Problem 32 - Precalculus with Calculus Previews

Question

Find the \(x\) - and \(y\) -intercepts of the given curves. $$ x=1+\sin t, y=\sin t-\cos t, 0 \leq t \leq 2 \pi $$

Step-by-Step Solution

Answer

X-intercepts: \((1 + \frac{\sqrt{2}}{2}, 0)\), \((1 - \frac{\sqrt{2}}{2}, 0)\); Y-intercept: \((0, -1)\).

Step 1: Understanding the Problem

We need to find the x- and y-intercepts of the given parametric equations. The equations are given as functions of the parameter \(t\): \(x = 1 + \sin t\) and \(y = \sin t - \cos t\). The range of \(t\) is from 0 to \(2\pi\).

Step 2: Finding the x-intercept

The x-intercept occurs where \(y = 0\). Set the equation for \(y\) to zero: \(\sin t - \cos t = 0\). Solving for \(t\), we get \(\sin t = \cos t\). This happens at \(t = \frac{\pi}{4}, \frac{5\pi}{4}\) within the given interval. Use these values to find corresponding \(x\) values: \(x = 1 + \sin \frac{\pi}{4} = 1 + \frac{\sqrt{2}}{2}\) and \(x = 1 + \sin \frac{5\pi}{4} = 1 - \frac{\sqrt{2}}{2}\).

Step 3: Calculating y-intercept

The y-intercept occurs where \(x = 0\). Set the equation for \(x\) to zero: \(1 + \sin t = 0\). Solving for \(t\) gives \(\sin t = -1\), which happens at \(t = \frac{3\pi}{2}\). Find the corresponding \(y\) value: \(y = \sin \frac{3\pi}{2} - \cos \frac{3\pi}{2} = -1\).

Step 4: Summarize Intercepts

The x-intercepts are \((1 + \frac{\sqrt{2}}{2}, 0)\) and \((1 - \frac{\sqrt{2}}{2}, 0)\). The y-intercept is \((0, -1)\).

Key Concepts

x-intercept

When dealing with parametric equations, finding the x-intercept involves setting the y-component equal to zero. This is because, on the x-axis, the value of y is always zero.
To find where the curve intercepts the x-axis, look at the parametric equation for y. For our exercise, we have:

• \( y = \sin t - \cos t \)

Setting the y equation to zero gives:

• \( \sin t - \cos t = 0 \)
• This implies \( \sin t = \cos t \)

Solve this equation to find potential values of t where the curve hits the x-axis. Within the interval \(0 \leq t \leq 2\pi\), we find \(t = \frac{\pi}{4}\) and \(t = \frac{5\pi}{4}\).
Using these t-values, substitute back into the x-component:

• \(x = 1 + \sin \frac{\pi}{4} = 1 + \frac{\sqrt{2}}{2} \)
• \(x = 1 + \sin \frac{5\pi}{4} = 1 - \frac{\sqrt{2}}{2} \)

These coordinates \((1 + \frac{\sqrt{2}}{2}, 0)\) and \((1 - \frac{\sqrt{2}}{2}, 0)\) represent the points where the curve meets the x-axis.

y-intercept

The y-intercept is found where the x-component is zero. This point is where the curve crosses the y-axis.
In our given parametric equations:

• \( x = 1 + \sin t \)

To determine the y-intercept, set \(x = 0\) since this is where the y-axis is intersected:

• \(1 + \sin t = 0\)
• This simplifies to \(\sin t = -1\)

The value of \(t\) that satisfies this equation within the interval \(0 \leq t \leq 2\pi\) is \(t = \frac{3\pi}{2}\).
Substitute \(t\) into the y-equation:

• \(y = \sin \frac{3\pi}{2} - \cos \frac{3\pi}{2}\)
• \(y = -1 - 0 = -1\)

Thus, our y-intercept is at the coordinate \((0, -1)\). It represents the point where the curve passes through the y-axis.

trigonometric functions

Trigonometric functions like sine and cosine play an essential role in parametric equations, especially when describing curves and their intersections with axes.
In the exercise, we use trigonometric identities to simplify and solve equations.

• For example, solving \(\sin t = \cos t\) involves understanding where these two trigonometric functions equal each other within the specified interval.
• Using \(\sin t\) and \(\cos t\), we can express components of the parametric equations to find intercepts.

Key trigonometric angles here include \(\frac{\pi}{4}\), \(\frac{5\pi}{4}\), and \(\frac{3\pi}{2}\), each corresponding to significant positions on the unit circle.
Understanding how \(\sin\) and \(\cos\) behave:

• \(\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}\)
• \(\sin \frac{5\pi}{4} = \cos \frac{5\pi}{4} = -\frac{\sqrt{2}}{2}\)
• \(\sin \frac{3\pi}{2} = -1\) and \(\cos \frac{3\pi}{2} = 0\)

Using these values appropriately helps in accurately plotting curves and determining intersections.