Calculus 1 Problem

Question

A boat leaves a dock at 2:00 PM and travels due south at a speed of 30 km/h. Another boat has been heading due east at 25 km/h and reaches the same dock at 4:00 PM. At what time were the two boats closest together?

Answer 1

The distance Boat #1 is from the dock (y) after 2:00 is 30t where t is the time in hours after 2:00.

The distance Boat #2 is from the dock (x) after 2:00 is 50 - 25t

Using the Pythagorean Theorem, d = √((30t)² + (50 - 25t)²)

Simplifying:

d = (900t² + (2500 - 2500t + 625t²))^1/2

d = (1525t² - 2500t + 2500)^1/2

d = (25(61t² - 100t + 100))^1/2

d(t) = 5(61t² - 100t + 100)^1/2

Taking the derivative:

d'(t) = 5/2(61t² - 100t + 100)^-1/2(122t - 100)

d'(t) = (5(122t - 100))/(2(61t² - 100t + 100)^1/2)

d'(t) = (610t - 500)/(2(61t² - 100t + 100)^1/2)

d'(t) = (305t - 250)/(61t² - 100t + 100)^1/2

Setting this equal to zero:

(305t - 250)/(61t² - 100t + 100)^1/2 = 0

This can only equal zero if the numerator equals zero, therefore:

305t - 250 = 0

305t = 250

t = 250/305

t = 0.82 hours

0.82 hours is (0,82)(60) = 49 minutes

Therefore, the boats are closest at 2:49 pm