Question

Find the slope of the tangent line to the given polar curve at the point given by the value of \(\theta\).\(r=\theta, \theta=\frac{\pi}{2}\)

Short answer

The slope of the tangent line is \(-\frac{2}{\pi}\).

Step-by-step solution

Express Polar to Cartesian Coordinates

The given polar equation is \( r = \theta \). At \( \theta = \frac{\pi}{2} \), substitute \( \theta \) into the equation to get the radius \( r = \frac{\pi}{2} \). The Cartesian coordinates can be derived from the polar coordinates using the formulas: \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). Substitute \( r = \frac{\pi}{2} \) and \( \theta = \frac{\pi}{2} \) to find that \( x = 0 \) and \( y = \frac{\pi}{2} \). Thus, the point is \((0, \frac{\pi}{2})\).

Differentiate with Respect to θ

To find the slope of the tangent line, we need to convert \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\). The formulas for derivatives in polar coordinates are \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \). Differentiate these with respect to \(\theta\): \( \frac{dx}{d\theta} = \cos(\theta) - \theta\sin(\theta) \) and \( \frac{dy}{d\theta} = \sin(\theta) + \theta\cos(\theta) \).

Substitute the Given θ

Now, substitute \( \theta = \frac{\pi}{2} \) into the derivatives. For \( \frac{dx}{d\theta} \), we get \( \frac{dx}{d\theta} = \cos\left(\frac{\pi}{2}\right) - \frac{\pi}{2} \sin\left(\frac{\pi}{2}\right) = 0 - \frac{\pi}{2}(1) = -\frac{\pi}{2} \). For \( \frac{dy}{d\theta} \), \( \frac{dy}{d\theta} = \sin\left(\frac{\pi}{2}\right) + \frac{\pi}{2} \cos\left(\frac{\pi}{2}\right) = 1 + \frac{\pi}{2} \times 0 = 1 \).

Calculate the Slope of the Tangent Line

The slope \( m \) of the tangent line to the curve is given by the formula \( m = \frac{dy/d\theta}{dx/d\theta} \). Using the derivatives we found: \( \frac{dy/d\theta}{dx/d\theta} = \frac{1}{-\frac{\pi}{2}} = -\frac{2}{\pi} \).

Key concepts

Tangent Line

A tangent line to a curve represents a straight line that touches the curve at just one point. It's crucial because it reflects the direction of the curve at that particular point. Imagine if you had a ball rolling on a track; at any point, its immediate direction is like the tangent line.

When dealing with curves in polar coordinates, the idea is similar. However, instead of using regular equations like in Cartesian coordinates, you start with polar equations. For instance, given a point on a polar curve, you can find the tangent line's slope by first converting polar coordinates to Cartesian coordinates. This transformation helps because we're more familiar with dealing with slopes when using the (x, y) plane.

The slope of the tangent line connects the rate at which the curve rises to the rate at which it moves horizontally, ultimately showing how steep the curve is at that particular spot. Understanding this concept is essential because it bridges curves in geometry with real-world slopes and directions.

Derivative

The derivative is a fundamental concept in calculus that lets us determine how a function changes. In simpler terms, it's like the velocity of a moving object; it tells us how fast something is changing at any given point and in which direction.

For a polar curve, determining the slope of the tangent line involves finding derivatives with respect to the angle (\(\theta\)). In the specific exercise provided, the curve was given in polar form as \(r = \theta\). The goal was to convert this to more familiar terms (Cartesian) and find the derivatives using the formulas:

• \( x = r \cos(\theta) \)
• \( y = r \sin(\theta) \)

By differentiating these expressions with respect to \(\theta\), we get \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \), which reveal how both x and y change in respect to the angle. These two parts are the key to calculating the slope of the tangent line in polar coordinates. By understanding derivatives, we essentially unlock a deeper insight into the motion and change of curves, not only in polar coordinates but universally in various kinds of functions.

Cartesian Coordinates

Cartesian coordinates are one of the most common and straightforward ways to describe a point in space. These coordinates use the familiar (x, y) system, where 'x' and 'y' define the position of a point on a flat, two-dimensional plane.

In problems involving polar coordinates, like the one we are discussing, often we need to convert between polar and Cartesian systems. This makes calculations like finding slopes or points more intuitive because many mathematical concepts, like derivatives and tangent lines, are well-established in Cartesian coordinates.

For instance, given polar coordinates \((r, \theta)\), you can convert to Cartesian using:

• \( x = r \cos(\theta) \)
• \( y = r \sin(\theta) \)

This transformation is critical to solve the problem since it moves us to a space where we can easily differentiate and find a curve's slope. This helps us apply foundational rules of calculus to new forms and equations in polar coordinates. By mastering these conversions, we effectively bridge two worlds of mathematical understanding, expanding our ability to solve complex problems with greater ease.