A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 8 m from the dock?
If the rope is of length h, and the boat is x away from the dock,
h^2 = 1+x^2
When x=8, h = √65
2h dh/dt = 3x dx/dt
Since dh/dt = -1,
2√65 (-1) = 3*8 dx/dt
dx/dt = -√65/12
Since dx/dt < 0, the boat is approaching the dock at about 0.67 m/s
To solve this problem, we can use related rates and the Pythagorean theorem. Let's call the distance between the boat and the dock "x" and the distance between the pulley and the dock "y".
We are given that the rope is being pulled in at a rate of 1 m/s, which means that dx/dt = 1 (the rate of change of x with respect to time is 1).
To find the rate at which the boat is approaching the dock, we need to find the rate of change of x with respect to time, which is dx/dt.
By the Pythagorean theorem, we have:
x^2 + y^2 = (x + y)^2.
Differentiating both sides with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 2(x + y)(dx/dt + dy/dt).
We can simplify this equation by substituting the values we have:
2(8)(1) + 2(1)(dy/dt) = 2(8 + 1)(1 + dy/dt).
Simplifying further:
16 + 2(dy/dt) = 18(1 + dy/dt).
Now we can solve for dy/dt by rearranging the equation:
2(dy/dt) - 18(dy/dt) = 18 - 16,
-16(dy/dt) = 2,
dy/dt = -2/16.
Therefore, the boat is approaching the dock at a rate of -1/8 m/s (negative because it is getting closer).