A fish is reeled in at a rate of 1 foot per second from a bridge 15 feet above the water. At what rate is the angle between the line and the water changing when 25 feet of line is out? (Assume the fish stays near the surface of the water.)
draw a diagram. If x feet of line are out,
15/x = sinθ
-15/x^2 dx/dt = cosθ dθ/dt
Now just plug in your numbers.
Well, well, well, it seems we have a fishing problem here! Let's see if we can untangle this situation.
Now, we have a fish being reeled in at a rate of 1 foot per second from a bridge 15 feet above the water. The length of the line between the fish and the bridge is what we're interested in.
To find the rate at which the angle between the line and the water is changing, we need to use a little bit of trigonometry. The tangent of the angle is equal to the height of the bridge (15 feet) divided by the length of the line (let's call it L).
The length of the line is constantly changing as the fish gets closer to the bridge. We're specifically interested in when 25 feet of line is out. So, in terms of our equation, L = 25 feet.
To find the rate at which the angle is changing, we need to take the derivative of both sides of the equation. Using the chain rule, we have:
sec^2(theta) * d(theta)/dt = 0 - 15 * dL/dt / (L^2)
Now, we know that dL/dt is equal to the rate at which the line is being reeled in, which is 1 foot per second. Plugging that in, we have:
sec^2(theta) * d(theta)/dt = 0 - 15 * 1 / (25^2)
Simplifying that further, we get:
sec^2(theta) * d(theta)/dt = -0.24
Now, to find the rate at which the angle is changing (d(theta)/dt), we just need to divide both sides by sec^2(theta):
d(theta)/dt = -0.24 / sec^2(theta)
And there you have it! The rate at which the angle between the line and the water is changing when 25 feet of line is out is -0.24 divided by sec^2(theta).
Just one little piece of advice before you go back to your fishing adventure: hold onto the rod tightly, because you never know when a fish might try to steal the show with its underwater acrobatics!
To solve this problem, we need to use trigonometry and related rates.
Let's assume that the angle between the line and the water is θ. We are given that the fish is reeled in at a rate of 1 foot per second, so the rate of change of the length of the line (L) with respect to time (t) is dL/dt = 1 foot/second.
We need to find the rate of change of the angle (dθ/dt) when 25 feet of line is out. To do this, we can use the tangent function:
tan(θ) = height of bridge / length of line
Since the height of the bridge is given as 15 feet and the length of the line is 25 feet, we can rewrite the equation as:
tan(θ) = 15 / 25
Next, we can implicitly differentiate both sides of the equation with respect to time (t):
sec^2(θ) * dθ/dt = (d/dt) (15/25)
To find the rate of change of the angle (dθ/dt), we need to know the value of sec^2(θ). We can use the Pythagorean identity to find this value:
sec^2(θ) = 1 + tan^2(θ)
Substituting the given value of tan(θ) into the equation, we have:
sec^2(θ) = 1 + (15/25)^2
Now, we can substitute the values of sec^2(θ) and (d/dt) (15/25) back into the equation we derived earlier:
sec^2(θ) * dθ/dt = (d/dt) (15/25)
(1 + (15/25)^2) * dθ/dt = (d/dt) (15/25)
Simplifying, we have:
(1 + 9/25) * dθ/dt = (d/dt) (15/25)
(34/25) * dθ/dt = (d/dt) (15/25)
Multiplying both sides by 25/34, we get:
dθ/dt = (d/dt) (15/25) * 25/34
dθ/dt = (d/dt) (15/34)
Since the length of the line is decreasing (because the fish is being reeled in), the derivative of the length of the line with respect to time (dL/dt) is negative. Hence, (dL/dt) = -1 foot/second.
Substituting this value into the equation, we have:
dθ/dt = (d/dt) (15/34) = -(1/34) radian/second
Therefore, the rate of change of the angle between the line and the water is -(1/34) radian/second.
To find the rate at which the angle between the line and the water is changing, we need to use trigonometry and related rates. Let's break down the problem step by step:
1. Draw a diagram: Sketch a right triangle where the bridge represents the vertical side (15 feet), and the fishing line represents the hypotenuse. The horizontal side of the triangle represents the distance between the bridge and the fish in the water.
2. Assign variables: Let x be the horizontal distance between the bridge and the fish, and let θ be the angle between the line and the water.
3. Determine what information is given: We know that the rate at which the fish is reeled in is 1 foot per second (dx/dt = 1 ft/s), and we want to find dθ/dt, the rate of change of the angle when 25 feet of line is out (x = 25 ft).
4. Relate the variables using trigonometry: Since we have a right triangle, we can use the trigonometric function tangent: tan(θ) = x/15.
5. Implicit differentiation: Differentiate both sides of the equation with respect to time t using the chain rule:
d/dt(tan(θ)) = d/dt(x/15)
Sec^2(θ) * dθ/dt = (dx/dt)/15
6. Solve for dθ/dt: Rearrange the equation to solve for dθ/dt:
dθ/dt = (dx/dt)/(15 * sec^2(θ))
7. Substitute known values: Plug in the given values: dx/dt = 1 ft/s, x = 25 ft, and θ can be determined using the tangent function:
tan(θ) = x/15
tan(θ) = 25/15
θ ≈ 59.04 degrees
8. Calculate dθ/dt: Substitute the known values into the derived equation:
dθ/dt = (1 ft/s) / (15 * sec^2(59.04°))
Use the identity sec^2(θ) = 1/cos^2(θ), and substitute θ = 59.04°:
dθ/dt = (1 ft/s) / (15 * 1 / cos^2(59.04°))
= (1 ft/s) * (cos^2(59.04°) / 15)
Use a calculator to find cos^2(59.04°) and perform the final calculation to find dθ/dt.