A boat leaves a dock at 3:00 P.M. 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 4:00 P.M. How many minutes after 3:00 P.M. were the two boats closest together?
kinda lost on this one
2007-11-08 06:16:15 · 2 answers · asked by syem64 2 in Science & Mathematics ➔ Mathematics
If we define the dock to be (0,0), then the position of the
first boat can be said to be:
Y1 = -20t X1 = 0
and the position of the 2nd boat is:
X2= -15 + 15t Y2 = 0
where t is in hours.
The distance between the boats is then:
D = sqrt( (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) )
or because x1 is 0 and y2 is 0
D = sqrt( x2*x2 + y1*y1) (the - on the x2's cancel out)
or
D = sqrt( (-15+15t)(-15+15t) + (-20t)*(-20t) )
D = sqrt( 225 -225t - 225t + 225t*t + 400t*t )
D = sqrt( 225 -450t +625t*t )
we want to know when D is the smallest. If D turns out > 1 this
will turn out to be the same time as D squared is smallest,
so we want to find the minimum of
225 - 450t + 625t*t
which we can find graphically, or....
If we know calculus, we know that this will happen when the
derivative is 0, so we can solve for:
-450 + 1250*t = 0
which happens when t = .36
remember that t is in hours, so to get the minutes
multiply by 60, and the answer is 21.6 minutes
2007-11-08 06:51:12 · answer #1 · answered by cryptogramcorner 6 · 0⤊ 0⤋
The person before me is right, except the derivative is -450 + 625t....there's no need to multiply 625 by 2.
2007-11-10 14:47:55 · answer #2 · answered by dh_settin_star 1 · 0⤊ 0⤋