Problem 25 - Calculus Early Transcendentals

Question

Show that the curve \(x=\cos t, y=\sin t \cos t\) has two tangents at \((0,0)\) and find their equations. Sketch the curve.

Step-by-Step Solution

Answer

The tangents at (0,0) are y = x and y = -x.

Step 1: Recognize parametric equations

The given curve is defined by the parametric equations: \( x = \cos t \) and \( y = \sin t \cos t \). Our task is to find the tangents at the point \((0,0)\).

Step 2: Express the point \((0,0)\) using \(t\)

For a point \((0,0)\) to belong to the curve, both \(x = \cos t = 0\) and \(y = \sin t \cos t = 0\) must be true. The condition \(\cos t = 0\) gives us \(t = \frac{\pi}{2}\) or \(t = \frac{3\pi}{2}\).

Step 3: Find derivatives \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \)

Calculate the derivatives for parametric equations to find the slopes of tangents. We have\[ \frac{dx}{dt} = -\sin t \]\[ \frac{dy}{dt} = \cos^2 t - \sin^2 t \]

Step 4: Calculate tangent slopes at \( t = \frac{\pi}{2} \)

Substitute \( t = \frac{\pi}{2} \) to find the tangent slope:\[ \frac{dx}{dt} = -\sin \left(\frac{\pi}{2}\right) = -1 \]\[ \frac{dy}{dt} = \cos^2\left(\frac{\pi}{2}\right) - \sin^2\left(\frac{\pi}{2}\right) = -1 \]So, the slope is \( \frac{dy}{dx} = \frac{-1}{-1} = 1 \).

Step 5: Calculate tangent slopes at \( t = \frac{3\pi}{2} \)

Substitute \( t = \frac{3\pi}{2} \) to find the tangent slope:\[ \frac{dx}{dt} = -\sin \left(\frac{3\pi}{2}\right) = 1 \]\[ \frac{dy}{dt} = \cos^2\left(\frac{3\pi}{2}\right) - \sin^2\left(\frac{3\pi}{2}\right) = -1 \]So, the slope is \( \frac{dy}{dx} = \frac{-1}{1} = -1 \).

Step 6: Determine equations of tangents

Using the point-slope formula \( y - y_0 = m(x - x_0) \), the equations are:- For slope \(1\): \( y = x \)- For slope \(-1\): \( y = -x \).

Step 7: Sketch the curve and tangents

When sketching, the parametric curve given by \( x = \cos t \), \( y = \sin t \cos t \) resembles the path of a cycloid where intersections at the origin have two distinct tangents, \( y = x \) and \( y = -x \).

Key Concepts

Parametric Equations

Parametric equations are a way of expressing geometric curves using one or more variables, known as parameters. In traditional algebraic equations, the relationship between variables is given directly. But with parametric equations, each coordinate of the point along the curve is expressed as a separate function of a parameter.
Consider the example given in the exercise:

• The equations are defined as:
  • \( x = \cos t \)
  • \( y = \sin t \cos t \)

Here, both \( x \) and \( y \) are expressed in terms of \( t \), providing a flexible way to define the curve's path by varying \( t \). This form is particularly helpful to describe paths that are not easily defined by simple algebraic equations.
By iterating over various values of \( t \), you can observe how the curve forms and at which specific values it reaches notable positions, like intersections with axes or turning points. For example, when \( t = \frac{\pi}{2} \) or \( t = \frac{3\pi}{2} \), the curve is confirmed to pass through the origin \( (0,0) \). This demonstrates how parametric equations provide a detailed way to explore and define curves.

Derivative Calculation

Derivative calculations in the context of parametric equations describe how one coordinate changes with respect to the other even when both are expressed in terms of a parameter. In simpler terms, derivatives help us find the slopes of tangent lines to these curves.
To find these derivatives, we usually perform the following steps:

• First, calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
• Then compute \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).

Let's look at the derivatives in our example.

• Given:
  • \( \frac{dx}{dt} = -\sin t \)
  • \( \frac{dy}{dt} = \cos^2 t - \sin^2 t \)

By substituting specific values for \( t \), like \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \), you can find the slopes of the tangents at those points. Hence, you'll notice how derivatives provide insight into the behavior of the curve at any given point based on your parameter. This not only tells us about the direction of the tangent but also about the curvature and other dynamic properties of the curve.

Tangent Lines

Tangent lines are straight lines that touch a curve at a single point without crossing over. They reflect the instantaneous direction of the curve at that point. In the context of parametric equations, tangent lines give a snapshot of the curve's behavior.
The procedure to find tangent lines involves understanding the slope derived from the derivatives and applying it to the point-slope form of a line:

• The formula: \( y - y_0 = m(x - x_0) \), where \( y_0 \) and \( x_0 \) are known coordinates, and \( m \) is the slope.

From the exercise at hand:

• For \( t = \frac{\pi}{2} \) at \( (0,0) \), the slope \( m = 1 \), leading to the tangent line equation \( y = x \).
• For \( t = \frac{3\pi}{2} \), the slope is \( m = -1 \), yielding \( y = -x \) as the tangent line equation.

Tangent lines are crucial in understanding curves, especially in identifying turning points, intersections, or visualizing motion dynamics. Additionally, the presence of two distinct tangent lines at a simple point such as \( (0,0) \) in our curve suggests a change in direction which is typical for more complex parametric paths.