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 1 m/s, how fast is the boat approaching the dock when it is 7 m from the dock? (Round your answer to two decimal places.)
m/s
Well, that's a tricky situation for our boat friend! Let's calculate the speed at which it's approaching the dock when it's 7 meters away.
We can use the concept of similar triangles here. We have a right triangle with one leg being the distance between the boat and the dock (7 m) and the other leg being the height of the pulley above the boat (1 m). The hypotenuse represents the rope.
Now, since the rope is being pulled in at a rate of 1 m/s, that means the length of the hypotenuse is changing at a rate of 1 m/s. But we want to find out how fast the distance between the boat and the dock is changing, which is the rate of change of the leg adjacent to the angle.
Using the Pythagorean theorem, we can express the relationship between the three sides of the triangle:
(7 m)^2 + (1 m)^2 = (distance between the boat and the dock)^2
Simplifying this equation:
49 m^2 + 1 m^2 = (distance between the boat and the dock)^2
50 m^2 = (distance between the boat and the dock)^2
Now, let's differentiate both sides of this equation with respect to time:
d/dt (50 m^2) = d/dt ((distance between the boat and the dock)^2)
0 = 2 * (distance between the boat and the dock) * (d (distance between the boat and the dock) / dt)
Now, we can plug in the values we know:
0 = 2 * (7 m) * (d (distance between the boat and the dock) / dt)
Simplifying:
0 = 14 m * (d (distance between the boat and the dock) / dt)
To find the rate of change of the distance between the boat and the dock (d (distance between the boat and the dock) / dt), we divide both sides by 14 m:
(d (distance between the boat and the dock) / dt) = 0 m/s
So, it turns out that when the boat is 7 m away from the dock, it is not approaching the dock at all! The rate of change is zero. I guess the boat decided to take a break and enjoy the view instead of hurrying to the dock. Lazy boat!
To solve this problem, we can use the concept of related rates. Let's denote the following variables:
- x: the horizontal distance between the boat and the dock
- y: the vertical distance between the bow of the boat and the dock
From the problem statement, we know that the rope is being pulled in at a rate of 1 m/s. Since the pulley is 1 m higher than the bow of the boat, the rate at which y is changing is also 1 m/s.
We are asked to find the rate at which the boat is approaching the dock, i.e., we need to find dx/dt when x = 7 m.
Now, let's find a relationship between x and y. We can use the Pythagorean theorem:
x^2 + y^2 = 7^2
Differentiating both sides of the equation with respect to time, we get:
2x * dx/dt + 2y * dy/dt = 0
Since we are given that dy/dt = 1 m/s, we can substitute this value into the equation:
2x * dx/dt + 2y * 1 = 0
Simplifying the equation, we get:
2x * dx/dt = -2y
Now, we need to find y when x = 7 m. Using the Pythagorean theorem, we can calculate:
7^2 + y^2 = x^2 + y^2
49 = x^2 + y^2
Rearranging the equation:
x^2 = 49 - y^2
Substituting this expression into our previous equation, we have:
2(7) * dx/dt = -2y
Now, we can substitute the known values and solve for dx/dt:
2(7) * dx/dt = -2y
14 * dx/dt = -2y
14 * dx/dt = -2(1)
14 * dx/dt = -2
dx/dt = -2/14
dx/dt = -1/7
Therefore, the boat is approaching the dock at a rate of -1/7 m/s when it is 7 m from the dock.
To find the speed at which the boat is approaching the dock, we can use the concept of related rates.
Let's define some variables:
- Let x represent the distance between the boat and the dock (in meters).
- Let y represent the height of the rope from the dock floor (in meters).
We are given that the rope is being pulled in at a rate of 1 m/s. This means that the rate at which x is changing is -1 m/s since the boat is approaching the dock.
We need to find the rate at which the boat is approaching the dock, which is the rate of change of x with respect to time (dx/dt) when x = 7 m.
From the given information, we know that y is always 1 m higher than x. Therefore, we can express y in terms of x: y = x + 1.
Differentiating both sides of the equation with respect to time (t), we get:
(dy/dt) = (d/dt)(x + 1)
Since (dy/dt) represents the rate at which y is changing, and it's given that the rope is being pulled in at a rate of 1 m/s, we can substitute (dy/dt) with -1 m/s:
-1 = (d/dt)(x + 1)
Now, we can find dx/dt by differentiating both sides of the equation with respect to time (t):
0 = (d/dt)(dx/dt)
Simplifying, we have:
0 = (d^2x/dt^2)
This equation tells us that the acceleration of the boat is zero since the rate at which the distance between the boat and the dock is changing remains constant.
Therefore, when x = 7 m, the speed at which the boat is approaching the dock (dx/dt) is also 0 m/s.
Hence, the boat is not approaching the dock at a speed of 0 m/s when it is 7 m from the dock.