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# Calculus - Related rates problem?

Two boats leave the same port at the same time with one boat travelling North at 15 knots/hour and the other boat travelling west at 12 knots/hour. How fast is this distance between the two boats cahnging after 2 hours?

Thanks

### 3 Answers

- Anonymous1 decade agoFavorite Answer
First of all, the units that you have "knots/hour" is not an accurate unit. "Knots" is already a unit of speed and this is equivalent to 1.15 miles/hour.

What you have here is a right triangle configuration, i.e.,

C^2 = A^2 + B^2

where

C = distance between the boats at any time

A = distance travelled by the north-bound boat

B = distance travelled by the west-bound boat

Differentiating the above,

2C(dC/dt) = 2A(dA/dt) + 2B(dB/dt)

Modifying the above,

C(dC/dt) = A(dA/dt) + B(dB/dt)

After two hours,

A = 15*2 = 30

B = 12*2 = 24

and

C = sqrt(30^2 + 24^2)

C = 38.42

and since dA/dt = 15 and dB/dt = 12, then

38.42(dC/dt) = 30(15) + 24(12)

Solving for "dC/dt"

dC/dt = 19.21 knots

Hope this helps.

- Scrander berryLv 71 decade ago
Boats A goes north such that dA/dt = 15

Boats B goes west such that dB/dt = 12

After 2 hours

A has travelled 30 knots and boats B has travelled 24 knots

the distance "r" between them is

r² = 30² + 24²

r² = 1476

r = sqrt1476

at any point in time

r² = A² + B²

so

2r dr/dt = 2A dA/dt + 2B db/dt

r dr/dt = A dA/dt + B db/dt

sqrt1476 dr/dt = (30)(15) + (24)(12)

sqrt1476 dr/dt = 738

dr/dt = 738/sqrt1476

dr/dt = 19.21knots/hour (2dp)

- CDALv 61 decade ago
Simple pythagoras, relative speed is sqrt(15^2 + 12^2)

knots. This is sqrt(369) knots or 19.21 knots. The boats are receding from each other at 19.21 knots. The "after 2 hours" is irrelevant.