Mathematics: 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.

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The curves r1(t) = 2t, t2, t4 and r2(t) = sin t, sin

agarcia24101993

19.01.2022

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 $\theta =79.48^\circ$ .

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