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.

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

Use identity for angle between 2 vectors:
[tex]u*v = |u| |v| cos \theta[/tex]

Evaluate dot product and unit vectors:
[tex]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}[/tex]

Sub into identity and solve for theta:
[tex]4 = 4 \sqrt{30} cos \theta \\ \\ cos\theta = \frac{1}{\sqrt{30}} \\ \\ \theta = 79.48[/tex]

Answer:
Angle of intersection is about 79 degrees.

Cricetus Cricetus · Answer 2 · 2021-10-14T18:12:36+02:00

The Angle of intersection will be "[tex]79.48^{\circ}[/tex]".

According to the question,

[tex]r_1(t) = (4t, t^2, t^3)[/tex]
[tex]r_2(t) = (Sin (t), Sin (5t), 2t)[/tex]

Now,

The tangents of these two curves at the origin will be:

→ [tex]r_1' (t) = (4, 2t, 3t^2)[/tex]

[tex]= (4, 0,0)[/tex]

→ [tex]r_2'(t) = (Cos \ t, 5 Cos \ 5t, 2)_{t=0}[/tex]

[tex]= (1,5,2)[/tex]

As we know,

→ [tex]r_1' (0) , r_2'(0) = |r_1'(0)|. |r_2'(0)|. Cos \Theta[/tex]

→ [tex](4,0,0).(1,5,2) = \sqrt{4^2+0+0} . \sqrt{1^2+5^2+2^2}. Cos\Theta[/tex]

→ [tex]4 = 40\sqrt{30}. Cos\Theta[/tex]

→ [tex]Cos\Theta = \frac{1}{\sqrt{30} }[/tex]

→ [tex]\Theta = 79.48^{\circ}[/tex]

Thus the above answer is correct.

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