Question
Give your answer accurate to 3 decimal places.
A man on a dock is pulling in a boat by means of a rope attached to the bow of the boat at a point that is 1 ft above water
level. The rope goes from the bow of the boat to a pulley located at the edge of the dock 7 ft above water level. If he pulls
in the rope at a rate of 2 ft/sec, how fast (in feet per second) is the boat approaching the dock when the point of
attachment is 14 ft from the dock?
A man on a dock is pulling in a boat by means of a rope attached to the bow of the boat at a point that is 1 ft above water
level. The rope goes from the bow of the boat to a pulley located at the edge of the dock 7 ft above water level. If he pulls
in the rope at a rate of 2 ft/sec, how fast (in feet per second) is the boat approaching the dock when the point of
attachment is 14 ft from the dock?
Answers
Bot
We can use the Pythagorean theorem to find the length of the rope:
sqrt((14)^2 + (7)^2) = sqrt(245) = 7sqrt(5)
Let x be the distance between the boat and the dock. Then we have:
x^2 + (7)^2 = (7sqrt(5))^2
x^2 = 196
x = 14
Taking the derivative of both sides of the equation above, we get:
2x(dx/dt) = 0
dx/dt = 0 (when x = 14)
So when the point of attachment is 14 ft from the dock, the boat is not approaching or moving away from the dock. Therefore, the boat's speed is 0 ft/sec.
sqrt((14)^2 + (7)^2) = sqrt(245) = 7sqrt(5)
Let x be the distance between the boat and the dock. Then we have:
x^2 + (7)^2 = (7sqrt(5))^2
x^2 = 196
x = 14
Taking the derivative of both sides of the equation above, we get:
2x(dx/dt) = 0
dx/dt = 0 (when x = 14)
So when the point of attachment is 14 ft from the dock, the boat is not approaching or moving away from the dock. Therefore, the boat's speed is 0 ft/sec.