Question

The curves \({r_1}(t) = \left\langle {t,{t^2},{t^3}} \right\rangle \) and \({r_2}(t) = \langle sint,sin2t,t\rangle \) intersect at the origin. Find their angle of intersection correct to the nearest degree.

Short answer

The angle between the two given curves is \({\cos ^{ - 1}}\frac{1}{{\sqrt 6 }} \approx {66^^\circ }{\rm{. }}\)

Step-by-step solution

Step 1: To find the angle of intersection

Remember that Angle between two curves at their point of intersection is the angle between their tangent vectors at that points.

\({{\bf{r}}_1}(t) = t,{t^2},{t^3}\)

Therefore

\({\bf{r}}_1^\prime (t) = 1,2t,3{t^2}\)

The parameter value corresponding to the point \((0,0,0)\) is \(t = 0\).

Therefore, the tangent vector there is

\({\bf{r}}_1^\prime (0) = \langle 1,0,0\rangle \)

\({{\bf{r}}_2}(t) = \langle \sin t,\sin 2t,\;\;\;t\rangle \)

Therefore

\({\bf{r}}_2^\prime (t) = \langle \cos t,2\cos 2t,\;\;\;1\rangle \)

The parameter value corresponding to the point \((0,0,0)\) is \(t = 0\)

Therefore, the tangent vector there is

\({\bf{r}}_2^\prime (0) = \langle 1,2,\;\;\;1\rangle \)

Step 2: Final Proof

If \(\theta \) is the angle between the two curves then

\(\cos \theta = \frac{{\left( {{\bf{r}}_1^\prime (0)} \right) \cdot \left( {{\bf{r}}_2^\prime (0)} \right)}}{{\left| {{\bf{r}}_1^\prime (0)} \right|\left| {{\bf{r}}_2^\prime (0)} \right|}}\;\;\;{\rm{ since }}{\bf{a}} \cdot {\bf{b}} = |a| \cdot |b| \cdot \cos \theta {\rm{ }}\)

\(\cos \theta = \frac{{\langle 1,0,0\rangle \cdot \langle 1,2,1\rangle }}{{\sqrt {1 + 0 + 0} \sqrt {1 + 4 + 1} }}{\rm{ }}\)

\(\cos \theta = \frac{1}{{\sqrt 6 }}\)

Therefore, angle between the two curves is \({\cos ^{ - 1}}\frac{1}{{\sqrt 6 }} \approx {66^^\circ }\)