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The vector function given is:

\[
\mathbf{r}(t) = \langle \sin(2t), \cos(2t), \sin(2t)\cos(4t) \rangle
\]

To find the point where \(\mathbf{r}(t)\) intersects the xy-plane, we need to identify when the z-coordinate is 0, since the xy-plane is defined by \(z = 0\).

The z-component of \(\mathbf{r}(t)\) is:

\[
z(t) = \sin(2t)\cos(4t)
\]

Set \(z(t) = 0\):

\[
\sin(2t) \cos(4t) = 0
\]

This equation will be true if either \(\sin(2t) = 0\) or \(\cos(4t) = 0\).

### Case 1: \(\sin(2t) = 0\)

\(\sin(2t) = 0\) when:

\[
2t = n\pi \quad \Rightarrow \quad t = \frac{n\pi}{2}
\]

We are restricted to the interval:

\[
\frac{2\pi}{7} < t < \frac{4\pi}{9}
\]

Check possible values of \(t\) in this interval. For \(t = \frac{\pi}{2}\), we have:

\[
\frac{\pi}{2} \approx 1.57 \quad \text{and} \quad \frac{2\pi}{7} \approx 0.897, \, \frac{4\pi}{9} \approx 1.396
\]

Since \(t = \frac{\pi}{2}\) is not in the interval, this case does not give a solution.

### Case 2: \(\cos(4t) = 0\)

\(\cos(4t) = 0\) when:

\[
4t = \frac{\pi}{2} + n\pi \quad \Rightarrow \quad t = \frac{\pi}{8} + \frac{n\pi}{4}
\]

Now, check for values in the given interval. Let’s start with \(n = 0\):

\[
t = \frac{\pi}{8} \approx 0.392
\]

This is not in the interval. Next, try \(n = 1\):

\[
t = \frac{\pi}{8} + \frac{\pi}{4} = \frac{3\pi}{8} \approx 1.178
\]

This value of \(t = \frac{3\pi}{8}\) lies in the interval \(\frac{2\pi}{7} < t < \frac{4\pi}{9}\).

### Coordinates of the Point

Now, substitute \(t = \frac{3\pi}{8}\) into the x and y components of \(\mathbf{r}(t)\):

\[
x = \sin(2t) = \sin\left(2 \times \frac{3\pi}{8}\right) = \sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}
\]

\[
y = \cos(2t) = \cos\left(2 \times \frac{3\pi}{8}\right) = \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}
\]

Thus, the point where \(\mathbf{r}(t)\) intersects the xy-plane is:

\[
\left( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}, 0 \right)
\]

Would you like further clarification or details?

Here are five related questions to expand your understanding:
1. How do we determine when a vector intersects the xz-plane or yz-plane?
2. Why does setting \(z = 0\) help find intersections with the xy-plane?
3. Can we solve for intersections in a similar way for curves in 3D space?
4. How does the period of trigonometric functions affect the intervals where solutions occur?
5. Could we use numerical methods if no exact solution is obvious?

**Tip**: When solving for intersections with planes in vector functions, focus on the relevant coordinate component (e.g., z-component for the xy-plane) and solve accordingly.

Let r(t) = ⟨sin(2t), cos(2t), sin(2t)cos(4t)⟩. Find the point where r(t) intersects the xy-plane on the interval 2π/7 < t < 4π/9. (Use symbolic notation and fractions where needed. Give your answer as point coordinates in the form (*, *, *).)

Explore how to find the intersection of the vector function r(t) = ⟨sin(2t), cos(2t), sin(2t)cos(4t)⟩ with the xy-plane. Learn about trigonometric solutions and parametric intersections. Detailed step-by-step solution provided for students in Grades 11-12 or college-level courses.

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