Final answer:
To find the angle of intersection between the curves r1(t) = 2t, t^2, t^3 and r2(t) = sin(t), sin(3t), 5t at the origin, we need to find their tangent vectors at that point and then use the dot product to find the angle between them. The tangent vectors are v1(t) = (2, 2t, 3t^2) and v2(t) = (cos(t), 3cos(3t), 5) respectively. By taking the dot product of these vectors at the origin, we find that the angle of intersection is approximately 60 degrees.
Step-by-step explanation:
To find the angle of intersection between two curves, we first need to find the tangent vectors of each curve at the point of intersection. The tangent vectors represent the direction of each curve at that point. To find the tangent vector of a curve, we take the derivatives of each component function with respect to t, the parameter.
For r1(t) = 2t, t^2, t^3, the tangent vector is v1(t) = (2, 2t, 3t^2).
For r2(t) = sin(t), sin(3t), 5t, the tangent vector is v2(t) = (cos(t), 3cos(3t), 5).
The angle of intersection between the two curves at the origin is the angle between their tangent vectors at that point. We can use the dot product to find this angle. The dot product of two vectors a and b is given by a · b = |a| * |b| * cos(θ), where θ is the angle between the vectors. So, we can write v1(0) · v2(0) = |v1(0)| * |v2(0)| * cos(θ).
Substituting the values, we have (2, 0, 0) · (1, 0, 5) = |(2, 0, 0)| * |(1, 0, 5)| * cos(θ). Simplifying, 2 * 1 * cos(θ) = 2 * 5 * cos(θ). Cancelling out the common terms, we have cos(θ) = 5/2. To find θ, we take the inverse cosine of 5/2, giving us θ ≈ 60 degrees.