Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

find \(d y / d x\) and \(d^{2} y / d x^{2}\) without eliminating the parameter. $$ x=3 \tan t-1, y=5 \sec t+2 ; t \neq \frac{(2 n+1) \pi}{2} $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{5}{3} \sin t; \frac{d^2y}{dx^2} = \frac{5}{9} \cos^3 t\)."

Step 1: Differentiate x and y with respect to t

First, differentiate both parametric equations, \(x = 3 \tan t - 1\) and \(y = 5 \sec t + 2\), with respect to the parameter \(t\). - \(\frac{d x}{d t} = 3 \sec^2 t\) since the derivative of \(\tan t\) is \(\sec^2 t\).- \(\frac{d y}{d t} = 5 \sec t \tan t\), using the derivative of \(\sec t\) which is \(\sec t \tan t\).

Step 2: Find dy/dx using derivatives from Step 1

Since \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\), substitute the derivatives from Step 1:- \(\frac{dy}{dx} = \frac{5 \sec t \tan t}{3 \sec^2 t}\)- Simplify this to \(\frac{dy}{dx} = \frac{5 \tan t}{3 \sec t} = \frac{5 \sin t}{3}\).

Step 3: Differentiate dy/dx with respect to t to find d²y/dx²

Calculate \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\):- First, differentiate \(\frac{dy}{dx} = \frac{5 \sin t}{3}\) with respect to \(t\), yielding \(\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{5}{3} \cos t\).- Use the chain rule: \(\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{5 \cos t}{3 \cdot 3 \sec^2 t}\).- Simplify to get \(\frac{d^2y}{dx^2} = \frac{5 \cos^3 t}{9}\).

Key Concepts

Calculus

Calculus is a branch of mathematics focused on change and motion. It is used to find things like gradients of curves and rates of change. Primarily, calculus is divided into two fundamental branches: differentiation and integration.
Each of these branches helps solve problems involving changing quantities.

• Differentiation: This deals with finding the rate at which something changes. It's like finding the speed of a car, which tells us how fast the car is moving in relation to time.
• Integration: This is about finding the total quantity. Think of it as calculating the total distance covered by the car.

In this exercise, the focus is on differentiation. More specifically, parametric differentiation is used. Parametric equations involve variables that are expressed in terms of a third variable, often called the parameter. In our example, both variables, x and y, are expressed in terms of t, which makes them parametric equations.

Derivatives

Derivatives allow us to understand how a function changes at any given point. They provide the "instantaneous rate of change," the rate at which a variable changes with respect to another.
In parametric equations, finding derivatives involves differentiating each parametric equation individually and then relating them.The steps in the exercise show how this is done:

• The derivative of x with respect to t, shown as \( \frac{dx}{dt} \), is calculated first. Here, \( x = 3 \tan t - 1 \) is differentiated to get \( \frac{dx}{dt} = 3 \sec^2 t \).
• Next, for the equation \( y = 5 \sec t + 2 \), the derivative of y with respect to t is found to be \( \frac{dy}{dt} = 5 \sec t \tan t \).

These derivatives form the basis for finding \( \frac{dy}{dx} \), which describes how y changes in relation to x by using the parametric derivatives.

Chain Rule

The Chain Rule is a fundamental technique used in calculus to differentiate composite functions. Its significance shines in parametric differentiation, especially when dealing with derivatives of parametric forms. The rule is essential when functions are interconnected through an intermediary variable, like a parameter.
In the exercise, the relationship between derivatives \( \frac{dy}{dx} \) is established using the chain rule. This is because both x and y are given as functions of t. The Chain Rule allows us to link these derivatives:

• First, \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \) is derived, transforming our ability to calculate rates of change from one variable to another without removing the parameter.
• Additionally, the second derivative \( \frac{d^2y}{dx^2} \) involves the chain rule to differentiate \( \frac{dy}{dx} \) again.

Mastering the chain rule is critical for solving problems where multiple variables are interrelated, making it an indispensable tool in calculus.