Question

The curves r1(t) = 2t, t2, t4 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

Answer:

Therefore the angle of intersection is $\theta =79.48^\circ$

Step-by-step explanation:

Angle at the intersection point of two carve is the angle of the tangents at that point.

Given,

$r_1(t)=(2t,t^2,t^4)$

and $r_2(t)=(sin t , sin5t, 2t)$

To find the tangent of a carve , we have to differentiate the carve.

$r'_1(t)=(2,2t,4t^3)$

The tangent at (0,0,0) is     [ since the intersection point is (0,0,0)]

$r'_1(0)=(2,0,0)$      [ putting t= 0]

$|r'_1(0)|=\sqrt{2^2+0^2+0^2} =2$

Again,

$r'_2(t)=(cos t ,5 cos5t, 2)$

The tangent at (0,0,0) is    

$r'_2(0)=(1 ,5, 2)$        [ putting t= 0]

$|r'_1(0)|=\sqrt{1^2+5^2+2^2} =\sqrt{30}$

If θ is angle between tangent, then

$cos \theta =\frac{r'_1(0).r'_2(0)}{|r'_1(0)|.|r'_2(0)|}$

$\Rightarrow cos \theta =\frac{(2,0,0).(1,5,2)}{2.\sqrt{30} }$

$\Rightarrow cos \theta =\frac{2}{2\sqrt{30} }$

$\Rightarrow cos \theta =\frac{1}{\sqrt{30} }$

$\Rightarrow \theta =cos^{-1}\frac{1}{\sqrt{30} }$

$\Rightarrow \theta =79.48^\circ$

Therefore the angle of intersection is $\theta =79.48^\circ$.