Problem 93 - Calculus Volume 3

Question

Find \(d y / d x\) at the value of the parameter. $$ x=\cos t, \quad y=\sin t, \quad t=\frac{3 \pi}{4} $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx}\) is 1 at \(t = \frac{3\pi}{4}\).

Step 1: Differentiate x with respect to t

Given that \( x = \cos t \), find the derivative of \( x \) with respect to \( t \).\[\frac{d x}{d t} = \frac{d}{d t} \cos t = -\sin t\]

Step 2: Differentiate y with respect to t

Given that \( y = \sin t \), find the derivative of \( y \) with respect to \( t \).\[\frac{d y}{d t} = \frac{d}{d t} \sin t = \cos t\]

Step 3: Find \(\frac{dy}{dx}\) using the Chain Rule

To find \( \frac{dy}{dx} \), use the chain rule, noting that \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).\[\frac{dy}{dx} = \frac{\cos t}{-\sin t} = -\cot t\]

Step 4: Evaluate \(\frac{dy}{dx}\) at \(t = \frac{3\pi}{4}\)

Substitute \( t = \frac{3\pi}{4} \) into the equation for \( \frac{dy}{dx} \):\[\frac{dy}{dx} = -\cot \left(\frac{3\pi}{4}\right)\]Since \( \cot \left(\frac{3\pi}{4}\right) = -1 \),\[\frac{dy}{dx} = -(-1) = 1\]

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus used to find the derivative of a composite function. It is particularly useful when dealing with parametric equations. In this context, the Chain Rule allows you to differentiate indirectly by finding expressions for how one variable changes with respect to another.
In our problem, we have expressions for both \( x \) and \( y \) in terms of a parameter \( t \). When applying the Chain Rule here, we need to find \( \frac{dy}{dx} \), the derivative of \( y \) with respect to \( x \).

• First, differentiate \( x \) and \( y \) with respect to \( t \) to get \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
• Then, apply the Chain Rule formula: \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).

By rearranging the derivatives in this way, the Chain Rule connects the rates of change of \( x \) and \( y \) through their common parameter \( t \). This makes it versatile for solving problems involving variables defined parametrically.

Trigonometric Functions

Trigonometric functions play a crucial role in calculus, especially when dealing with motion and waves. In the given problem, both \( x \) and \( y \) are expressed using trigonometric functions: \( x = \cos t \) and \( y = \sin t \).
Each function has straightforward rules for differentiation:

• The derivative of \( \cos t \) with respect to \( t \) is \( -\sin t \).
• The derivative of \( \sin t \) with respect to \( t \) is \( \cos t \).

These derivatives are essential in applying the Chain Rule to find \( \frac{dy}{dx} \). Trigonometric identities, such as the relationship \( \cot t = \frac{\cos t}{\sin t} \), allow simplification of expressions, making it easier to evaluate derivatives at specific parameters, like \( t = \frac{3\pi}{4} \). Understanding these trig functions and their applications helps in efficiently solving parametric calculus problems.

Calculus Problem Solving

Successfully solving a calculus problem requires careful application of mathematical concepts and steps. Let's break down this particular problem to see how these principles are applied.
First, you identify the problem's framework; \( x \) and \( y \) are given as parametric equations involving \( t \). Then, follow these steps:

• Differentiate \( x \) and \( y \) with respect to \( t \) to find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
• Use the Chain Rule, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \), to express the derivative of \( y \) in terms of \( x \).
• Simplify the expression, taking advantage of trigonometric identities where necessary.
• Substitute the given parameter \( t = \frac{3\pi}{4} \) to find the specific value of \( \frac{dy}{dx} \).

Through these steps, calculus problems become more approachable. Accurate use of calculus tools and strategies ensures a clear path to find the solution. Practicing these methods builds strong problem-solving skills, which are critical in tackling real-world mathematical challenges.