A boat is held at a dock by a bow line which is wound about a circular windlass 3 feet higher than the bow of the boat. How fast is the bow line increasing its length at the instant the boat is 4 feet from the dock if the boat is drifting at a rate of 7 feet per second from the dock?
let the length of line be L ft
let the horizontal distance of the boat from the dock be x ft
You have a right-angled triangle, thus
x^2 + 3^2 = L^2
2x dx/dt = 2L dL/dt
for the given case: x = 4
L^2 = 4^2 + 3^2
L = 5
2(4)(7) = 2L dL/dt
dL/dt = 2(5)/(56) = 5/28 ft/s
Hi,
Thank you but why are we plugging in the 4 afterwards when it is already given so why don'y we just use it in the beginning?
Oh nevermind I think its because the initial x isn't given and they want us to find the x when it is 4 feet away right?
To find the rate at which the bow line is increasing its length, we need to use the concept of related rates.
Let's define the following variables:
- Let x be the distance between the boat and the dock (in feet).
- Let y be the length of the bow line (in feet).
- Let z be the distance from the windlass to the dock (in feet). Since the windlass is 3 feet higher than the bow of the boat, z is a constant value of 3 feet.
We are given that the boat is drifting at a rate of 7 feet per second from the dock. This means that dx/dt = 7 ft/s.
We want to find dy/dt, the rate at which the bow line is increasing its length when the boat is 4 feet from the dock.
We can establish a right triangle where the horizontal line represents x, the vertical line represents y, and the hypotenuse represents the bow line (which has constant length z = 3 feet).
Using the Pythagorean theorem, we have:
x^2 + y^2 = z^2
Differentiating both sides of the equation with respect to time (t), we have:
2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)
Now we can substitute the given values and solve for dy/dt:
2(4)(7) + 2y(dy/dt) = 2(3)(0)
8(7) + 2y(dy/dt) = 0
56 + 2y(dy/dt) = 0
2y(dy/dt) = -56
dy/dt = -56 / (2y)
Since we want to find dy/dt when the boat is 4 feet from the dock, we can substitute x = 4 into the equation x^2 + y^2 = z^2:
(4)^2 + y^2 = (3)^2
16 + y^2 = 9
y^2 = 9 - 16
y^2 = -7
Since we are dealing with a physical measurement, we discard the negative solution, so y = √7.
Substituting y = √7 into the equation dy/dt = -56 / (2y), we have:
dy/dt = -56 / (2√7)
dy/dt = -28 / √7 ft/s
dy/dt ≈ -11.832 ft/s
Therefore, the bow line is decreasing in length at a rate of approximately 11.832 feet per second when the boat is 4 feet from the dock.