calculus
posted by LALA .
a boat leaves a dock at 2:00 pm and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15 km/h and reaches the same dock at 3:00 pm. at what time were the two boats closest together

make a diagram.
It took the eastbound ship 1 hr to reach the dock, so when the southbound ship was at the dock, the eastbound ship was 15 km from the dock
So after t hrs, the southbound ship went 20t
and the eastbound ship went 15t
I see a rightangled triangle with a vertical of 20t and a horizontal of 1515t
let d be the distance between them
d^2 = (20t)^2 + (1515t)^2
2d dd/dt = 2(20t)(20) + 2(1515t)(15)
at a minimum of d, dd/dt = 0
800t  30(1515t) = 0
800t  450 + 450t = 0
1250t = 450
t = .36 hrs or 21.6 minutes
so they were closest at 2:21:36 pm
check:
when t = .36 , d^2 = 51.84 + 92.16 = 144, d = 12
take a value slightly higher and smaller
t = .37 , d^2 = 54.76 + 89.3025 = 144.0625 , d = 12.0026 , slightly farther
t = .35 , d^2 = 49 + 95.0625 = 144.0625 , d = 12.0026 , again slightly farther
my answer is correct
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