Here's an interesting Calc problem. Maybe someone would like to give their input as to a 'simpler method'. If I remember correctly, years ago I had a calc professor who worked this problem in a simpler fashion which did not require such a monstrous integral and was solvable without technology. Besides, I just downloaded TexAide and I want to try it out.
An amusement park ride has cars that move up and down on a track.
The cars are attached to arms that are 20 feet long and connected to a common point on the axis of rotation(z-axis). The arms swing up and down as the cars move along the track. Their elevation is given by \(\displaystyle z=4sin^{2}{\theta}.\)
Find the length of the track.
Here's my approach with spherical coordinates:
\(\displaystyle \begin{array}{l}
\int\limits_0^{2\pi } {\sqrt {({\textstyle{{d\rho } \over {d\theta }}})^2 + \rho ^2 \sin ^2 \phi + \rho ^2 ({\textstyle{{d\phi } \over {d\theta }}})^2 } d\theta } \\
20\int\limits_0^{2\pi } {\sqrt {\sin ^2 \phi + ({\textstyle{{d\phi } \over {d\theta }}})^2 } d\theta } \\
20\int\limits_0^{2\pi } {\sqrt {1 - \cos ^2 \phi + ({\textstyle{{d\phi } \over {d\theta }}})^2 } d\theta } \\
20\int\limits_0^{2\pi } {\sqrt {1 - (\sin ^4 \theta )/25 + ({\textstyle{{d\phi } \over {d\theta }}})^2 } d\theta } \\
20\int\limits_0^{2\pi } {\sqrt {1 - {\textstyle{{\sin ^4 \theta } \over {25}}} + ({\textstyle{{(4\sin ^2 \theta )(\cos ^2 \theta )} \over {(\sin ^2 \theta + 5)(\cos ^2 \theta + 4)}}})} d\theta } \\
= 125.99 \\
\end{array}\)
I also done this with polar and arrived at the same answer. My question is, would anyone like to offer input as to a simpler method that does not result in a monstrous integral in which technology is necessary. It took my TI-92 about 20 minutes to spit out the answer.
An amusement park ride has cars that move up and down on a track.
The cars are attached to arms that are 20 feet long and connected to a common point on the axis of rotation(z-axis). The arms swing up and down as the cars move along the track. Their elevation is given by \(\displaystyle z=4sin^{2}{\theta}.\)
Find the length of the track.
Here's my approach with spherical coordinates:
\(\displaystyle \begin{array}{l}
\int\limits_0^{2\pi } {\sqrt {({\textstyle{{d\rho } \over {d\theta }}})^2 + \rho ^2 \sin ^2 \phi + \rho ^2 ({\textstyle{{d\phi } \over {d\theta }}})^2 } d\theta } \\
20\int\limits_0^{2\pi } {\sqrt {\sin ^2 \phi + ({\textstyle{{d\phi } \over {d\theta }}})^2 } d\theta } \\
20\int\limits_0^{2\pi } {\sqrt {1 - \cos ^2 \phi + ({\textstyle{{d\phi } \over {d\theta }}})^2 } d\theta } \\
20\int\limits_0^{2\pi } {\sqrt {1 - (\sin ^4 \theta )/25 + ({\textstyle{{d\phi } \over {d\theta }}})^2 } d\theta } \\
20\int\limits_0^{2\pi } {\sqrt {1 - {\textstyle{{\sin ^4 \theta } \over {25}}} + ({\textstyle{{(4\sin ^2 \theta )(\cos ^2 \theta )} \over {(\sin ^2 \theta + 5)(\cos ^2 \theta + 4)}}})} d\theta } \\
= 125.99 \\
\end{array}\)
I also done this with polar and arrived at the same answer. My question is, would anyone like to offer input as to a simpler method that does not result in a monstrous integral in which technology is necessary. It took my TI-92 about 20 minutes to spit out the answer.
