Optimum values and w/ chain rule

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The problem involves two boats departing from a dock at different times and speeds, with one heading west and the other south. The first boat leaves at 2:00 PM traveling at 15 km/h, while the second boat departs at 3:00 PM traveling at 12 km/h. The distances of both boats from the dock are expressed using parametric equations, allowing for the calculation of the distance between them over time. By finding the derivative of the distance function and applying the first derivative test, the closest point in time was determined to be 21 minutes and 36 seconds after the first boat's departure. It's advised to verify the calculations, especially given the complexity of the problem.
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A boat leaves dock at 2:00 PM, heading west at 15 km/h. Another boat heads south at 12 km/h and reaches the same dock at 3:00 PM. When were the boats closest to each other?

I've solved other problems similar to this one, but I can't seem to figure this one out. I'm not sure what I have to do with the times that are given... I know you guys like to see work done to show that I've tried it, but all I have is a crappy diagram...

thanks.
 
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On the assumption that this is parametric, that would make boat A:
x(t) = -15t
y(t) = 0

And Boat B:
x(t) = 0
y(t) = -12t+12

therefore,
d = ((x_a(t)-x_b(t))^2+(x_a(t)-x_b(t))^2)^0.5

Plug in the equations and simplify this(I have pity on you if you don't have a graphing calculator for this) and find the derivative of the distance with respect to time, which you can graph as you would any other derivative function. At that point you should know how to use the first derivative test for local extrema.

I got 21 minutes, 36 seconds. I'd definitely doublecheck, though, since I'm still in AP Cal AB and could very well be pulling a great deal of this out of my ass.
 
Last edited:
Since the derivative of x2 with respect to t is 2x(dx/dt), minimizing the distance is the same as minimizing the square of the distance. You don't need the 1/2 power.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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