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 1m/s,how fast is the boat approaching the dock when it is 8 m from the dock?
draw the triangle
vertical distance
horizontal distance= sqrt(x^2-1)
hypotenuse= x
given dx/dt=1m/s
what is d(sqrt(x^2-1))/dt
d(hyp)/dt=1/2 * 1/sqrt(x^2-1) * 2xdx/dt
so when sqrt(x^2-1)=8, how fast then
1/2 (1/8)*2(sqrt(64+1)*1
= sqrt(65)/8 m/s
check my work.
To solve this problem, we can use the concept of related rates. Let's denote the distance between the boat and the dock as "x" and the height of the rope above the boat as "h". We are given that the rope is being pulled in at a rate of 1 m/s.
Now, let's set up some equations:
1. The Pythagorean theorem: x^2 + h^2 = (x + 1)^2
This equation represents the relationship between the distance, x, and the height, h.
2. Differentiate both sides of equation 1 with respect to time (t):
d/dt (x^2 + h^2) = d/dt ((x + 1)^2)
Using the chain rule, we get:
2x(dx/dt) + 2h(dh/dt) = 2(x + 1)(dx/dt)
Simplifying gives us:
x(dx/dt) + h(dh/dt) = (x + 1)(dx/dt)
We are interested in finding the rate of change of x (how fast the boat is approaching the dock) when x = 8 m. We can substitute the given values into the equation and solve for (dx/dt).
When x = 8:
8(dx/dt) + h(dh/dt) = (8 + 1)(1) ... substituting x = 8 and dx/dt = 1
8(dx/dt) + h(dh/dt) = 9
To solve for (dx/dt), we need to find the value of h when x = 8. We can use equation 1 to find h:
8^2 + h^2 = (8 + 1)^2
64 + h^2 = 81
h^2 = 81 - 64
h^2 = 17
h = sqrt(17)
Now we have h = sqrt(17) and x = 8:
8(dx/dt) + sqrt(17)(dh/dt) = 9
We can rearrange the equation to solve for (dx/dt):
8(dx/dt) = 9 - sqrt(17)(dh/dt)
(dx/dt) = (9 - sqrt(17)(dh/dt))/8
We know that (dh/dt) = -1 m/s since the rope is being pulled in at a rate of 1 m/s and the distance is decreasing. Thus, (dx/dt) = (9 - sqrt(17)(-1))/8
Simplifying further:
(dx/dt) = (9 + sqrt(17))/8
Therefore, when the boat is 8 m from the dock, it is approaching the dock at a rate of (9 + sqrt(17))/8 m/s.