- #1
rocomath
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A runner sprints around a circular track of radis 100m at a constant speed of 7m/s. The runner's friend is standing at a distance 200m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200m.
http://img213.imageshack.us/img213/6400/37gb7.jpg
Equations: Area of a Triangle, Law of Cosines, and Sector of a Circle.
[tex]A = \frac {1}{2}bh \ (\sin\theta = hr) \rightarrow A = \frac {1}{2}br\sin\theta[/tex]
[tex]A = \frac {1}{2}r^{2}\theta \ (s = r\theta) \rightarrow A = \frac {s^2}{2\theta}[/tex]
Setting Area of the Triangle and Sector of the Circle equal to each other ...
[tex]\frac {1}{2}br\sin\theta = \frac {s^2}{2\theta} \rightarrow 200\sin\theta = s[/tex]
Using Law of Cosines to relate l with theta ...
[tex]l^2 = b^2 + r^2 - 2br\cos\theta \rightarrow l^2 = 50,000 - 40,000\cos\theta[/tex]
[tex]\theta = \cos^{ - 1}\left(\frac {b^2 + r^2 - l^2}{2br}\right)[/tex]
Taking the derivative ...
[tex]\frac {d}{dt}(200\sin\theta) = \frac {d}{dt}(s) \rightarrow \frac {d\theta}{dt} = \frac {1}{200\cos\theta}\frac {ds}{dt}[/tex]
[tex]\frac {d}{dt}(l^2) = \frac {d}{dt}(50,000 - 40,000\cos\theta) \rightarrow \frac {dl}{dt} = 100\sin\theta\frac {d\theta}{dt}[/tex]
Substituting ...
[tex]\frac {dl}{dt} = \frac {1}{2}\tan\theta\frac {ds}{dt} = \frac {1}{2}\tan\left[\cos^{ - 1}\left(\frac {b^2 + r^2 - l^2}{2br}\right)\right]\frac {ds}{dt} \approx 13.56m/s[/tex]
Actual answer = 6.78m/s
http://img213.imageshack.us/img213/6400/37gb7.jpg
Equations: Area of a Triangle, Law of Cosines, and Sector of a Circle.
[tex]A = \frac {1}{2}bh \ (\sin\theta = hr) \rightarrow A = \frac {1}{2}br\sin\theta[/tex]
[tex]A = \frac {1}{2}r^{2}\theta \ (s = r\theta) \rightarrow A = \frac {s^2}{2\theta}[/tex]
Setting Area of the Triangle and Sector of the Circle equal to each other ...
[tex]\frac {1}{2}br\sin\theta = \frac {s^2}{2\theta} \rightarrow 200\sin\theta = s[/tex]
Using Law of Cosines to relate l with theta ...
[tex]l^2 = b^2 + r^2 - 2br\cos\theta \rightarrow l^2 = 50,000 - 40,000\cos\theta[/tex]
[tex]\theta = \cos^{ - 1}\left(\frac {b^2 + r^2 - l^2}{2br}\right)[/tex]
Taking the derivative ...
[tex]\frac {d}{dt}(200\sin\theta) = \frac {d}{dt}(s) \rightarrow \frac {d\theta}{dt} = \frac {1}{200\cos\theta}\frac {ds}{dt}[/tex]
[tex]\frac {d}{dt}(l^2) = \frac {d}{dt}(50,000 - 40,000\cos\theta) \rightarrow \frac {dl}{dt} = 100\sin\theta\frac {d\theta}{dt}[/tex]
Substituting ...
[tex]\frac {dl}{dt} = \frac {1}{2}\tan\theta\frac {ds}{dt} = \frac {1}{2}\tan\left[\cos^{ - 1}\left(\frac {b^2 + r^2 - l^2}{2br}\right)\right]\frac {ds}{dt} \approx 13.56m/s[/tex]
Actual answer = 6.78m/s
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