Question

In Exercises \(41-48,\) find the equation of the line tangent to the curve at the point defined by the given value of \(t\) . $$x=t-\sin t, \quad y=1-\cos t, \quad t=\pi / 3$$

Short answer

The equation of the line tangent to the point defined by \(t = \pi/3\) on the curve defined by \(x = t - \sin(t)\) and \(y = 1 - \cos(t)\) can be obtained using the slope point form of the line equation \(y - y1 = m (x - x1)\) where m is the slope of the tangent line at \(t = \pi/3\) and \((x1, y1)\) is the point on the curve when \(t = \pi/3\)

Step-by-step solution

Find the values of x and y at \(t = \pi/3\)

Substitute \(t = \pi/3\) into the functions \(x(t)\) and \(y(t)\) to get the point of tangency. \[x=\pi/3 - \sin (\pi/3)\] \[y=1-\cos (\pi/3)\]

Find the derivatives of x and y with respect to t

We find the derivative of \(x(t) = t - \sin(t)\) which gives us \(x'(t) = 1 - \cos(t)\), and the derivative of \(y(t) = 1- \cos(t)\) which gives us \(y'(t) = \sin(t)\)

Substitute the given t value into the derivatives

Substitute \(t = \pi/3\) into the derivatives. We have \(x'(\pi/3) = 1 - \cos(\pi/3)\) and \(y'(\pi/3) = \sin(\pi/3)\)

Find the slope of the tangent line

The slope of the curve is given by \(y'/x'\). Substitute the given values of \(y'\) and \(x'\) to find the slope. Therefore, the slope \(m = y'(\pi/3) / x'(\pi/3)\)

Find the equation of the tangent line

Use the slope and point obtained in the first step to write the equation of the tangent line in slope-point form as \(y - y1 = m(x - x1)\)

Key concepts

Derivatives

To comprehend the basics of a tangent line equation, mastering derivatives is crucial. A derivative of a function measures how the function's output value changes as its input value changes. In simple terms, it's the function's rate of change or the slope of the function at any point.
For instance, given a position-time graph for an object in motion, the derivative at any point represents the object's instantaneous velocity. Similarly, when finding the tangent line to a curve at a specific point, we're essentially seeking the derivative at that point. The derivative informs us about the steepness or slope of the function precisely at that point, and this slope is used to define the tangent line.
In our exercise, we calculate the derivatives with respect to 't' for the parametric equations that describe the curve. We find that the derivative of the x-coordinate function, noted as \(x'(t)\), equals \(1 - \text{cos}(t)\), and the derivative of the y-coordinate function, \(y'(t)\), equals \(\text{sin}(t)\). These derivatives are necessary for determining the slope of the tangent line at the point where \(t = \frac{\text{\(\text{pi}\)}}{3}\).

Slope of a Curve

The concept of the slope of a curve at a point is a natural extension of the slope of a straight line. While a straight line has a constant slope, the slope of a curve varies at different points.
To find this varying slope, we use derivatives. The slope of the curve at any given point is essentially the tangent of the angle that the tangent line makes with the positive direction of the x-axis. It reflects how steep or flat the curve is at the particular point.
In our exercise, after determining the derivatives \(x'(t)\) and \(y'(t)\), we calculate the slope of the curve at \(t = \frac{\text{\(\text{pi}\)}}{3}\) by dividing \(y'(\text{\(\text{pi}\)}}/3)\) by \(x'(\text{\(\text{pi}\)}}/3)\). The quotient yields the slope of the tangent line, which we then utilize in the point-slope formula to find the equation of the tangent line.

Parametric Equations

Parametric equations are a powerful way to describe curves by expressing the coordinates of the points on the curve as functions of a single parameter. Often, 't' is used as this parameter. Unlike traditional y = f(x) functions, parametric equations represent both x and y in terms of t: x = f(t) and y = g(t).
This approach allows us to describe more complex curves that may not be functions themselves, like circles or loops. In the context of our exercise, the curve is described parametrically with \(x(t) = t - \text{sin}(t)\) and \(y(t) = 1 - \text{cos}(t)\), providing a relationship between x and y through the parameter t. Calculating the derivatives of these equations with respect to t gives us the rate of change of each coordinate, which ultimately helps us understand the slope of the tangent at any point along the curve. After evaluating these derivatives at \(t = \frac{\text{\(\text{pi}\)}}{3}\), we use them to find the tangent line's equation.

Trigonometric Functions

Trigonometric functions such as sine, cosine, and tangent are relationships between the angles and sides of triangles, but they also have applications to periodic phenomena, such as waves. These functions aid in describing and analyzing oscillating systems.
Within calculus, trigonometric functions are often differentiated and integrated to solve problems related to angles and rates of change. In our exercise, we encounter sine and cosine functions within the parametric equations for the curve. Understanding how these functions behave and how their derivatives look is imperative.
We know that the derivatives of sine and cosine are cosine and negative sine, respectively. This knowledge is directly applied when finding the derivatives of our parametric equations, \(x'(t)\) and \(y'(t)\), and thereby assists in calculating the slope of the tangent line and its equation. Knowing trigonometric derivatives allows us to navigate through problems involving motion, waves, or any phenomena exhibiting periodicity.