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?

---

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.