angle of intersection of 3-dimensional curves

[mathstresser]

The curves r1(t)=<t,t^2,t^3> and r2(t)=<sint, sin2t, t> intersect at the origin. Find their angle of intersection correct to the nearest degree.

I know when you are changing from rectangluar to cylindrical, tan(theta)=y/x

But otherwise, I have no idea what to use. And even with that I don't know where to start.

 

[soroban]

Hello, mathstresser!

The curves \(\displaystyle \,r_1(t)\:=\L\langle t,\,t^2,\,t^3\rangle\,\) and \(\displaystyle \,r_2(t)\:=\:\langle\sin t,\,\sin2t,\,t\rangle\.\) intersect at the origin.
Find their angle of intersection correct to the nearest degree.

You need to find the tangent vectors at their intersection.

Then find the angle between the vectors with this formula:

\(\displaystyle \L\;\;\;\cos\theta \;=\;\frac{\vec{u}\cdot \vec{v}}{|\vec{u}||\vec{v}|}\)


The derivatives are: \(\displaystyle \,r'_1(t)\:=\:\langle 1,\,2t\,3t^2\rangle\;\;\;r'_2(t)\:=\:\langle \cos t,\,2\cos2t,\,1\rangle\)

The tangent vectors are: \(\displaystyle \,r'_1(0)\:=\:\langle 1,\,0,\,0\rangle\;\;\;r'_2(0)\:=\:\langle 1,\,2,\,1\rangle\)

Hence: \(\displaystyle \L\,\cos\theta \;= \;\frac{\langle1,\,0,\,0\rangle\cdot\langle1,\,2,\,1\rangle}{\sqrt{1^2+0^2+0^2}\,\sqrt{1^2+2^2+1^2}} \;=\;\frac{1\cdot1\,+\,0\cdot2\,+\,0\cdot1}{\sqrt{1+0+0}\,\sqrt{1+4+1}} \;=\;\frac{1\,+\,0\,+\,0}{\sqrt{1}\,{\sqrt{6}}}\;=\;\frac{1}{\sqrt{6}}\)

Therefore: \(\displaystyle \,\theta\;=\;\cos^{-1}\left(\frac{1}{\sqrt{6}}\right)\;=\;56.90515745\;\approx\;66^o\)

 

[mathstresser]

Thanks!