Question

The curves r1(t) = 4t, t2, t3 and r2(t) = sin(t), sin(5t), 2t intersect at the origin. find their angle of intersection, θ, correct to the nearest degree.

Answer 1

First get the tangent vectors by differentiating r1 and r2 
$r_1 '(t) = (4,2t,3t^2) \\ \\ r_2'(t) = (cos (t), 5 cos(5t), 2)$
Evaluate at t=0
$r_1' = (4,0,0) \\ \\ r_2' = (1,5,2)$

Use identity for angle between 2 vectors:
$u*v = |u| |v| cos \theta$

Evaluate dot product and unit vectors:
$u*v = (4,0,0)*(1,5,2) = 4 \\ \\ |u| = \sqrt{4^2 +0^2+0^2} = 4 \\ \\ |v| = \sqrt{1^2 + 5^2+2^2} = \sqrt{30}$

Sub into identity and solve for theta:
$4 = 4 \sqrt{30} cos \theta \\ \\ cos\theta = \frac{1}{\sqrt{30}} \\ \\ \theta = 79.48$

Answer:
Angle of intersection is about 79 degrees.