Math Problem Statement

A kite glides horizontally at an altitude of 30 m while we unspool the string. We need to determine the rate at which the angle between the string and the horizontal decreases over time once 60 m of string has been unspooled. The kite's horizontal velocity at that instant is 3 m/s.

Solution

This problem involves a kite gliding horizontally while unspooling a string, and we need to determine how the angle between the string and the horizontal decreases over time.

Let's break down each part of the problem:

Given information:

  • Altitude h=30h = 30 m (the height of the kite).
  • The string is being unspooled such that 60 meters of string have been unspooled at a specific moment.
  • The kite's horizontal velocity is dxdt=3 m/s\frac{dx}{dt} = 3 \text{ m/s}, where xx is the horizontal distance of the kite from the point where the string is attached to the ground.

We need to find the angle θ\theta between the string and the horizontal line at that moment, as well as the rate of change of this angle, dθdt\frac{d\theta}{dt}.


(a) Expressing θ\theta as a function of xx:

From the problem, we have a right triangle formed by:

  • The height h=30h = 30 meters (vertical side).
  • The horizontal distance xx meters (horizontal side).
  • The length of the string L=60L = 60 meters (the hypotenuse of the triangle).

From basic trigonometry: cos(θ)=xL\cos(\theta) = \frac{x}{L} Thus, the angle θ\theta can be written as: θ=cos1(xL)\theta = \cos^{-1} \left( \frac{x}{L} \right) Substituting L=60L = 60 m: θ=cos1(x60)\theta = \cos^{-1} \left( \frac{x}{60} \right)


(b) Finding the value of xx:

At the moment in question, we can apply the Pythagorean theorem to solve for xx. The relationship between the height, the horizontal distance, and the length of the string is: x2+h2=L2x^2 + h^2 = L^2 Substituting h=30h = 30 m and L=60L = 60 m: x2+302=602x^2 + 30^2 = 60^2 x2+900=3600x^2 + 900 = 3600 x2=2700x^2 = 2700 x=2700=303mx = \sqrt{2700} = 30\sqrt{3} \, \text{m}


(c) Finding dθdt\frac{d\theta}{dt}:

To find the rate of change of θ\theta, we implicitly differentiate the trigonometric relationship cos(θ)=xL\cos(\theta) = \frac{x}{L} with respect to time tt. We get: sin(θ)dθdt=1Ldxdt-\sin(\theta) \frac{d\theta}{dt} = \frac{1}{L} \frac{dx}{dt} Solving for dθdt\frac{d\theta}{dt}: dθdt=1Lsin(θ)dxdt\frac{d\theta}{dt} = -\frac{1}{L \sin(\theta)} \frac{dx}{dt} Now, we substitute known values:

  • L=60L = 60 m,
  • dxdt=3\frac{dx}{dt} = 3 m/s,
  • sin(θ)=hL=3060=12\sin(\theta) = \frac{h}{L} = \frac{30}{60} = \frac{1}{2}.

Thus: dθdt=160×12×3=130×3=110rad/s\frac{d\theta}{dt} = -\frac{1}{60 \times \frac{1}{2}} \times 3 = -\frac{1}{30} \times 3 = -\frac{1}{10} \, \text{rad/s}


Summary of results:

(a) θ=cos1(x60)\theta = \cos^{-1} \left( \frac{x}{60} \right).

(b) x=303mx = 30\sqrt{3} \, \text{m}.

(c) dθdt=110rad/s\frac{d\theta}{dt} = -\frac{1}{10} \, \text{rad/s}.

Would you like further clarification on any part of this solution?

Here are 5 related questions you could explore next:

  1. How would the solution change if the altitude of the kite were different?
  2. How do different rates of horizontal velocity affect dθdt\frac{d\theta}{dt}?
  3. What would happen if the string length increased faster than it currently does?
  4. How would the trigonometric relationships change if the kite's altitude varied over time?
  5. How could you solve this problem if the kite’s trajectory were not horizontal?

Tip:

Whenever you're working with rates of change (like dθdt\frac{d\theta}{dt}), implicit differentiation is a powerful tool for connecting related variables!

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Rates of Change
Implicit Differentiation
Pythagorean Theorem

Formulas

cos(θ) = x / L
θ = cos⁻¹(x / L)
x² + h² = L²
-sin(θ) * dθ/dt = (1 / L) * dx/dt
dθ/dt = -(1 / (L * sin(θ))) * (dx/dt)

Theorems

Pythagorean Theorem
Trigonometric Identities

Suitable Grade Level

Grades 10-12 (High School)