A man pulls a rope over a simple pulley attached to the bow of a rowboat. The man’s hands are 8 feet above the water and the bow of the boat is 1 foot above the water. How fast must the man pull the rope to tow the boat in at 3 feet per second when it is 25 feet from the dock? How fast must he pull the rope to tow the boat at that same speed when it is 10 feet from the dock? If he pulls the rope at 3 feet per second when the boat is 18 feet from the dock, how fast is the boat moving?
Your situation can be described by a right-angled triangle, where the base is along the water.
let the boat be x ft from the dock, let the length of rope from the man to the boat be y ft
then y^2 = x^2 + 7^2
2y dy/dt = 2x dx/dt
y dy/dt = x dx/dt
case #1, dx/dt = -3 ft/sec, x = 25
y^2 = 25^2 + 7^ = 674 , ----> y = √674
√674 dy/dt = 25(-3)
dy/dt = -75/√674 = appr -2.89 ft/sec
case 2, when x = 10, dx/dt = -3
y^2 = 100+49 = 149 , ---> y = √149
√149 dy/dt = 10(-3)
dy/dt = --2.46 ft/s
case3 , when x = 28 , dy/dt = -3
y^2 = 784+49 = 833 , -----> y = √833
√833(-3 = 28 dx/dt
dx/dt = √833(-3)/28 = -3.09 ft/s
To solve these questions, we can use the concept of similar triangles.
Let's consider the first scenario where the boat is 25 feet from the dock. We have a triangle formed by the man's hands, the bow of the boat, and a point on the dock directly below the boat. The two vertical sides of this triangle are the man's hand height (8 feet) and the boat height (1 foot), respectively. The base of the triangle is the horizontal distance between the boat and the dock (25 feet).
Since the triangles formed by the man's hands, the boat, and the dock are similar, we can set up a proportion to find the rate at which the man needs to pull the rope to tow the boat at 3 feet per second. Let's call this rate "r1."
8 feet / 1 foot = 25 feet / r1
Simplifying this proportion, we get:
r1 = (1 foot * 25 feet) / 8 feet
= 25/8
= 3.125 feet per second
Therefore, the man needs to pull the rope at a rate of 3.125 feet per second to tow the boat at 3 feet per second when it is 25 feet from the dock.
Now, let's consider the second scenario where the boat is 10 feet from the dock. Again, we can set up a proportion to find the rate at which the man needs to pull the rope to tow the boat at 3 feet per second. Let's call this rate "r2."
8 feet / 1 foot = 10 feet / r2
Simplifying this proportion, we get:
r2 = (1 foot * 10 feet) / 8 feet
= 10/8
= 1.25 feet per second
Therefore, the man needs to pull the rope at a rate of 1.25 feet per second to tow the boat at 3 feet per second when it is 10 feet from the dock.
Finally, let's find the speed of the boat when the man pulls the rope at 3 feet per second and the boat is 18 feet from the dock. In this case, we know the rate at which the rope is being pulled (3 feet per second) and the base of the triangle formed by the man's hands, the boat, and the dock (18 feet). Let's call the speed of the boat "v."
3 feet per second / v = 18 feet / 1 foot
Simplifying this proportion, we get:
v = (3 feet per second * 1 foot) / 18 feet
= 3/18
= 1/6 feet per second
Therefore, the boat is moving at a speed of 1/6 feet per second when the man pulls the rope at 3 feet per second and the boat is 18 feet from the dock.