A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 15 km/h. Another boat has been heading due east at 20 km/h and reaches the same dock at 3:00 P.M. How many minutes past 2:00 P.M. were the boats closest together?
as usual, draw a diagram. The distance z between the boats at time t hrs after 2:00 is
z^2 = (15t)^2 + (20-20t)^2
2z dz/dt = 2(15t)(15) + 2(20-20t)(-20)
z dz/dt = 1250t - 800
clearly dz/dt=0 when t=8/1250 hours
Now just change that to a time of day.
To find out how many minutes past 2:00 P.M. the boats were closest together, we need to determine the time when they were closest together.
Let's start by analyzing the situation. One boat is traveling due south, while the other is heading due east. We can think of their movements as two perpendicular lines on a grid. The boat heading south is a straight vertical line, and the boat heading east is a straight horizontal line.
Since the boat heading east reaches the dock at 3:00 P.M., we know that it took one hour to reach that point. What this means is that the boat heading east traveled for one hour while the boat heading south traveled for a portion of that time.
To find out how much time the boat heading south traveled, we need to determine the distance between the dock and the point where the two boats were closest together.
The distance between two points can be determined by using the Pythagorean theorem: a² + b² = c². In this case, the two sides are the distances traveled by each boat, and the hypotenuse is the distance between the two boats when they were closest together.
Let's assume that the distance traveled by the boat heading east is 'x' km. Since it was traveling at a speed of 20 km/h for one hour, we have x = 20 km.
The distance traveled by the boat heading south is a bit trickier to determine. Since we don't know when the boats were closest together, let's assume they were closest together after 't' hours from 2:00 P.M.
Since the boat was traveling at a speed of 15 km/h, its distance traveled after 't' hours would be 15t km.
So, using the Pythagorean theorem, the distance between the two boats is √(15t)² + (20)² = √(225t² + 400).
To find the time when the boats were closest together, we need to find the value of 't' that minimizes the distance between them. This can be achieved by minimizing the expression inside the square root, which is (225t² + 400).
To find the minimum value of this expression, we take its derivative with respect to 't' and set it equal to zero. Let's differentiate the expression and solve for 't':
d/dt (225t² + 400) = 2(225t) = 0
450t = 0
t = 0
Since the derivative is zero at t = 0, it means this is a minimum point. Therefore, the boats were closest together at t = 0, which means right at the start, at 2:00 P.M.
So, the boats were closest together immediately at 2:00 P.M., and no minutes had passed.
Now we can answer the original question: No minutes past 2:00 P.M. were the boats closest together.