A boat being pulled toward a wharf by a rope attached to the boats deck from a point 8.0 m above the deck. If the rope is being pulled in at a rate of 2.0 m/min, how fast is the boat approaching the wharf when it is 12.0 m away?
12
To solve this problem, we can use the concept of related rates.
Let's consider the situation and label the given information:
- Let x be the distance between the boat and the wharf.
- The rate at which the rope is being pulled in is dx/dt = 2.0 m/min.
- The height of the attachment point of the rope from the deck is h = 8.0 m.
We want to find dy/dt, the rate at which the boat is approaching the wharf when it is 12.0 m away.
We can form a right triangle with the distance between the boat and the wharf, the height of the attachment point of the rope, and the length of the rope itself. From this triangle, we have the relationship:
x^2 + h^2 = L^2, where L is the length of the rope.
To find dy/dt, we need to differentiate both sides of the equation with respect to time:
2x(dx/dt) + 2h(dh/dt) = 2L(dL/dt).
Since we know dx/dt = 2.0 m/min and h = 8.0 m, we can solve for dL/dt.
We can find L by using the Pythagorean theorem:
L = sqrt(x^2 + h^2).
Differentiating L with respect to time, we get:
dL/dt = (1/2)(x^2 + h^2)^(-1/2) * (2x(dx/dt)).
Plugging in the known values, we have:
dL/dt = (1/2)(x^2 + 8.0^2)^(-1/2) * (2x * 2.0).
Simplifying this expression, we get:
dL/dt = 4x / sqrt(x^2 + 64.0).
Now, we substitute this value back into the first equation:
2x(dx/dt) + 2h(dh/dt) = 2L(dL/dt).
Plugging in the values, we have:
2(x)(2.0) + 2(8.0)(dh/dt) = 2(sqrt(x^2 + 64.0))(4x / sqrt(x^2 + 64.0)).
Simplifying further:
4x + 16.0(dh/dt) = 8x.
Now, we can solve for dh/dt in terms of x:
16.0(dh/dt) = 4x.
dh/dt = (4x) / 16.0.
Substituting x = 12.0 m:
dh/dt = (4 * 12.0) / 16.0.
dh/dt = 3.0 m/min.
Therefore, the boat is approaching the wharf at a rate of 3.0 m/min when it is 12.0 m away.