Math Problem Statement

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 (*, *, *).)

Solution

The vector function given is:

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

To find the point where r(t)\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=0z = 0.

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

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

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

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

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

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

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

2t=nπt=nπ22t = n\pi \quad \Rightarrow \quad t = \frac{n\pi}{2}

We are restricted to the interval:

2π7<t<4π9\frac{2\pi}{7} < t < \frac{4\pi}{9}

Check possible values of tt in this interval. For t=π2t = \frac{\pi}{2}, we have:

π21.57and2π70.897,4π91.396\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=π2t = \frac{\pi}{2} is not in the interval, this case does not give a solution.

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

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

4t=π2+nπt=π8+nπ44t = \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=0n = 0:

t=π80.392t = \frac{\pi}{8} \approx 0.392

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

t=π8+π4=3π81.178t = \frac{\pi}{8} + \frac{\pi}{4} = \frac{3\pi}{8} \approx 1.178

This value of t=3π8t = \frac{3\pi}{8} lies in the interval 2π7<t<4π9\frac{2\pi}{7} < t < \frac{4\pi}{9}.

Coordinates of the Point

Now, substitute t=3π8t = \frac{3\pi}{8} into the x and y components of r(t)\mathbf{r}(t):

x=sin(2t)=sin(2×3π8)=sin(3π4)=22x = \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(2×3π8)=cos(3π4)=22y = \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 r(t)\mathbf{r}(t) intersects the xy-plane is:

(22,22,0)\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=0z = 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.

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Math Problem Analysis

Mathematical Concepts

Parametric Equations
Vector Functions
Trigonometry
Plane Intersections

Formulas

sin(2t)cos(4t) = 0 to determine the intersection with the xy-plane
x = sin(2t), y = cos(2t), z = sin(2t)cos(4t)

Theorems

Trigonometric Identity Solutions
Plane Intersections in Parametric Equations

Suitable Grade Level

Grades 11-12 or College