A barge, whose deck is 10 ft below the level of a Dock, is being drawn in by means of a cable attached to the deck and passing through a ring on the dock. When the barge is 24ft away and. Approaching the Dock at 3/4 per second. How fast is the cable being pulled?
To find the speed at which the cable is being pulled, we can use the concept of related rates.
Let's assign some variables:
- Let x be the distance between the barge and the dock.
- Let y be the length of the cable.
- Let t be time in seconds.
Given information:
- x = 24 ft
- dx/dt = -3/4 ft/s (negative sign because the barge is approaching the dock)
We want to find dy/dt, the rate at which the cable is being pulled.
Now, we can set up the equation using the Pythagorean theorem:
x^2 + y^2 = (10ft)^2
Differentiating both sides of the equation with respect to time (t):
2x(dx/dt) + 2y(dy/dt) = 0
Substituting the given values:
2(24 ft)(-3/4 ft/s) + 2y(dy/dt) = 0
Simplifying the equation:
-36 ft/s + 2y(dy/dt) = 0
2y(dy/dt) = 36 ft/s
dy/dt = 36/2y ft/s
Now, we need to find the value of y, the length of the cable. Using the Pythagorean theorem:
x^2 + y^2 = (10ft)^2
(24 ft)^2 + y^2 = 100 ft^2
576 ft^2 + y^2 = 100 ft^2
y^2 = 100 ft^2 - 576 ft^2
y^2 = -476 ft^2 (discard since y cannot be negative in this context)
y = √476 ft ≈ 21.82 ft
Now, we can substitute the value of y into the equation:
dy/dt = 36/2y ft/s
dy/dt = 36/(2√476) ft/s
dy/dt ≈ 36/42.72 ft/s
dy/dt ≈ 0.844 ft/s
Therefore, the cable is being pulled at a speed of approximately 0.844 ft/s.
To find the speed at which the cable is being pulled, we need to use the concept of related rates.
Let's assign some variables:
- Let's call the distance between the barge and the dock as "x" (in ft).
- The rate at which the distance between the barge and the dock is changing is given as dx/dt = 3/4 ft/s (since it is approaching at 3/4 per second).
- Let's call the length of the cable as "y" (in ft).
Now, let's consider a right triangle formed by the deck of the barge, the ring on the dock, and the cable connecting them. The triangle's sides are x, y, and the hypotenuse, which is the distance between the deck and the dock (10 ft).
Using the Pythagorean theorem, we can relate x, y, and the hypotenuse:
x^2 + y^2 = 10^2.
Differentiating both sides of the equation with respect to time t:
2x(dx/dt) + 2y(dy/dt) = 0.
Now, we need to solve for dy/dt, the rate at which the cable is being pulled. Plugging in the given values:
2(24)(3/4) + 2y(dy/dt) = 0,
48(3/4) + 2y(dy/dt) = 0,
36 + 2y(dy/dt) = 0,
2y(dy/dt) = -36,
dy/dt = -36 / (2y).
We need to find y, the length of the cable. Let's use the Pythagorean theorem equation we derived earlier:
x^2 + y^2 = 10^2.
Plugging in the value of x (24 ft) given in the problem:
24^2 + y^2 = 100.
576 + y^2 = 100,
y^2 = 100 - 576,
y^2 = -476.
Since we are considering positive lengths, this means that there is no real solution for y. Hence, the length of the cable cannot be determined, and consequently, the rate at which the cable is pulled (dy/dt) cannot be determined.
d = distance to barge , c = length of cable
d^2 + 10^2 = c^2
implicit differentiation ... 2 d dd/dt = 2 c dc/dt ... d dd/dt = c dc/dt
24 * 3/4 = √(24^2 + 10^2) * dc/dt