We are tasked with finding the point where the parametric curve $\mathbf{r}(t) = \langle \sin(9t), \cos(9t), \sin(9t)\cos(18t) \rangle$ intersects the $xy$-plane, within the given interval $\frac{\pi}{9} < t < \frac{3\pi}{18}$.

Step 1: Condition for Intersection with the $xy$-Plane

For the curve to intersect the $xy$-plane, the $z$-coordinate must be zero. From the vector $\mathbf{r}(t)$, the $z$-coordinate is given by: $z(t) = \sin(9t) \cos(18t)$ We need to solve: $\sin(9t) \cos(18t) = 0$

Step 2: Solving for $t$

There are two cases that can make the product zero:

1. $\sin(9t) = 0$
2. $\cos(18t) = 0$

Case 1: $\sin(9t) = 0$

The general solution for $\sin(9t) = 0$ is: $9t = n\pi \quad \text{where} \quad n \in \mathbb{Z}$ Thus: $t = \frac{n\pi}{9}$ Now, we check which $t$ values lie within the interval $\frac{\pi}{9} < t < \frac{3\pi}{18}$, which simplifies to $\frac{\pi}{9} < t < \frac{\pi}{6}$.

For $n = 1$: $t = \frac{\pi}{9}$ But this is not within the open interval $\left(\frac{\pi}{9}, \frac{\pi}{6}\right)$.

For $n = 2$: $t = \frac{2\pi}{9}$ This value is within the interval.

Case 2: $\cos(18t) = 0$

The general solution for $\cos(18t) = 0$ is: $18t = \frac{\pi}{2} + n\pi$ Thus: $t = \frac{\pi}{36} + \frac{n\pi}{18}$ Now, we check which $t$ values lie within the interval $\frac{\pi}{9} < t < \frac{\pi}{6}$.

For $n = 1$: $t = \frac{\pi}{36} + \frac{\pi}{18} = \frac{3\pi}{36} = \frac{\pi}{12}$ This value is within the interval.

Step 3: Find the Intersection Points

We have two possible values for $t$ where $z(t) = 0$:

1. $t = \frac{2\pi}{9}$
2. $t = \frac{\pi}{12}$

Now, we plug these values into $\mathbf{r}(t) = \langle \sin(9t), \cos(9t), \sin(9t) \cos(18t) \rangle$ to find the corresponding points on the curve.

For $t = \frac{2\pi}{9}$:

$\sin(9t) = \sin(2\pi) = 0, \quad \cos(9t) = \cos(2\pi) = 1$ Thus, the point is: $\mathbf{r}\left(\frac{2\pi}{9}\right) = \langle 0, 1, 0 \rangle$

For $t = \frac{\pi}{12}$:

$\sin(9t) = \sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}, \quad \cos(9t) = \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$ Thus, the point is: $\mathbf{r}\left(\frac{\pi}{12}\right) = \left\langle \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}, 0 \right\rangle$

Final Answer:

The curve intersects the $xy$-plane at the points: $\mathbf{r}\left(\frac{2\pi}{9}\right) = (0, 1, 0) \quad \text{and} \quad \mathbf{r}\left(\frac{\pi}{12}\right) = \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}, 0\right)$

Would you like further details on any step?

Here are 5 related questions:

1. How do you find parametric equations for curves in 3D space?
2. What is the significance of a curve intersecting the xy-plane in vector functions?
3. How do sine and cosine functions affect parametric curves?
4. What are the general conditions for finding intersections of parametric curves with planes?
5. Can vector functions be used to describe physical motions in space?

Tip: To understand the behavior of parametric curves, graphing them can provide valuable insights into their intersections and shapes.