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?
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.
To solve this problem, we can use the chain rule in calculus. Let's consider the following variables:
- x: represents the horizontal distance between the boat and the dock.
- y: represents the height of the boat above the water (y = 0 when the boat is level with the dock).
- h: represents the horizontal distance from the dock to the point of attachment of the rope.
We are given that the boat is being pulled in at a rate of 2 ft/sec, so dx/dt = 2. We want to find dy/dt, the rate at which the boat is approaching the dock.
Using the Pythagorean theorem, we can see that:
x^2 + y^2 = h^2
Differentiating both sides of this equation with respect to time (t), we have:
2x(dx/dt) + 2y(dy/dt) = 2h(dh/dt)
Given that h = 14 ft and dh/dt = 0 (since the pulley is fixed), we can substitute these values into the equation:
2x(2) + 2y(dy/dt) = 2(14)(0)
Simplifying the equation, we have:
4x + 2y(dy/dt) = 0
Now, we can solve for dy/dt:
2y(dy/dt) = -4x
dy/dt = -2x/y
To find the value of dy/dt when x = 14 ft, we need to find the value of y at that point. Using the Pythagorean theorem, we can calculate:
(14)^2 + y^2 = 7^2
196 + y^2 = 49
y^2 = 49 - 196
y^2 = -147 (Since y represents a measurement of distance, y^2 cannot be negative. So, there is no solution for y^2 = -147.)
Therefore, the boat is not approaching the dock at any point when the point of attachment is 14 ft from the dock.