1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Related Rates - Sector/Cosine

  1. Jan 3, 2008 #1
    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.

    [​IMG]

    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
     
    Last edited: Jan 3, 2008
  2. jcsd
  3. Jan 3, 2008 #2

    dynamicsolo

    User Avatar
    Homework Helper

    Why not just start from the Law of Cosines?
    I'll give you a tip: implicit differentiation is the heart of related rates problems. If you differentiate your result here with respect to time, you have

    [tex]\frac {d}{dt}(l^2) = \frac {d}{dt}(50,000 - 40,000\cos\theta) \rightarrow 2l \frac {dl}{dt} = 40,000\sin\theta\frac {d\theta}{dt}[/tex] .

    They were nice to you in this problem and told you l = 200 m right off, giving you

    [tex]\frac {dl}{dt} = 100\sin\theta\frac {d\theta}{dt}[/tex] ,

    as you found. The Law of Cosines gives us

    [tex]l^2 = 40,000 = 50,000 - 40,000\cos\theta[/tex],

    so [tex]\cos\theta = \frac{1}{4}[/tex] , making [tex]\sin\theta[/tex] equal to [sqrt(15)]/4 .

    The arclength of the runner's path is increasing at 7 m/sec, so the angular rate is

    [tex]\frac {d\theta}{dt} = \frac{7}{100} rad/sec[/tex].

    So at last we have

    [tex]\frac {dl}{dt} = 100 (srqt 15)(1/4)(7/100) = 6.78 m/sec[/tex].

    I'd avoid dealing with the angle explicitly, since the derivatives get messy and leave one more open to error (especially on a timed exam).

    I think I see where the extra factor of 2 got into your analysis. I don't think the areas argument is actually helpful in this problem; you will need the runner's angular speed around the circular path, but that comes directly from the definition of angular velocity. The trigonometric equation for the sides of the triangle is sufficient to work from
     
    Last edited: Jan 3, 2008
  4. Jan 3, 2008 #3
    Ah, you are a miracle worker :-]

    Question: Did you find the angular rate through the question? When it states that "the runner sprints around a circular track of radius 100m at a constant speed of 7m/s" ... so 7/100?

    nvm, you used D=rtheta
     
    Last edited: Jan 3, 2008
  5. Jan 4, 2008 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Not quite. The 7 m is measured around the circumference of a circle while the 100 m is a radius. You have to compare circumferences, not radii. Radian measure can be defined by "[itex]\theta[/itex]= "arc length" / circumference of circle" so a distance of 7 meters around a circle of radius 100m, diameter 200 m and so circumference [itex]2000\pi[/itex]. The radian measure of the angle is [itex]7/(200\pi)[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Related Rates - Sector/Cosine
  1. Related Rates (Replies: 2)

Loading...