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Math Help - Optimization

  1. #1
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    Optimization

    A Capybara needs to get from point A to point B around a circular pool (see diagram). If the Capybara swims half as fast as he can run, what angle should he jump in the pool to minimize the total travel time?
    Optimization-screenshot_1.png
    I know that to minimize time traveling you need to find the equation for time then derive and set to zero.
    So far I have that time swimming = 2t, time running = t, speed = distance/time
    I know that distance traveled on land is r*theta, but I have no idea how to find the distance traveled in water.
    Right now my equation is Time spent traveling = r*theta*t + (Distance traveled in water)*2t

    How can I find distance traveled by water? And is the rest of what I have done right?
    Last edited by nubshat; October 29th 2013 at 04:59 PM.
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  2. #2
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    Re: Optimization

    Quote Originally Posted by nubshat View Post
    So far I have that time swimming = 2t, time running = t
    Why do you think that?

    Quote Originally Posted by nubshat View Post
    I know that distance traveled on land is r*theta, but I have no idea how to find the distance traveled in water.
    Use the law of cosines.
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  3. #3
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    Re: Optimization

    Quote Originally Posted by emakarov View Post
    Why do you think that?
    I have that because the question says that the Capybara swims half as fast as he can run, so if it takes him 't' seconds to run a certain distance, it will take him twice as long to swim that distance(2t).
    Also, how can I use the cosine law if I don't know the angle to the right of theta, do I use 180-theta?
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  4. #4
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    Re: Optimization

    Quote Originally Posted by nubshat View Post
    I have that because the question says that the Capybara swims half as fast as he can run, so if it takes him 't' seconds to run a certain distance, it will take him twice as long to swim that distance(2t).
    Why do you think the Capybara runs and swims the same distance?

    Quote Originally Posted by nubshat View Post
    Also, how can I use the cosine law if I don't know the angle to the right of theta, do I use 180-theta?
    Yes.
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  5. #5
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    Re: Optimization

    ok I got that they don't travel the same distance, so now I have S_running=2S_swimming

    I used cosine law to find that the distance swimming = sqrt(r^2+r^2-2r^2cos(180-theta)) = rsqrt(2-2cos(180-theta)

    Then for for total time I have

    t=r*theta/2S_swimming + (rsqrt(2-2cos(180-theta)))/S_swimming . Can you please let me know if what I have is correct before I solve for the minimum time?
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  6. #6
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    Re: Optimization

    Quote Originally Posted by nubshat View Post
    Then for for total time I have

    t=r*theta/2S_swimming + (rsqrt(2-2cos(180-theta)))/S_swimming .
    Yes, this is correct. Note that \cos(180-\theta)=-\cos\theta.
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  7. #7
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    Re: Optimization

    Quote Originally Posted by emakarov View Post
    Yes, this is correct. Note that \cos(180-\theta)=-\cos\theta.
    ok thanks that will help a lot:P
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  8. #8
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    Re: Optimization

    I got theta = pi/3 when I solve it, but I plugged in theta=0 and theta = pi and I'm getting the minimum time to be when theta = 0(Capybara swims the while way), but I think the right answer is that the Capybara is suppose to run the whole way. Can anyone confirm/deny this?
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  9. #9
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    Re: Optimization

    Quote Originally Posted by nubshat View Post
    I got theta = pi/3 when I solve it
    Can you show your work?

    Quote Originally Posted by nubshat View Post
    but I plugged in theta=0 and theta = pi and I'm getting the minimum time to be when theta = 0(Capybara swims the while way), but I think the right answer is that the Capybara is suppose to run the whole way. Can anyone confirm/deny this?
    We have t=\frac{r}{S}(\theta/2 + \sqrt{2+2\cos\theta}. So, t(0) = 2r/S, t(\pi)=(\pi/2)r/S and \pi/2<2. The minimum for \theta\ge0 is indeed attained at \theta=\pi: see the graph.
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  10. #10
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    Re: Optimization

    Okay I got it now thanks a lot for all the help!
    I was being stupid and thought cos(pi) was 0 for some reason and I was getting that as the max value rather then the min value.
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