A fish is reeled in at a rate of 3.4 feet per second from a point 10 feet above the water. At what rate is the angle between the line and the water changing when there is a total of 25 feet of line out?
To find the rate at which the angle between the fishing line and the water is changing, we can use trigonometry.
Let's assume the angle between the line and the water is θ. We need to find dθ/dt, the rate of change of θ with respect to time.
Given:
- The fish is reeled in at a rate of 3.4 feet per second.
- The initial height of the fish is 10 feet.
- The total length of the line is 25 feet.
We can start by finding the length of the fishing line remaining in the water, L, as a function of time t.
Since the fish is being reeled in at a constant rate, we can write L = 25 - 3.4t.
Next, let's draw a right triangle to represent the situation.
The length of the line out of the water (vertical side) is given by the equation y = 10 - L = 10 - (25 - 3.4t) = 3.4t - 15.
Now, the horizontal side of the triangle (x) represents the distance traveled by the fish along the water's surface. It is related to y by x = y * tan(θ).
Differentiating both sides with respect to time t, we get:
dx/dt = dy/dt * tan(θ) + y * sec^2(θ) * dθ/dt.
We know that dx/dt is the rate at which the fish travels along the water's surface, which is given by dx/dt = 3.4 ft/s.
Substituting y = 3.4t - 15 and dy/dt = 3.4, we have:
3.4 = 3.4 * tan(θ) + (3.4t - 15) * sec^2(θ) * dθ/dt.
Now, we need to solve for dθ/dt.
To isolate dθ/dt, rearrange the equation:
3.4t - 15 = 3.4 * (tan(θ) + (3.4t - 15) * sec^2(θ) * dθ/dt.
Move the last term to the right side:
(3.4 * (tan(θ) + (3.4t - 15) * sec^2(θ)) * dθ/dt = 3.4t - 15.
Finally, divide both sides by 3.4 * (tan(θ) + (3.4t - 15) * sec^2(θ)) to solve for dθ/dt:
dθ/dt = (3.4t - 15) / (3.4 * (tan(θ) + (3.4t - 15) * sec^2(θ)).
Now we have the formula to find the rate at which the angle between the line and the water is changing. Plug in the values for t and θ to find the specific rate at a given moment.
To solve this problem, we need to find the rate at which the angle between the line and the water is changing. This can be done by finding the derivative of the angle with respect to time.
Let's define some variables:
- Let h be the height of the fisherman's hand above the water.
- Let x be the horizontal distance from the fisherman's hand to the fish.
- Let θ be the angle between the line (the fishing line) and the water.
From the problem, we know:
- h = 10 feet (constant) since it is given that the point is 10 feet above the water.
- dx/dt = 3.4 feet/second, which is the rate at which the fish is being reeled in.
- We need to find dθ/dt, the rate at which the angle θ is changing.
Using trigonometry, we can find the relationship between x, h, and θ. In a right triangle, the tangent of an angle is defined as the ratio of the opposite side to the adjacent side:
tan(θ) = h/x
To solve for x, we rearrange the equation:
x = h/tan(θ)
Now, differentiate both sides of the equation with respect to time (t):
dx/dt = dh/dt * (1/tan(θ)) * dθ/dt
We know dx/dt and h are constants, so we can substitute them with their given values:
3.4 = 10 * (1/tan(θ)) * dθ/dt
We need to solve for dθ/dt, so rearrange the equation:
dθ/dt = (3.4 * tan(θ))/10
Substituting the given value of x = 25 feet at the moment when there is a total of 25 feet of line out, we can find the value of θ. Using the equation x = h/tan(θ):
25 = 10/tan(θ)
Solving for θ:
tan(θ) = 10/25
tan(θ) = 2/5
Taking the inverse tangent (arctan) on both sides:
θ = arctan(2/5)
Finally, substitute the value of θ into the equation to find dθ/dt:
dθ/dt = (3.4 * tan(arctan(2/5)))/10
Solving this expression will give us the rate at which the angle between the line and the water is changing when there is a total of 25 feet of line out.
When the line length is x,
sinθ = 10/x
cosθ dθ/dt = -10/x^2 dx/dt
so plug in your numbers when x=25