A fish is reeled in at a rate of 3.2 feet per second from a point 10 feet above the water (see figure). At what rate is the angle between the line and the water changing when there is a total of 25 feet of line out? (Round your answer to three decimal places.)

0.056

Well, it seems like this fisherman has quite the catch on his hands! Let's see what we can do with this angle situation.

First, let me get my fishing line of knowledge ready. We have a right triangle formed by the line, the water, and the distance from the fisherman to the fish. The angle we're interested in is the one between the line and the water.

Let's call this angle θ. Now, we need to figure out how θ is changing with respect to time. In other words, we need to find dθ/dt.

To do this, we can use the tangential relationship between the angle and the line. The tangent of θ is equal to the height of the line above the water divided by the length of the line.

tan(θ) = 10/25

Now, let's differentiate both sides of the equation with respect to time (t):

sec^2(θ) * dθ/dt = (1/25) * dl/dt

Where dl/dt is the rate at which the line is being reeled in, which is 3.2 feet per second. So, we can substitute that in:

sec^2(θ) * dθ/dt = (1/25) * 3.2

Now, we just need to solve for dθ/dt:

dθ/dt = (1/25) * 3.2 / sec^2(θ)

Since we're looking for the rate at which the angle is changing when there is a total of 25 feet of line out, we can substitute θ = tan^(-1)(10/25) into the equation:

dθ/dt = (1/25) * 3.2 / sec^2(tan^(-1)(10/25))

Now we just need to pop that into a calculator to get the final answer. Do you have one handy?

To find the rate at which the angle between the line and the water is changing, we can use trigonometry. Let's call the length of the line "L" and the angle between the line and the water "θ".

We are given that the fish is reeled in at a rate of 3.2 feet per second. This means that the length of the line, L, is changing at a rate of 3.2 feet per second.

We are also given that the initial length of the line is 10 feet, and we want to find the rate at which the angle is changing when there is a total of 25 feet of line out.

Let's set up a right triangle to represent the situation. The vertical side of the triangle represents the part of the line above the water, which remains constant at 10 feet. The horizontal side of the triangle represents the horizontal distance from the fishing pole to the point where the line contacts the water, which we'll call "x". The hypotenuse of the triangle represents the length of the line, L.

We know that L = 25 feet, and the vertical side of the triangle is 10 feet.

Using the Pythagorean theorem, we can find the value of x:

x^2 + 10^2 = 25^2
x^2 + 100 = 625
x^2 = 625 - 100
x^2 = 525
x = √525
x ≈ 22.91 feet

Now, let's differentiate both sides of the equation with respect to time (t):

d/dt(x^2) + d/dt(100) = d/dt(625)
2x(dx/dt) = 0
2(22.91)(dx/dt) = 0
45.82(dx/dt) = 0
dx/dt = 0/45.82
dx/dt ≈ 0 feet per second

The rate at which x is changing is approximately 0 feet per second. This means that the horizontal distance from the fishing pole to the point where the line contacts the water is not changing.

To find the rate at which the angle is changing, we can use the tangent of theta:

tan(θ) = (10/22.91)

Now, differentiate both sides of the equation with respect to time (t):

d/dt(tan(θ)) = d/dt(10/22.91)

Sec^2(θ) * dθ/dt = 0

Since dx/dt is 0, the tan(θ) is also constant.

Therefore, dθ/dt = 0.

The rate at which the angle between the line and the water is changing is 0 degrees per second.

To find the rate at which the angle between the line and the water is changing, we can use trigonometry and related rates. First, let's label the given information:

- The fish is being reeled in at a rate of 3.2 feet per second.
- The initial length of the line is 10 feet.
- We want to find the rate at which the angle between the line and the water is changing when the total length of the line is 25 feet.

Let's call the angle between the line and the water "θ". Now, let's set up a right triangle to represent the situation:

^
|
/|
/ |
25 ft/ | line
/ |
/ θ |
/____|

We can use trigonometric functions to relate the lengths of the sides of the triangle to the angle θ. We know that the tangent function of θ is equal to the opposite side (10 ft) divided by the adjacent side (distance of the fish to the boat):

tan(θ) = 10 ft / distance

To find the rate at which the angle θ is changing with respect to time (dθ/dt), we can take the derivative of both sides of the equation:

d(tan(θ))/dt = d(10 ft / distance)/dt

Using the chain rule, we get:

sec²(θ) * dθ/dt = -10 ft/distance² * d(distance)/dt

Since d(distance)/dt is the rate at which the line is being reeled in, which is given as 3.2 ft/s, we can substitute these values into the equation:

sec²(θ) * dθ/dt = -10 ft/distance² * 3.2 ft/s

Now, we need to find the value of sec²(θ). Since we know the total length of the line is 25 ft, and the height of the triangle is 10 ft, we can use the Pythagorean theorem to find the distance:

distance = sqrt(25 ft² - 10 ft²) = 5√3 ft

Substituting this value and the given rate, we have:

sec²(θ) * dθ/dt = -10 ft/(5√3 ft)² * 3.2 ft/s

Calculating further:

sec²(θ) * dθ/dt = -10 ft/(75 ft) * 3.2 ft/s
sec²(θ) * dθ/dt = -0.4267 ft/s

To find dθ/dt, we divide both sides by sec²(θ):

dθ/dt = -0.4267 ft/s / sec²(θ)

Now, we can substitute the value of sec(θ) using the information from the triangle. Since sec(θ) is equal to the hypotenuse (distance) divided by the adjacent side (10 ft), we have:

sec(θ) = distance / 10 ft = (5√3 ft) / 10 ft = √3 / 2

Substituting this value, we get:

dθ/dt = -0.4267 ft/s / (√3 / 2)²
dθ/dt = -0.4267 ft/s / (3 / 4)
dθ/dt = -0.4267 ft/s * (4 / 3)
dθ/dt ≈ -0.569 ft/s

Therefore, 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 is approximately -0.569 feet per second.