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One ship is sailing south at a rate of 5 knots, an and another is sailing east at a rate of 10 knots. At 2 P.M. the second ship was at the place occupied by the first ship one hour before. At what time was the distance between the ships not changing?

If we draw a (traiangular) diagram as depicted in the ASCII art below:

x

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y| / z

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Knowing the ship speed (at any time) and its positions at 2PM, we can say

x = 10(t-2) and y = 5(t-1), (z = distance between the two ships)

where t is the time in hours after 1PM ( so that at t=2, x = 5, y = 0 and dx/dt=10, dy/dt=5).

The variables x, y and z obey the Pythagoras theorem (at all time t)

x^2 + y^2 = z^2

Implicitly differentiating the equation with respect to t, we get

2x dx/dt + 2y dy/dt = 2z dz/dt.

So asking when the distance between the shops not changing (albeit momentarily) is asking for the moment

when dz/dt = 0. Now if we substitute dz/dt=0 and all the other information we know we ended up with

2* 10(t-2)*10 + 2* 5(t-1)* 5 = 2 * z * (0)

=> 200 (t-2) + 50 (t-1) = 0

=> 4 (t-2) + (t-1) = 0 (after diving by 50 on the equation)

=> 4t - 8 + t - 1 = 0 (remove parentheses)

=> 5t = 9

=> t = 9/5

t=0 corresponds to the time 1:00PM, so t=9/5 = 1 hour and 48 minutes corresponds to the time 2:48PM.

Hence at exactly 2;48PM the distance between the ships was not changing.

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Comment | thanks for your efforts, but the answer on my book is 1:48 PM?how is it?can you look it where its wrong? |

Calculus

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