Math Problem Statement

Budek is flying a kite. The kite is initially directly over her head at a height of 30 m. Then the wind starts carrying the kite in one direction at a rate of 40 m/min while Budek starts running in the opposite direction at a rate of 200 m/min. Assuming that the height of the kite remains constant and that the string forms a straight line, find the rate at which the string is paying out after 10 s.

Solution

This problem involves related rates and can be analyzed geometrically. The key is to determine the rate at which the length of the kite string (ss) is increasing, given the height of the kite and the combined horizontal motion.

Step-by-step Solution:

Step 1: Define the problem geometry

  1. The height of the kite is constant at h=30mh = 30 \, \text{m}.
  2. Budek is running in one direction at vB=200m/minv_{\text{B}} = 200 \, \text{m/min}.
  3. The wind carries the kite horizontally in the opposite direction at vK=40m/minv_{\text{K}} = 40 \, \text{m/min}.
  4. The total horizontal distance xx between Budek and the kite increases at a rate of: dxdt=vB+vK=200+40=240m/min.\frac{dx}{dt} = v_{\text{B}} + v_{\text{K}} = 200 + 40 = 240 \, \text{m/min}.

Step 2: Relate the length of the string (ss) to the geometry

The string forms the hypotenuse of a right triangle, where: s2=x2+h2.s^2 = x^2 + h^2.

Differentiating both sides with respect to time tt: 2sdsdt=2xdxdt.2s \frac{ds}{dt} = 2x \frac{dx}{dt}.

Simplify: sdsdt=xdxdt.s \frac{ds}{dt} = x \frac{dx}{dt}.

Solve for dsdt\frac{ds}{dt}, the rate at which the string is paying out: dsdt=xsdxdt.\frac{ds}{dt} = \frac{x}{s} \frac{dx}{dt}.

Step 3: Calculate xx, ss, and rates at t=10st = 10 \, \text{s}

  1. Convert dxdt\frac{dx}{dt} into meters per second: dxdt=240m/min=4m/s.\frac{dx}{dt} = 240 \, \text{m/min} = 4 \, \text{m/s}.

  2. The horizontal distance xx after 10 seconds: x=dxdt×t=4m/s×10s=40m.x = \frac{dx}{dt} \times t = 4 \, \text{m/s} \times 10 \, \text{s} = 40 \, \text{m}.

  3. The length of the string ss is: s=x2+h2=402+302=1600+900=2500=50m.s = \sqrt{x^2 + h^2} = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, \text{m}.

Step 4: Calculate dsdt\frac{ds}{dt}

Using the formula: dsdt=xsdxdt,\frac{ds}{dt} = \frac{x}{s} \frac{dx}{dt}, substitute x=40mx = 40 \, \text{m}, s=50ms = 50 \, \text{m}, and dxdt=4m/s\frac{dx}{dt} = 4 \, \text{m/s}: dsdt=40504=0.84=3.2m/s.\frac{ds}{dt} = \frac{40}{50} \cdot 4 = 0.8 \cdot 4 = 3.2 \, \text{m/s}.

Final Answer:

The rate at which the string is paying out after 10 seconds is: 3.2m/s\boxed{3.2 \, \text{m/s}}


Would you like further details or clarification? Here are related questions to explore:

  1. What happens if Budek stops running after 10 seconds?
  2. How does the string length change if the kite's height increases over time?
  3. How would the result change if Budek and the kite moved in the same direction?
  4. What is the string's rate of change if Budek changes direction halfway?
  5. Can this problem be solved using parametric equations instead?

Tip: Always ensure the units are consistent when solving related rates problems.

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Math Problem Analysis

Mathematical Concepts

Related Rates
Right Triangle Geometry
Differentiation

Formulas

s^2 = x^2 + h^2
2s(ds/dt) = 2x(dx/dt)
ds/dt = (x/s)(dx/dt)

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 10-12