Problem 31 - Precalculus with Calculus Previews

Question

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

Step-by-Step Solution

Answer

X-intercept: \(x = -1 - \sqrt{3}\); Y-intercepts: \(y = 1 + \sqrt{3}\) and \(y = 1 - \sqrt{3}\).

Step 1: Understanding the Parametric Equations

The given parametric equations are \(x = -1 + 2\cos t\) and \(y = 1 + 2\sin t\). We need to find the points where the curve intersects the x-axis and y-axis. This means solving for \(t\) when \(y=0\) for the x-intercept and when \(x=0\) for the y-intercept.

Step 2: Find the X-Intercept

For the x-intercept, set \(y = 0\): \[ 0 = 1 + 2\sin t \] \[ -1 = 2\sin t \] \[ \sin t = -\frac{1}{2} \] \(\sin t = -\frac{1}{2}\) at two standard positions in the interval \(0 \leq t \leq 2\pi\): \(t = \frac{7\pi}{6}\) and \(t = \frac{11\pi}{6}\). For these values of \(t\), plug them back into the x-equation: - For \(t = \frac{7\pi}{6}\): \[ x = -1 + 2 \cos\left(\frac{7\pi}{6}\right) = -1 + 2(-\frac{\sqrt{3}}{2}) = -1 - \sqrt{3} \]- For \(t = \frac{11\pi}{6}\): \[ x = -1 + 2 \cos\left(\frac{11\pi}{6}\right) = -1 + 2(-\frac{\sqrt{3}}{2}) = -1 - \sqrt{3} \]Thus, the x-intercept is at \(x = -1 - \sqrt{3}\).

Step 3: Find the Y-Intercept

For the y-intercept, set \(x = 0\):\[ 0 = -1 + 2\cos t \]\[ 1 = 2\cos t \]\[ \cos t = \frac{1}{2} \]\(\cos t = \frac{1}{2}\) also occurs at two standard positions in \(0 \leq t \leq 2\pi\): \(t = \frac{\pi}{3}\) and \(t = \frac{5\pi}{3}\).For these values of \(t\), substitute back into the y equation:- For \(t = \frac{\pi}{3}\): \[ y = 1 + 2\sin\left(\frac{\pi}{3}\right) = 1 + 2\left(\frac{\sqrt{3}}{2}\right) = 1 + \sqrt{3} \]- For \(t = \frac{5\pi}{3}\): \[ y = 1 + 2\sin\left(\frac{5\pi}{3}\right) = 1 + 2\left(-\frac{\sqrt{3}}{2}\right) = 1 - \sqrt{3} \]Therefore, the y-intercepts are \(y = 1 + \sqrt{3}\) and \(y = 1 - \sqrt{3}\).

Step 4: Conclusion

The x-intercept of the curve is \(x = -1 - \sqrt{3}\). The y-intercepts are \(y = 1 + \sqrt{3}\) and \(y = 1 - \sqrt{3}\), occurring at different values of \(t\).

Key Concepts

X-Intercept

In the context of parametric equations, the x-intercept is the point where the curve crosses the x-axis on a graph. To find this, we need to determine the value of the parameter, in this case, \(t\), that makes the y-coordinate equal to zero.

Given the parametric equations \(x = -1 + 2\cos t\) and \(y = 1 + 2\sin t\), we set \(y = 0\) to find when the curve touches the x-axis. This leads to the equation \(1 + 2\sin t = 0\).

• First, solve \(1 + 2\sin t = 0\) to find that \(\sin t = -\frac{1}{2}\).
• Within the interval \(0 \leq t \leq 2\pi\), \(\sin t = -\frac{1}{2}\) at \(t = \frac{7\pi}{6}\) and \(t = \frac{11\pi}{6}\).
• Substituting these values into the x-equation shows the x-intercept at \(x = -1 - \sqrt{3}\).

Thus, the x-intercept of the curve represented by these parametric equations is \(x = -1 - \sqrt{3}\). This value confirms the point where the curve hits the x-axis.

Y-Intercept

To find the y-intercept in parametric equations, we need to identify the value of the parameter when the x-coordinate equals zero. This point is where the curve intersects the y-axis.

Given \(x = -1 + 2\cos t\), set \(x = 0\) to achieve \(-1 + 2\cos t = 0\), which simplifies to \(2\cos t = 1\).

• Solving \(2\cos t = 1\) gives us \(\cos t = \frac{1}{2}\).
• Within \(0 \leq t \leq 2\pi\), this condition is satisfied at \(t = \frac{\pi}{3}\) and \(t = \frac{5\pi}{3}\).
• Plug these into the y-equation \(y = 1 + 2\sin t\).
• For \(t = \frac{\pi}{3}\), \(y = 1 + \sqrt{3}\); and for \(t = \frac{5\pi}{3}\), \(y = 1 - \sqrt{3}\).

The y-intercepts occur at these points, illustrating the curve's entry and exit points on the y-axis.

Trigonometric Functions

Trigonometric functions like sine and cosine are foundational in parametric equations, especially when these describe circular or wave-like patterns. The given parametric equations make use of these trigonometric functions to determine positions along a curve.

In the exercise, the cosine function, expressed in \(x = -1 + 2\cos t\), influences how far left or right along the x-axis the point travels.

• As \(\cos t\) varies between -1 and 1, it dictates the horizontal movement of the curve.
• For \(x = 0\), \(\cos t = \frac{1}{2}\), resulting in specific positions like \(t = \frac{\pi}{3}\) and \(t = \frac{5\pi}{3}\).

Similarly, the sine function in \(y = 1 + 2\sin t\) affects the vertical shift of the point along the y-axis.

• When \(\sin t = -\frac{1}{2}\), \(y\) equals zero, indicating the curve's irruption onto the x-axis.
• This trigonometric behavior is crucial for finding x- and y-intercepts in parametric forms.

Overall, trigonometric functions represent cyclical changes and oscillations, critical for understanding the motion and intercepts of parametric curves.