Problem 15 - Essential Calculus

Question

Find an equation of the tangent line at each given point on the curve. $$ \begin{array}{l} x=2 \cot \theta \\ y=2 \sin ^{2} \theta \end{array} $$

Step-by-Step Solution

Answer

The equation of the tangent line to the given curve at a specific point can be found by taking the derivative of the given equations (to find \(dy/dx\)) and then substitiating this result along with the known coordinates of the point into general equation for the tangent line. The final concrete equation depends on the specified point.

Step 1: Find dx/dtheta and dy/dtheta

Take the first derivative with respect to \(\theta\) for given x and y:\(dx/d\theta = -2 csc^2 \theta, dy/d\theta = 4 \sin \theta \cos \theta\)

Step 2: Find dy/dx

In general, \( dy/dx = (dy/d\theta) / (dx/d\theta) \). Substituting the values from Step 1, we obtain \( dy/dx = -2 \sin \theta \cos \theta/ \cot^2 \theta \)

Step 3: Find the Equation of Tangent Line

The general equation for the tangent line is \(y - y1 = m(x - x1)\), where m is the slope of the tangent line (that is, \(dy/dx\)), and \((x1, y1)\) are the coordinates of the given point on the curve. So, substitute \( m = -2 \sin \theta \cos \theta/ \cot^2 \theta \) and the given point into this equation to find the equation of the tangent line.

Key Concepts

Parametric Equations

In mathematics, parametric equations are a way of defining a curve or a surface using parameters. For curves, this means that both coordinates of a point on the curve are functions of a third variable, usually denoted as \(t\) or \(\theta\).
These equations are essential because they allow us to describe complex shapes more easily than with standard Cartesian coordinates.

• Each coordinate in the plane is expressed as a function of a parameter.
• For example, in our exercise, \(x = 2 \cot \theta\) and \(y = 2 \sin^2 \theta\).
• This means that for any value of \(\theta\), we can find corresponding \(x\) and \(y\) coordinates on the curve.

This representation is particularly useful for curves that cannot be easily described using \(y = f(x)\) or \(x = g(y)\). Parametric equations also facilitate finding derivatives, as we can differentiate both equations with respect to the parameter \(\theta\).

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function is changing at any given point. This concept is key when working with curves defined parametrically.
In our exercise, we use differentiation to find the derivatives \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\).

• For the x-equation, \(x = 2 \cot \theta\), the derivative \(\frac{dx}{d\theta}\) is calculated as \(-2 \csc^2 \theta\).
• For the y-equation, \(y = 2 \sin^2 \theta\), the derivative \(\frac{dy}{d\theta}\) turns out to be \(4 \sin \theta \cos \theta\).

These derivatives describe how \(x\) and \(y\) change with respect to \(\theta\). Once these derivatives are known, they can be combined to find the slope of the curve at a particular point by computing \(\frac{dy}{dx}\).

Slope of a Curve

The slope of a curve at any point is a measure of how steep the curve is at that location, and it can be found using differentiation.
For curves expressed using parametric equations, the slope \(\frac{dy}{dx}\) is determined by the formula \(\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\).

• This ratio effectively describes the rate at which \(y\) changes relative to \(x\) as the parameter \(\theta\) changes.
• In our example, substituting the parametric derivatives gives \(\frac{dy}{dx} = \frac{-2 \sin \theta \cos \theta}{\cot^2 \theta}\).

Calculating the slope is crucial for determining the equation of the tangent line to the curve at a given point.
With the slope known, the canonical tangent line equation \(y - y_1 = m(x - x_1)\) is used to express the tangent line, where \(m\) is the slope \(\frac{dy}{dx}\) and \((x_1, y_1)\) are the coordinates of the specific point on the curve.