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

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

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)|=√(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)|=√(1^2+5^2+2^2) =√(30)`

If θ is angle between tangent, then

`cos \theta =(r'_1(0).r'_2(0))/(|r'_1(0)|.|r'_2(0)|)`

`\Rightarrow cos \theta =((2,0,0).(1,5,2))/(2.√(30) )`

`\Rightarrow cos \theta =(2)/(2√(30) )`

`\Rightarrow cos \theta =(1)/(√(30) )`

`\Rightarrow \theta =cos^(-1)(1)/(√(30) )`

`\Rightarrow \theta =79.48^\circ`

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