t}{(1+t^2)^2} \] \[ \frac{d y}{d t} = \frac{-2(1+3t^2)}{t^2(1+t^2)^2} \] Now, use these to find \( \frac{d y}{d x} \):\[ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{\frac{-2(1+3t^2)}{t^2(1+t^2)^2}}{\frac{-4t}{(1+t^2)^2}} = \frac{-2(1+3t^2)(1+t^2)^2}{-4t} \]Simplify:\[ \frac{d y}{d x} = \frac{2(1+3t^2)}{4t} \Rightarrow \frac{1+3t^2}{2t} \]

Step 4: Find d²y/dx² using implicit differentiation

We have:\[ \frac{d y}{d x} = \frac{1+3t^2}{2t} \]Differentiate with respect to \( t \) for \( \frac{d^2y}{dx^2} \):\[ \frac{d}{dt} \left( \frac{d y}{d x} \right) = \frac{d}{dt} \left( \frac{1+3t^2}{2t} \right) = \frac{(6t)(2t) - (1+3t^2)(2)}{(2t)^2} = \frac{12t^2 - 2 - 6t^2}{4t^2} \]Simplify:\[ \frac{6t^2 - 2}{4t^2} -> \frac{3t^2 - 1}{2t^2} \]

Step 5: Express final results

The first derivative of \( y \) with respect to \( x \) is:\[ \frac{d y}{d x} = \frac{1+3t^2}{2t} \]The second derivative of \( y \) with respect to \( x \) is:\[ \frac{d^2y}{dx^2} = \frac{3t^2 - 1}{2t^3} \]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, particularly when handling derivatives of composite functions. Imagine it as a way to "chain" together multiple functions to find the rate of change with respect to another variable. In simpler terms, it helps us differentiate complex, layered functions step by step.

When applying the chain rule to parametric equations like those given for \( x = \frac{2}{1+t^2} \) and \( y = \frac{2}{t(1+t^2)} \), we start by differentiating each component with respect to the parameter \( t \). This method is crucial in finding how one dependent variable changes with respect to the other.

Here's how the chain rule works in this context:

• First, we differentiate \( x \) with respect to \( t \) to get \( \frac{dx}{dt} \).
• Next, we differentiate \( y \) and find \( \frac{dy}{dt} \).
• Finally, by dividing these results, we apply the chain rule again to discover \( \frac{dy}{dx} \), expressing the rate of change of \( y \) with respect to \( x \).

Recognizing when and how to apply the chain rule simplifies the process of handling complicated curves and rates of change.

Second Derivative

Finding the second derivative is like peeling another layer off the calculus onion. It allows us to explore the curvature and concavity of a curve, providing deeper insights on how the function behaves.When we talk about the second derivative in parametric terms, it involves using the derivative of the first derivative. Let's unravel what this means:

• First, determine \( \frac{dy}{dx} \), which tells us the slope or "steepness" of the curve.
• To find \( \frac{d^2y}{dx^2} \), differentiate \( \frac{dy}{dx} \) with respect to \( t \), the parameter.
• This requires a chain of differentiation, involving both the numerator and denominator of \( \frac{dy}{dx} \).

Through this process, the second derivative \( \frac{d^2y}{dx^2} \) acts as a window into the acceleration or the rate of change of the rate of change. It answers whether a function is curving upwards, downwards, or remains linear—with increasing or decreasing steepness.

Calculus Techniques

Mastering calculus is like learning the tools of a trade; you need the right technique for each job. In the world of parametric equations, several techniques come in handy.

• Product Rule: When a function involves multiplication, such as in the given parametric equation of \( y \), the product rule enables us to differentiate effectively. It involves differentiating each component (product) individually, then combining them.
• Simplification: After applying the chain rule and product rule, you'll often encounter expressions that look complex. Simplification is crucial here to make the results more approachable and interpretable.
• Substitution: This technique helps in managing cumbersome terms. Substituting equivalent expressions simplifies differentiation operations, especially when obtaining \( \frac{d^2y}{dx^2} \).

Each of these techniques requires a good grasp of basic calculus rules, along with practice. By applying these methods, we not only find solutions but also gain a clearer understanding of how changes in one variable affect the others in a relationship.