A kite 100 ft above the ground moves horizontally at a speed of 9 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?
let the angle be Ø, and the horizontal distance be x
given : dx/dt = 9
find dØ/dt
we have a right angled triangle, and
tanØ =100/x
xtanØ = 100
x sec^2 Ø dØ/dt + tanØ (dx/dt) = 0 ***
when 200 ft are let out,
x^2 + 100^2 = 200^2
x^2 = 300
x = 10√3
secØ = 10√3/100 = √3/10
sec^2 Ø = 3/100
tanØ = 100/10√3 = 10/√3
in ***
(10/√3)(3/100) dØ/dt + (10/√3)(9) = 0
solve for dØ/dt
sorry I didnt get whats the answer
Just noticed my "late-night" calculations have an errors and typos, sorry about that.
corrected version:
when 200 ft are let out,
x^2 + 100^2 = 200^2
x^2 = 30,000
x = 100√3
secØ = 200/ (100√3) = 2/√3
sec^2 Ø = 4/3
tanØ = 100/100√3 = 1/√3
then my equation should be
(100√3)(4/3) dØ/dt + (1/√3)(9) = 0
I get dØ/dt = -.0225 radians/s
check my work again carefully, still on my first coffee.
To find the rate at which the angle between the string and the horizontal is changing, we can use the chain rule of differentiation.
Let's denote:
- h: height above the ground (in feet)
- s: length of the string (in feet)
- θ: angle between the string and the horizontal (in radians)
- 𝑑θ/𝑑𝑡: rate at which the angle is changing with respect to time (in radians per second)
- 𝑑s/𝑑𝑡: rate at which the length of the string is changing with respect to time (in feet per second)
We have the following information:
- Initially, the kite is 100 ft above the ground, so h = 100 ft.
- The kite is moving horizontally at a speed of 9 ft/s, which means the rate of change of height is 𝑑h/𝑑𝑡 = 0 ft/s.
- We need to find 𝑑θ/𝑑𝑡 when 200 ft of string have been let out, so s = 200 ft.
To solve for 𝑑θ/𝑑𝑡, we need to express 𝑑θ/𝑑𝑡 in terms of 𝑑s/𝑑𝑡.
From the given information, we can relate the variables using the Pythagorean theorem:
s² = h² + 9²
Differentiating this equation implicitly with respect to time (t), we get:
2s (𝑑s/𝑑𝑡) = 2h (𝑑h/𝑑𝑡)
Since 𝑑h/𝑑𝑡 is zero, we have:
𝑑s/𝑑𝑡 = 0
Now, we can relate θ and s using trigonometry:
tan(θ) = h / (s - 9𝑡)
Applying the tangent function to both sides of the equation, we get:
tan(θ) = h / (s - 9𝑡)
1 / tan(θ) = (s - 9𝑡) / h
Differentiating this equation implicitly with respect to time (t), we get:
(sec²(θ))(𝑑θ/𝑑𝑡) = ((𝑑s/𝑑𝑡) - 9) / h
We can substitute the known values into this equation. We know 𝑑s/𝑑𝑡 = 0, h = 100, and s = 200.
(sec²(θ))(𝑑θ/𝑑𝑡) = (0 - 9) / 100
(sec²(θ))(𝑑θ/𝑑𝑡) = -9/100
Multiplying both sides by (1/cos²(θ)):
(𝑑θ/𝑑t) = (-9/100) cos²(θ)
To calculate the rate of change of the angle 𝑑θ/𝑑𝑡, we need to know the value of cos²(θ).
Since we are given the length of the string (s = 200 ft) and the height of the kite (h = 100 ft), we can use the Pythagorean theorem to find cos(θ).
cos(θ) = h / (s - 9𝑡)
cos(θ) = 100 / (200 - 9𝑡)
cos²(θ) = (100²) / (200 - 9𝑡)²
Substituting cos²(θ) into the previous equation for 𝑑θ/𝑑𝑡:
(𝑑θ/𝑑t) = (-9/100) * [(100²) / (200 - 9𝑡)²]
Now, we can solve for 𝑑θ/𝑑𝑡 when 200 ft of string have been let out by substituting t = 200/9 into the equation:
(𝑑θ/𝑑t) = (-9/100) * [(100²) / (200 - 9 * (200/9))²]
Simplifying this expression will give us the rate at which the angle is decreasing when 200 ft of string have been let out.
oh, please -- do some of the work, okay?
(10/√3)(3/100) dØ/dt + (10/√3)(9) = 0
solve for dØ/dt
Now it's just Algebra I ...