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?
- 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
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
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
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.