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A boat leaves a dock at 2:00 PM and travels due south at a speed of $ 20 km/h $. Another boat has been heading due east at $ 15 km/h $ and reaches the same dock at 3:00 PM. At what time were the boats closest together?

$2 : 21 : 36 \mathrm{PM}$

03:52

Wen Z.

01:57

Amrita B.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

Volume

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

Lectures

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

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