A boat is pulled into a dock by a string attached to the bow of the boat and passing through a pulley on the dock that is1ft higher than the bow of the boat. If the string is pulled at a rate of 3 ft/second, how fast is the boat approaching the dock when it is 7 ft from the dock?
Make a diagram, you should have a right-angled triangle.
let the horizontal distance of the boat from the dock be x ft
let the hypotenuse, the length of the rope, be y ft.
then x^2 + 1^2 = y^2
2xdx/dt = 2ydy/dt
when x=7
y^2 = 7^2 + 1^ = 50
y = √50
and you are given dy/dt = -3 ft/s (I made it negative since y is decreasing)
sub in and you are on your way.
To solve this problem, we need to use related rates, specifically the concept of rates of change. Let's start by assigning some variables:
Let x represent the distance between the boat and the dock.
Let y represent the length of the string.
Let t represent time.
From the problem, we know that the string is being pulled at a rate of 3 ft/second, so dy/dt = 3 ft/second.
We are given that the string is passing through a pulley on the dock that is 1 ft higher than the bow of the boat. This means that the vertical distance between the string and the dock can be represented by y - x = 1.
Now, we need to find dx/dt, the rate at which the boat is approaching the dock when it is 7 ft from the dock.
To find dx/dt, we need to find an equation that relates x, y, and their rates of change.
From the Pythagorean theorem, we know that x^2 + (y - x)^2 = y^2.
Differentiate both sides of this equation with respect to time t:
2x(dx/dt) + 2(y - x)(dy/dt) = 2y(dy/dt)
Now, plug in the given values:
2(7)(dx/dt) + 2(1)(3) = 2(8)(3)
14(dx/dt) + 6 = 48
14(dx/dt) = 42
dx/dt = 42/14
dx/dt = 3 ft/second
Therefore, the boat is approaching the dock at a rate of 3 ft/second when it is 7 ft from the dock.