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 9 m from the dock?
I am getting "-[(82)^(1/2)]/9", but the computer isn't liking it.
I also get -√82/9 m/s
The negative signs shows me that the distance between the dock and the boat is decreasing.
Try entering the positive result of that.
or the decimal equivalent of 1.006
To solve this problem, we can use related rates, which involves differentiating a given equation with respect to time. Let's denote the distance between the boat and the dock as "x" and the height of the pulley above the boat as "y".
We are given that the rope is being pulled in at a rate of 1 m/s, which means dx/dt = -1 (since the distance is decreasing). We need to find the rate at which the boat is approaching the dock, which is dy/dt.
Using the Pythagorean theorem, we have the equation:
x^2 + y^2 = (x + 1)^2
To solve for dy/dt, we need to differentiate this equation implicitly with respect to time:
2x(dx/dt) + 2y(dy/dt) = 2(x + 1)(dx/dt)
Substituting the given values, we have:
2(9)(-1) + 2y(dy/dt) = 2(9 + 1)(-1)
Simplifying the equation:
-18 + 2y(dy/dt) = -20
Now, let's solve for dy/dt:
2y(dy/dt) = -20 + 18
2y(dy/dt) = -2
dy/dt = -2/(2y)
dy/dt = -1/y
To find the value of y when the boat is 9 m from the dock, we can use the Pythagorean theorem again:
9^2 + y^2 = (9 + 1)^2
81 + y^2 = 100
y^2 = 100 - 81
y^2 = 19
y = √19
Substituting this value back into the equation for dy/dt, we get:
dy/dt = -1/√19
So, when the boat is 9 m from the dock, the boat is approaching the dock at a rate of -1/√19 m/s.
Note: Please check your calculations again, as the negative sign might have been missed when entering the answer.