Optimum values and w/ chain rule

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SUMMARY

The problem involves two boats departing from a dock at different times and speeds, with one boat heading west at 15 km/h and the other heading south at 12 km/h. The closest distance between the boats occurs 21 minutes and 36 seconds after the first boat departs. The distance function is derived using parametric equations for both boats, and the first derivative test is applied to find local extrema. The calculations confirm that minimizing the square of the distance simplifies the process.

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  • Understanding of parametric equations
  • Knowledge of derivatives and the first derivative test
  • Familiarity with distance formulas in coordinate systems
  • Basic graphing calculator skills
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  • Study parametric equations in calculus
  • Learn about the first derivative test for local extrema
  • Explore distance minimization techniques in coordinate geometry
  • Practice using graphing calculators for calculus problems
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Students in AP Calculus, educators teaching optimization problems, and anyone interested in applying calculus to real-world scenarios involving motion and distance.

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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.
 

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