a woman on a dock is using a rope to pull in a canoe. the rope is pulled at the rate of 2 ft/sec, 3ft above the point level with the connection of the rope to the canoe. How fast is the canoe approaching the dock when the length of the rope from her hands to the canoe is 10 ft?
similar to your previous question.
make sketch of right-angled triangle
height = 3, base = x, hypotenuse = y
y^2 = x^2 + 9
follow the same steps I showed you in previous post
Thanks Darling.
To solve this problem, we can use the concept of related rates. We need to find the rate at which the canoe is approaching the dock, given that the rope is being pulled at a constant rate.
Let's assign some variables to the quantities involved:
- Let x be the distance between the canoe and the dock.
- Let y be the length of the rope from the woman's hands to the canoe.
We are given that dx/dt = 2 ft/sec (the rate at which the rope is being pulled inward) and y = 10 ft.
We need to find dy/dt, the rate at which the length of the rope is changing, when y = 10 ft.
Using the Pythagorean theorem, we know that x^2 + y^2 = d^2, where d is the distance between the woman's hands and the dock (which remains constant). Taking the derivative of both sides with respect to time, we get:
2x(dx/dt) + 2y(dy/dt) = 2d(dd/dt)
Since d is constant, dd/dt = 0. Therefore, our equation becomes:
2x(dx/dt) + 2y(dy/dt) = 0
Substituting the given values dx/dt = 2 ft/sec and y = 10 ft, and rearranging the equation, we have:
2x(2) + 2(10)(dy/dt) = 0
Simplifying further, we get:
4x + 20(dy/dt) = 0
Now, we can solve for dy/dt:
20(dy/dt) = -4x
dy/dt = -4x/20
Plugging in x = 6 ft (since the rope is 3 ft above the point level with the connection of the rope to the canoe), we have:
dy/dt = -4(6)/20
dy/dt = -24/20
Simplifying, we get:
dy/dt = -6/5 ft/sec
Therefore, the canoe is approaching the dock at a rate of -6/5 ft/sec (negative sign indicating that it is moving inward).