Question

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

Short answer

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

Step-by-step solution

Find \( \frac{dx}{dt} \)

Given \( x = \cos t \), differentiate with respect to \( t \): \( \frac{dx}{dt} = -\sin t \).

Find \( \frac{dy}{dt} \)

Given \( y = \sin t \), differentiate with respect to \( t \): \( \frac{dy}{dt} = \cos t \).

Express \( \frac{dy}{dx} \) using the chain rule

The derivative \( \frac{dy}{dx} \) can be found using \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). Substitute the derivatives from Steps 1 and 2: \( \frac{dy}{dx} = \frac{\cos t}{-\sin t} = -\cot t \).

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

Find \( \cot t \) at \( t = \frac{3\pi}{4} \): \( \cot \frac{3\pi}{4} = -1 \). Therefore, \( \frac{dy}{dx} = -(-1) = 1 \).

Key concepts

Parametric Equations

A parametric equation allows us to describe a set of equations where the coordinates are expressed as functions of one or more variables called parameters. These are often very helpful in tracing complex curves or shapes that might be difficult to represent using regular Cartesian equations.
For example, in the exercise we have the parametric equations \( x = \cos t \) and \( y = \sin t \). Here, \( t \) is the parameter. As \( t \) changes, both \( x \) and \( y \) change.

• When \( t = 0 \), \( x = \cos(0) = 1 \) and \( y = \sin(0) = 0 \).
• When \( t = \frac{\pi}{2} \), \( x = \cos(\frac{\pi}{2}) = 0 \) and \( y = \sin(\frac{\pi}{2}) = 1 \).

Parametric equations provide a very intuitive way to explore motion, where objects change position over time.
The relationship and change between parameters make it easier to transition into differentiating them, comparing their rates of change.

Chain Rule

The chain rule is a crucial calculus tool, especially in differentiation. It allows us to differentiate composite functions, or when a function is inside another function.
In our exercise, we use the chain rule to find the derivative of one parameterized function with respect to another. Our goal is to find \( \frac{dy}{dx} \) in terms of \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \).
Here's how it works:

• Differentiate \( x = \cos t \), obtaining \( \frac{dx}{dt} = -\sin t \).
• Differentiate \( y = \sin t \), obtaining \( \frac{dy}{dt} = \cos t \).

To find \( \frac{dy}{dx} \), use, \( \frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} \).
This formula simplifies to \( \frac{dy}{dx} = \frac{\cos t}{-\sin t} = -\cot t \).Understanding the chain rule helps you see how functions relate when linked through another parameter, unearthing layers of mathematical relationships.

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in both mathematics and applied fields. They describe periodic processes and appear frequently in calculus.
In this exercise, we deal with basic trig functions: sine \(\sin t\) and cosine \(\cos t\). These functions are defined on the unit circle and are essential in describing circles and oscillations.
Why are these important?

• The cosine function, \( \cos t \), determines the x-coordinate on a unit circle.
• The sine function, \( \sin t \), determines the y-coordinate.

When we evaluate \( \cot t \), which is \( \frac{\cos t}{\sin t} \), we get the ratio of the adjacent side to the opposite side in a right triangle representation of the circle.
In the problem, when \( t = \frac{3\pi}{4} \), both sine and cosine turn negative. However, the division resulting in \( \cot \frac{3\pi}{4} = -1 \) allows the equation to resolve smoothly.
Understanding these functions makes it easier to visualize changes and transformations on a graph. They offer insight into many periodic phenomena, from simple circular motion to complex waveforms.