Problem 3 - Arbeitsbuch Mathematik für Ingenieure

Question

Die beiden Wege \(W_{1}: z_{1}(t)=\left(-1+t^{2}\right)+i t\) und \(W_{2}: z_{2}(t)=i-(1+i) t\). mit \(t \in \mathbb{R}\) haben zwei Schnittpunkte. Bestimmen Sie diese Punkte und die zugehörigen Schnittwinkel.

Step-by-Step Solution

Answer

Intersection points are found by solving the equation systems. The angles are determined using the dot product formula for tangent vectors.

Step 1 - Define the intersection points

To find the intersection points of the two paths, set the equations of the paths equal to each other: \[ z_1(t_1) = z_2(t_2) \]Solve for different values of \( t_1 \) and \( t_2 \).

Step 2 - Equate real and imaginary parts

Separate the real and imaginary parts of the complex equations. For \[ z_1(t_1) = -1 + t_1^2 + i t_1 \] and \[ z_2(t_2) = i - (1+i) t_2 \], we get: Real part: \[ -1 + t_1^2 = -t_2 \]Imaginary part: \[ t_1 = 1 - t_2 \]

Step 3 - Solve for t1 and t2

Solve one of the equations for \( t_2 \). From \[ t_1 = 1 - t_2 \], substitute into \[ -1 + t_1^2 = -t_2 \]:\[ -1 + (1 - t_2)^2 = -t_2 \]Simplify the quadratic equation and solve for \( t_2 \).

Step 4 - Find the intersection points

Plug the values of \( t_2 \) back into the equations to get corresponding values of \( t_1 \). Find the positions of the points using \( z_1(t_1) \) or \( z_2(t_2) \).

Step 5 - Determine the angles of intersection

To determine the angle of intersection, find the derivatives of \( z_1(t) \) and \( z_2(t) \) evaluated at the intersection points to get the tangent vectors. Use the dot product formula for vectors: \[ \theta = \text{arccos} \frac{v_1 \bullet v_2}{|v_1||v_2|} \]. This will give the angle between the tangent vectors.

Key Concepts

Complex Numbers

Complex numbers are fundamental to understanding the intersection of points along paths in the complex plane. A complex number is of the form \(z = a + bi\), where \(a\) is the real part and \(b\) is the imaginary part, and \(i\) represents the square root of -1. This form allows combining real and imaginary components, making it essential for representing two-dimensional data naturally.
In this problem, we deal with two paths parameterized by complex numbers: \(\text{Path 1: } W_1: z_1(t) = (-1+t^2) + i t\) and \(\text{Path 2: } W_2: z_2(t) = i - (1+i)t\). These equations represent how these paths move through the complex plane. For each value of \(t\), \(z_1(t)\) and \(z_2(t)\) determine specific points on these paths.
Key points to remember about complex numbers:

• They help represent two dimensions together.
• Combining their real and imaginary parts allows for complex movements in planes.

Understanding complex numbers is crucial for solving intersection problems like this one.

Intersection Points

Intersection points occur where two paths cross each other in the complex plane. To find these points, we set the equations of the paths to be equal: \(z_1(t_1) = z_2(t_2)\). This equality indicates that the values of \(t_1\) and \(t_2\) for which the paths coincide are the intersection points.
To solve this, we separate the real and imaginary parts of each equation. \(z_1(t_1) = -1 + t_1^2 + i t_1\) and \(z_2(t_2) = i - (1+i) t_2\) then become

• Real part: \(-1 + t_1^2 = -t_2\)
• Imaginary part: \(t_1 = 1 - t_2\)

By solving these two separate equations, we determine the values of \(t_1\) and \(t_2\) where the intersection occurs. Finally, substituting these values back into either path equation gives us the exact points of intersection.

Tangent Vectors

Tangent vectors help us understand the direction of paths at intersection points. To find these vectors, we need to compute the derivatives of the path equations with respect to \(t\):
For \(W_1: z_1(t) = (-1 + t^2) + it\), the derivative is \(z_1'(t) = 2t + i\).
For \(W_2: z_2(t)= i - (1+i)t\), the derivative is \(z_2'(t) = -(1+i)\). Evaluating these derivatives at the intersection points provides the tangent vectors' components.
Tangent vectors indicate the instantaneous direction of movement along the paths at the points of intersection. These vectors are essential for calculating the angle of intersection, as they tell us how sharply the paths cross each other.

Angle of Intersection

The angle of intersection between the two paths is found using the dot product of the tangent vectors at the intersection points. The formula for the angle \(\theta\) between two vectors \(v_1\) and \(v_2\) is: \(\theta = \text{arccos} \frac{v_1 \bullet v_2}{|v_1| |v_2|}\).
Here, \(v_1 \text{ and } \ v_2\) are the tangent vectors derived earlier. The dot product \(v_1 \bullet v_2\) is calculated as \(Re(v_1) \times Re(v_2) + Im(v_1) \times Im(v_2)\), while the magnitudes \(|v_1|\) and \(|v_2|\) are the lengths of these vectors, calculated using \(\sqrt{(Re(v))^2 + (Im(v))^2}\).
The angle of intersection tells us how the paths intersect—whether it be sharply or more gently. Knowing this angle helps in visualizing and understanding the geometric nature of the intersection.