Math Problem Statement

A man 2 metres tall walks towards a mast of height 26metres at the rate of 1.67m/s determine the rate of change of the angle of elevation of the top of man's head when he is 16 metres from the mast

Solution

To solve this problem, we will use related rates and trigonometry.

Let:

  • h=26h = 26 meters be the height of the mast.
  • y=2y = 2 meters be the height of the man.
  • x(t)x(t) be the distance between the man and the base of the mast at time tt.
  • θ(t)\theta(t) be the angle of elevation from the top of the man's head to the top of the mast.

Step-by-Step Solution

  1. Define the right triangle: The distance from the man to the base of the mast is x(t)x(t), and the height difference between the top of the mast and the man's head is hy=262=24h - y = 26 - 2 = 24 meters.

    The angle of elevation θ\theta is given by: tan(θ)=24x(t).\tan(\theta) = \frac{24}{x(t)}.

  2. Differentiate with respect to time tt:

    To find the rate of change of the angle dθdt\frac{d\theta}{dt}, we take the derivative of both sides with respect to tt: ddt(tan(θ))=ddt(24x(t)).\frac{d}{dt} \left( \tan(\theta) \right) = \frac{d}{dt} \left( \frac{24}{x(t)} \right).

    Using the chain rule on the left-hand side: sec2(θ)dθdt=24x(t)2dxdt.\sec^2(\theta) \cdot \frac{d\theta}{dt} = -\frac{24}{x(t)^2} \cdot \frac{dx}{dt}.

  3. Substitute given values:

    At the moment when x=16x = 16 meters, and the rate of change of distance is dxdt=1.67\frac{dx}{dt} = -1.67 m/s (negative because the distance is decreasing).

    Calculate tan(θ)\tan(\theta) when x=16x = 16: tan(θ)=2416=32.\tan(\theta) = \frac{24}{16} = \frac{3}{2}.

    Therefore: θ=tan1(32).\theta = \tan^{-1}\left( \frac{3}{2} \right).

    Also, we have: sec2(θ)=1+tan2(θ)=1+(32)2=1+94=134.\sec^2(\theta) = 1 + \tan^2(\theta) = 1 + \left( \frac{3}{2} \right)^2 = 1 + \frac{9}{4} = \frac{13}{4}.

  4. Solve for dθdt\frac{d\theta}{dt}:

    Substitute these values into the differentiated equation: 134dθdt=24162(1.67).\frac{13}{4} \cdot \frac{d\theta}{dt} = -\frac{24}{16^2} \cdot (-1.67).

    Simplify: 134dθdt=24×1.67256.\frac{13}{4} \cdot \frac{d\theta}{dt} = \frac{24 \times 1.67}{256}.

    Solve for dθdt\frac{d\theta}{dt}: dθdt=24×1.67256413.\frac{d\theta}{dt} = \frac{24 \times 1.67}{256} \cdot \frac{4}{13}.

    Now, calculate the numeric value: dθdt40.08256413.\frac{d\theta}{dt} \approx \frac{40.08}{256} \cdot \frac{4}{13}.

    Simplify: dθdt40.08×4256×13=160.3233280.0482radians per second.\frac{d\theta}{dt} \approx \frac{40.08 \times 4}{256 \times 13} = \frac{160.32}{3328} \approx 0.0482 \, \text{radians per second}.

Final Answer:

The rate of change of the angle of elevation of the top of the man's head is approximately 0.04820.0482 radians per second when he is 16 meters from the mast.

Would you like more details or have any questions?


Here are 5 related questions for further exploration:

  1. How would the rate change if the height of the mast were different?
  2. What would be the rate of change if the man were walking away from the mast?
  3. How does the height of the man affect the rate of change of the angle?
  4. What is the significance of the rate of change being positive or negative in this context?
  5. How would you apply related rates to a situation involving a moving car and a stationary observer?

Tip: Understanding the relationship between angles and distances can help solve various real-life problems, from navigation to architectural design.

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Math Problem Analysis

Mathematical Concepts

Related Rates
Trigonometry
Differentiation

Formulas

tan(θ) = (h - y) / x(t)
d/dt [tan(θ)] = sec^2(θ) * dθ/dt
Rate of change formula using chain rule: sec^2(θ) * dθ/dt = -24 / x(t)^2 * dx/dt

Theorems

Chain Rule
Trigonometric Derivatives

Suitable Grade Level

Grades 11-12