Question

A man is moving away from a tower 41.6 m high at a rate of 2 m/sIf the eye level of the man is 1.6 m above the ground, then the rate at which the angle of elevation of the top of the tower changes, when he is at a distance 30 m from the foot of the tower is:

• A. \( -\frac{4}{125} \) rad/sec
• B. \( \frac{4}{625} \) rad/sec
• C. \( -\frac{2}{125} \) rad/sec
• D. \( \frac{1}{625} \) rad/sec

Hint:

In related rates with angles, always express in terms of \( \tan \theta \) or \( \sin \theta \)Then differentiate implicitly and substitute values carefully.

Answer 1

The Correct Option is A

Step 1: Define variables.
Let distance from tower = \( x \), angle of elevation = \( \theta \).
Height difference:
\[ 41.6 - 1.6 = 40 \, \text{m} \]

Step 2: Form the relation.
\[ \tan \theta = \frac{40}{x} \]

Step 3: Differentiate implicitly.
\[ \sec^2 \theta \cdot \frac{d\theta}{dt} = -\frac{40}{x^2}\frac{dx}{dt} \]

Step 4: Substitute known values.
\[ x = 30,\quad \frac{dx}{dt} = 2 \]
\[ \sec^2 \theta \cdot \frac{d\theta}{dt} = -\frac{40}{900} \cdot 2 \]
\[ = -\frac{80}{900} = -\frac{4}{45} \]

Step 5: Find \( \sec^2 \theta \).
\[ \tan \theta = \frac{40}{30} = \frac{4}{3} \]
\[ \sec^2 \theta = 1 + \tan^2 \theta = 1 + \frac{16}{9} = \frac{25}{9} \]

Step 6: Solve for \( \frac{d\theta}{dt} \).
\[ \frac{25}{9} \cdot \frac{d\theta}{dt} = -\frac{4}{45} \]
\[ \frac{d\theta}{dt} = -\frac{4}{45} \cdot \frac{9}{25} \]
\[ = -\frac{36}{1125} = -\frac{4}{125} \]

Step 7: Final conclusion.
\[ \boxed{-\frac{4}{125} \text{ rad/sec}} \]