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We're told that a boat leaves the dock at 2
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p.m. And travels due south at a speed of 20
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kilometers per hour. Another boat has been heading due
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east at 15 kilometers per hour and reaches the same
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dock at 3 p.m. Were asked at what time where
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the boats closest together yourself, I'll consider the position
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of the dock as the origin. So the doctors
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at the 0.0 just look right now. Boat one
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is moving south at a speed of 20 kilometers per
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hour. So the position of but one is zero
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negative. 20 t where T is in ours and
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at 2 p.m. Both to is 15 kilometers due west
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of the dock and because it took the boat one
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hour to reach the dock at a speed of 15
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kilometers per hour. Its position then well, the
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position of boat to at 2 p.m. Is negative 15
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0, and we're told the position of the 02
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is changing with time because it is moving east with
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the speed of 15 kilometers per hour, therefore,
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its position at any time. T This is negative
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15 plus 15 t zero Yes, just And so
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now, since we know the positions of both boats
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at any time, you can use the distance between
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two points rule to write a function of the distance
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between these two boats. So this distance, which
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I'll call little D this is the square root of
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Let's see the X coordinate of boat too, which
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is negative. 15 plus 15 t minus the X
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coordinate of but one which is zero squared, plus
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the why coordinate about two which of course, is
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zero minus the Y coordinates of Boat one, which
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is negative. 20 tease. This is positive 20
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t squared and this simplifies to not work. Well
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, this sort of as a tricky question because in
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fact, our function d is minimized when the function
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F, which is D squared, is minimized so
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we can simplify this problem. So f is a
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function of t is negative. 15 plus 15 t
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squared plus 20 t squared. And if you foil
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this out, this is 625 t squared minus 450
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t mhm plus 225. So now to find the
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shortest distance or the sorry, the minimum of f
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. We'll take the derivative of F and set it
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equal to zero. So f prime of t is
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a two times 625. This is 12. 50
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T minus 4. 50 equals zero. Yes,
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and so we get the T is equal to 4
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. 50/12. 50 or t equals 9/25. It
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would be now using the first derivative test. We
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have that if t is less than nine. 25th
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, then it follows that f prime of T is
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less than zero. And that T is greater than
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nine. 25th than F Prime of T is greater
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than zero. Therefore, by the first derivative test
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, it follows that f has a minimum AT T
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equals nine 25th. Now what is this in terms
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of time? Well, this is in ours.
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We want to write this in clock notation. I
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guess you'd say so as an analog time. Okay
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, well, mhm nine, 25th hours. This
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is the same as well, there's 60 minutes in
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one hour. And so this is how many minutes
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? 108 5th minutes, which if you divide,
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we get 21 plus and We have a remainder of
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3/5 minutes now. What is this? 3/5 minutes
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in terms of seconds. Well, 3/5 minutes times
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. Once again, we have 60 seconds per one
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minute. This is 36 seconds and so T our
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time is equal to 21 minutes and 36 seconds.
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So adding this to the start of the journey at
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2 p.m. It follows the boats are closest at time
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. Two. 21 36 p. M. Yeah
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.