anonymous
  • anonymous
A man walks along a straight path at a speed of 4ft/s. A searchlight is located on the ground 20ft from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 15 ft from the point on the path closest to the searchlight.
Mathematics
jamiebookeater
  • jamiebookeater
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
Now note that tan(y) = x=20, so that sec^2y *dy/dt=1/20 dx/dt At the instant described in the problem, x = 15 and the hypotenuse of the above triangle thus has length 25, so we compute: (25/20)^2 dy/dt=1/20*4 ==>dy/dt=1/2*400/625 =16/125 rad/sec
anonymous
  • anonymous
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anonymous
  • anonymous
hope this helps

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anonymous
  • anonymous
Thank you so mcuh! :)
anonymous
  • anonymous
anytime :)

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