A kite 75\text{ft} above the ground moves horizontally at a speed of 4\text{ft/s}. At what rate is the angle between the string and the horizontal decreasing when 250\text{ft} of string has been let out?
Answer (in radians per second)
To find the rate at which the angle between the string and the horizontal is decreasing, we can use trigonometry and differentiation.
Let's denote the angle between the string and the horizontal as θ.
Given that the kite is 75 ft above the ground, we can use inverse trigonometry to find the length of the string, s, as follows:
s = 75 + h
where h is the height of the kite above the ground. Since the kite is moving horizontally at a speed of 4 ft/s, the horizontal distance covered, x, is given by:
x = 4t
where t is the time in seconds.
Using the Pythagorean theorem, we can write the relationship between x and s as:
s^2 = x^2 + h^2
Differentiating both sides of this equation with respect to time, t, we get:
2s(ds/dt) = 2x(dx/dt) + 2h(dh/dt)
We are given that ds/dt = 4 ft/s, dx/dt = 4 ft/s, and h = 75 ft. Substituting these values into the equation, we have:
2(75 + 75)(4) = 2(4)(4) + 2(75)(dh/dt)
Simplifying further:
300 = 32 + 150(dh/dt)
150(dh/dt) = 300 - 32
dh/dt = (300 - 32)/150
dh/dt = 268/150
dh/dt = 1.7867 ft/s
To find the rate in radians per second, we need to convert this rate to radians by dividing by the length of the string, which is 250 ft.
Therefore, the rate at which the angle between the string and the horizontal is decreasing is:
1.7867 ft/s / 250 ft = 0.0071468 radians per second (rounded to 6 decimal places)
To find the rate at which the angle between the string and the horizontal is decreasing, we can use the concept of related rates.
Let's denote the angle between the string and the horizontal as θ, and the length of the string as s. We are given that the kite is 75 ft above the ground and that the string is being let out at a rate of 4 ft/s.
To begin, we need to find an equation connecting the quantities θ and s. In this case, we can use the tangent function. Since the kite is 75 ft above the ground, we can consider the height of the kite as the opposite side of the right triangle and the length of the string as the hypotenuse. This gives us:
tan(θ) = height of the kite / length of the string
Using the given information, we have:
tan(θ) = 75 ft / s
Next, we need to differentiate both sides of the equation with respect to time t. This will help us find the rate at which the angle is changing with respect to time.
Differentiating the equation implicitly, we get:
sec^2(θ) * dθ/dt = 0 - 75 / s^2 * ds/dt
Since we are interested in finding dθ/dt, we can rearrange the equation:
dθ/dt = -75 / (s^2 * sec^2(θ)) * ds/dt
Finally, we can plug in the given values to find the rate of change of the angle:
s = 250 ft
ds/dt = 4 ft/s
θ is the angle we want to find
Using the values above, we can substitute into the derived equation:
dθ/dt = -75 / (250^2 * sec^2(θ)) * 4
Simplifying the expression, we have:
dθ/dt = -3 / (250^2 * sec^2(θ))
Now we need to find the value of sec(θ) at the given point. Since we're not provided with the exact value of θ, we cannot directly substitute into the equation. However, we can use the known values of the sides of the right triangle to find sec(θ).
In a right triangle, we have:
sec(θ) = hypotenuse / adjacent side
In this case, the adjacent side is the horizontal distance traveled by the kite. We know that the kite moves horizontally at a speed of 4 ft/s. So, the adjacent side is given by:
adjacent side = horizontal speed * time
The time it takes to let out 250 ft of string is:
time = length of string / speed of letting out = 250 ft / 4 ft/s = 62.5 s
Now we can calculate the adjacent side:
adjacent side = 4 ft/s * 62.5 s = 250 ft
So, substituting this value into the equation for sec(θ), we have:
dθ/dt = -3 / (250^2 * (250/250)^2)
Simplifying further:
dθ/dt = -3 / (250^4)
Therefore, the rate at which the angle between the string and the horizontal is decreasing is approximately -3 / (250^4) radians per second.