SigepBrandon
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- Joined
- Feb 17, 2011
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Anyone care to check my work? I've not done calculus in quite a while and going back to school taking a teach yourself distance learning course, so I'm sure I made some algebra mistakes or my logic could be wrong... Any help is greatly appreciated. Also, this is my first attempt at coding, so please let me know if it is too confusing to follow. Thank you.
Suppose a particle rolling down the helix shaped track given by \(\displaystyle r(t)=<2*cos(t), 2*sin(t), t>\)
flies off the track when \(\displaystyle t=\frac{\pi}{4}\). Determine the point where the particle will hit the xy-plane.
to find a direction vector at \(\displaystyle t=\frac{\pi}{4}\)
\(\displaystyle r'(t)=<cos(t)-2*sin(t), sin(t)+2*cos(t), 1>\)
\(\displaystyle r'(\frac{\pi}{4})=<cos(\frac{\pi}{4})-2*sin(\frac{\pi}{4}), sin(\frac{\pi}{4})+2*cos(\frac{\pi}{4})\)
\(\displaystyle = <\frac{1}{\sqrt{2}}-2*\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}+2\frac{1}{\sqrt{2}},1>\)
\(\displaystyle = <\frac{1}{\sqrt{2}},\frac{3}{\sqrt{2}},1>\)
to find the point on the track at \(\displaystyle t=\frac{\pi}{4}\)
\(\displaystyle r(\frac{\pi}{4})=<2*cos(\frac{\pi}{4}), 2*sin(\frac{\pi}{4}), (\frac{\pi}{4})>\)
\(\displaystyle =<2*\frac{1}{\sqrt{2}}, 2*\frac{1}{\sqrt{2}}, \frac{\pi}{4}>\)
\(\displaystyle =<\frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}}, \frac{\pi}{4}>\)
by parametrized tangent line equation:
\(\displaystyle L(t)=(\frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}}, \frac{\pi}{4})+t*<\frac{1}{\sqrt{2}},\frac{3}{\sqrt{2}},1>\)
\(\displaystyle =<\frac{2}{\sqrt{2}}-\frac{1}{\sqrt{2}}*t, \frac{2}{\sqrt{2}}-\frac{3}{\sqrt{2}}*t, \frac{\pi}{4}+t >\)
and the particle will intersect the xy-plane when z=o so:
\(\displaystyle (\frac{\pi}{4})+t=0\) which implies t=\(\displaystyle -(\frac{\pi}{4})\)
Plugging that in to L(t), the particle should intersect at \(\displaystyle (\frac{2\pi}{\sqrt2},\frac{6\pi}{\sqrt2},0)\)
Suppose a particle rolling down the helix shaped track given by \(\displaystyle r(t)=<2*cos(t), 2*sin(t), t>\)
flies off the track when \(\displaystyle t=\frac{\pi}{4}\). Determine the point where the particle will hit the xy-plane.
to find a direction vector at \(\displaystyle t=\frac{\pi}{4}\)
\(\displaystyle r'(t)=<cos(t)-2*sin(t), sin(t)+2*cos(t), 1>\)
\(\displaystyle r'(\frac{\pi}{4})=<cos(\frac{\pi}{4})-2*sin(\frac{\pi}{4}), sin(\frac{\pi}{4})+2*cos(\frac{\pi}{4})\)
\(\displaystyle = <\frac{1}{\sqrt{2}}-2*\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}+2\frac{1}{\sqrt{2}},1>\)
\(\displaystyle = <\frac{1}{\sqrt{2}},\frac{3}{\sqrt{2}},1>\)
to find the point on the track at \(\displaystyle t=\frac{\pi}{4}\)
\(\displaystyle r(\frac{\pi}{4})=<2*cos(\frac{\pi}{4}), 2*sin(\frac{\pi}{4}), (\frac{\pi}{4})>\)
\(\displaystyle =<2*\frac{1}{\sqrt{2}}, 2*\frac{1}{\sqrt{2}}, \frac{\pi}{4}>\)
\(\displaystyle =<\frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}}, \frac{\pi}{4}>\)
by parametrized tangent line equation:
\(\displaystyle L(t)=(\frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}}, \frac{\pi}{4})+t*<\frac{1}{\sqrt{2}},\frac{3}{\sqrt{2}},1>\)
\(\displaystyle =<\frac{2}{\sqrt{2}}-\frac{1}{\sqrt{2}}*t, \frac{2}{\sqrt{2}}-\frac{3}{\sqrt{2}}*t, \frac{\pi}{4}+t >\)
and the particle will intersect the xy-plane when z=o so:
\(\displaystyle (\frac{\pi}{4})+t=0\) which implies t=\(\displaystyle -(\frac{\pi}{4})\)
Plugging that in to L(t), the particle should intersect at \(\displaystyle (\frac{2\pi}{\sqrt2},\frac{6\pi}{\sqrt2},0)\)