Question

the curves r1(t) = 4t, t2, t3 and r2(t) = sin(t), sin(3t), 4t intersect at the origin. find their angle of intersection, , correct to the nearest degree.

Answer 1

The angle of intersection between the curves can be found using the dot product of their tangent vectors at the point of intersection. First, find the tangent vectors of the curves by taking their derivatives. Then, evaluate the limit of the dot product divided by the product of the magnitudes.

Step-by-step explanation:

The angle of intersection between the curves can be found using the dot product of their tangent vectors at the point of intersection. First, we find the tangent vectors of the curves by taking the derivatives of each curve. The tangent vectors are r1'(t) = 4, 2t, 3t^2 and r2'(t) = cos(t), 3cos(3t), 4. At the point of intersection, the tangent vectors are both zero, so we cannot use the dot product directly. Instead, we can find the angle between the curves by evaluating the limit as t approaches zero of the dot product of the tangent vectors divided by their magnitudes.

Let's calculate the dot product and the magnitudes:

r1'(t) · r2'(t) = 4cos(t) + 6tcos(3t) + 12t^2

|r1'(t)| = sqrt(16 + 4t^2 + 9t^4)

|r2'(t)| = sqrt(cos^2(t) + 9cos^2(3t) + 16)

Now, we can evaluate the limit:

lim(t->0) (r1'(t) · r2'(t)) / (|r1'(t)| · |r2'(t)|) .