# Calculus Problem 2

Discussion in 'Community Discussion' started by Quboid, Dec 7, 2006.

1. ### Quboid macrumors 6502

Joined:
Oct 16, 2006
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everywhere
#1
One again i have a tricky calculus question and an exam tomorrow. Any help would be appreciated.

At 7:00am One boat (A) was 60 miles east of another boat (B). If boat A is navigated 20 miles per hour to the west and boat (B) is navigated 30 miles oer hour to the south east, at what time would the two boats be closest to each other?

Ans. 8:09:03am. (from the back of the text book)

This one is difficult, but i think in the diagram one should obtain a triangle and work from there. I am stuck....help please.........

2. ### bearbo macrumors 68000

Joined:
Jul 20, 2006
#2
i love math problems

okay... let me try this

assumption: boat b is going at 45 deg. angel south east

D = distance between the two boat
t = time in hr

reference point = initial point of B

Vbx = 1/sqrt(2)*30
Vby = -1/sqrt(2)*30

Vax = -20
Vay = 0

D = sqrt( ( (1/sqrt(2)*30 -20)*t+60 )^2 + (-1/sqrt(2)*30*t)^2)
dD = 0, find t

do you need me to solve it?

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Oct 16, 2006
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#3

4. ### bearbo macrumors 68000

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Jul 20, 2006
#4
what level of calculus is this class you are taking

also, can you use any kind of aid like calculator or software aid?

5. ### Quboid thread starter macrumors 6502

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Oct 16, 2006
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#5
University level calculus. I am allowed to use calculator.

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Jan 1, 2006
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UK
#6

7. ### swiftaw macrumors 603

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Jan 31, 2005
Location:
Omaha, NE, USA
#7
You can actually make the problem a little easier. Since the minimum value of D occurs at the same time as the minimum value of D^2 we can just find the value of t that minimizes D^2 = ( (1/sqrt(2)*30 -20)*t+60 )^2 + (-1/sqrt(2)*30*t)^2

That is find the t that makes the first derivative 0 and the second derivative positive.

Saves having to deal with that extra sqrt.

8. ### bearbo macrumors 68000

Joined:
Jul 20, 2006
#8
right, that'd be part of solving technique

however when i did that... i didn't get the right answer, i was gonna check later to see where i made the mistake

here it is anyway, someone check for me

D = sqrt( ( (1/sqrt(2)*30 -20)*t+60 )^2 + (-1/sqrt(2)*30*t)^2)

D1 = D^2, which has the local minimum where D has its local minimum

D1 = ( (1/sqrt(2)*30 -20)*t+60 )^2 + (-1/sqrt(2)*30*t)^2)
dD1/dt = 2((1/sqrt(2)*30-20)*t+60)*(1/sqrt(2)*30 -20)+2*1/2*900*t = 0
= ((sqrt(2)*30-40)*t+120)*(1/sqrt(2)*30-20)+900t
= (sqrt(2)*30*t-40*t+120)(1/sqrt(2)*30-20)+900t
= 1/sqrt(2)*30(sqrt(2)*30*t-40*t+120)-20*(sqrt(2)*30*t-40*t+120)+900t
= 900t-1200/sqrt(2)*t+3600/sqrt(2)-600*sqrt(2)*t+800*t-2400+900t
=900t-600*sqrt(2)*t+1800sqrt(2)-600*sqrt(2)*t+800t-2400+900t
therefore
(1800-1200sqrt(2))*t=2400-1800sqrt(2)
t = (2400-1800*sqrt(2))/(1800-1200*(sqrt(2))