Find the point of intersection of the tangent lines to the curve 

**r**(*t*) = 2 sin(𝜋*t*), 4 sin(𝜋*t*), 5 cos(𝜋*t*)

 at the points where 

*t* = 0

 and 

*t* = 0.5

We are given the vector function \(\mathbf{r}(t) = (2 \sin(\pi t), 4 \sin(\pi t), 5 \cos(\pi t))\), and we need to find the point of intersection of the tangent lines to this curve at \( t = 0 \) and \( t = 0.5 \).

### Step 1: Find the position vectors at \( t = 0 \) and \( t = 0.5 \)
- When \( t = 0 \):
  \[
  \mathbf{r}(0) = (2 \sin(0), 4 \sin(0), 5 \cos(0)) = (0, 0, 5)
  \]

- When \( t = 0.5 \):
  \[
  \mathbf{r}(0.5) = (2 \sin(\pi \cdot 0.5), 4 \sin(\pi \cdot 0.5), 5 \cos(\pi \cdot 0.5)) = (2, 4, 0)
  \]

### Step 2: Find the velocity (tangent) vector \(\mathbf{v}(t)\)
The tangent to the curve at any point is given by the derivative \(\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt}\).

\[
\mathbf{v}(t) = \left( 2 \pi \cos(\pi t), 4 \pi \cos(\pi t), -5 \pi \sin(\pi t) \right)
\]

### Step 3: Calculate the tangent vectors at \( t = 0 \) and \( t = 0.5 \)

- When \( t = 0 \):
  \[
  \mathbf{v}(0) = (2 \pi \cos(0), 4 \pi \cos(0), -5 \pi \sin(0)) = (2 \pi, 4 \pi, 0)
  \]

- When \( t = 0.5 \):
  \[
  \mathbf{v}(0.5) = (2 \pi \cos(\pi \cdot 0.5), 4 \pi \cos(\pi \cdot 0.5), -5 \pi \sin(\pi \cdot 0.5)) = (0, 0, -5 \pi)
  \]

### Step 4: Parametrize the tangent lines
We now have the tangent lines at the points \( t = 0 \) and \( t = 0.5 \):

- The tangent line at \( t = 0 \) can be written as:
  \[
  \mathbf{r}_1(s) = \mathbf{r}(0) + s \mathbf{v}(0) = (0, 0, 5) + s(2 \pi, 4 \pi, 0) = (2 \pi s, 4 \pi s, 5)
  \]

- The tangent line at \( t = 0.5 \) can be written as:
  \[
  \mathbf{r}_2(s) = \mathbf{r}(0.5) + s \mathbf{v}(0.5) = (2, 4, 0) + s(0, 0, -5 \pi) = (2, 4, -5 \pi s)
  \]

### Step 5: Solve for the point of intersection
To find the point of intersection, set the parametric equations equal:
\[
(2 \pi s_1, 4 \pi s_1, 5) = (2, 4, -5 \pi s_2)
\]

This gives the following system of equations:
1. \( 2 \pi s_1 = 2 \)
2. \( 4 \pi s_1 = 4 \)
3. \( 5 = -5 \pi s_2 \)

From equation (1):
\[
s_1 = \frac{2}{2 \pi} = \frac{1}{\pi}
\]

From equation (3):
\[
s_2 = -\frac{5}{5 \pi} = -\frac{1}{\pi}
\]

### Step 6: Verify and find the intersection point
Using \( s_1 = \frac{1}{\pi} \) in the parametric equation of the first tangent line:
\[
\mathbf{r}_1\left( \frac{1}{\pi} \right) = (2 \pi \cdot \frac{1}{\pi}, 4 \pi \cdot \frac{1}{\pi}, 5) = (2, 4, 5)
\]

Using \( s_2 = -\frac{1}{\pi} \) in the parametric equation of the second tangent line:
\[
\mathbf{r}_2\left( -\frac{1}{\pi} \right) = (2, 4, -5 \pi \cdot \left( -\frac{1}{\pi} \right)) = (2, 4, 5)
\]

Thus, the point of intersection of the tangent lines is:
\[
\boxed{(2, 4, 5)}
\]

Would you like more details or have any questions?

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Here are 5 related questions to explore further:

1. How do you compute the tangent to a curve in 3D space?
2. What is the geometric significance of a tangent vector?
3. How do parametric curves behave when \( t \) varies?
4. What is the difference between the velocity vector and the acceleration vector for a curve?
5. How do you find points of intersection for two parametric lines in 3D?

**Tip:** Always check the consistency of parametric solutions by plugging them back into the original equations.

Find the point of intersection of the tangent lines to the curve r(t) = 2 sin(𝜋*t), 4 sin(𝜋*t), 5 cos(𝜋*t) at the points where t = 0 and t = 0.5.

Learn how to find the intersection of tangent lines to the parametric curve r(t) = 2 sin(πt), 4 sin(πt), 5 cos(πt) at t = 0 and t = 0.5. This problem covers key vector calculus concepts, including tangent vectors, parametric lines, and solving systems of equations in 3D space. Suitable for advanced high school and undergraduate students.

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