Questions: A woman standing on a cliff is watching a motorboat through binoculars as the boat approaches the shoreline directly below her. If she is standing 100 feet above the water level and if the boat is approaching at 10 feet per second, at what rate is the angle of her head decreasing when the boat is 200 feet from the shore? radians per second The problem asks for a rate of decrease, so enter a positive number.

Solution Steps

The problem involves a right triangle formed by the woman on the cliff, the boat, and the point on the shore directly below the woman. The height of the cliff is 100 feet, and the horizontal distance from the boat to the shore is changing as the boat approaches. We need to find the rate at which the angle of her head, denoted as \(\theta\), is decreasing when the boat is 200 feet from the shore.

Let:

• \( h = 100 \) feet (height of the cliff)
• \( x \) be the horizontal distance from the boat to the shore
• \(\theta\) be the angle of elevation from the boat to the woman

The relationship between these variables is given by the tangent function: \[ \tan(\theta) = \frac{h}{x} = \frac{100}{x} \]

To find the rate of change of \(\theta\), differentiate both sides of the equation with respect to time \( t \): \[ \frac{d}{dt}[\tan(\theta)] = \frac{d}{dt}\left(\frac{100}{x}\right) \]

Using the chain rule, the derivative of \(\tan(\theta)\) is: \[ \sec^2(\theta) \cdot \frac{d\theta}{dt} \]

The derivative of \(\frac{100}{x}\) is: \[

• \frac{100}{x^2} \cdot \frac{dx}{dt} \]

Equating the derivatives, we have: \[ \sec^2(\theta) \cdot \frac{d\theta}{dt} = - \frac{100}{x^2} \cdot \frac{dx}{dt} \]

We know \(\frac{dx}{dt} = -10\) feet per second (since the boat is approaching the shore, \(x\) is decreasing). When the boat is 200 feet from the shore, \(x = 200\).

First, find \(\theta\) when \(x = 200\): \[ \tan(\theta) = \frac{100}{200} = \frac{1}{2} \]

Thus, \(\sec^2(\theta) = 1 + \tan^2(\theta) = 1 + \left(\frac{1}{2}\right)^2 = \frac{5}{4}\).

Substitute these values into the differentiated equation: \[ \frac{5}{4} \cdot \frac{d\theta}{dt} = - \frac{100}{200^2} \cdot (-10) \]

Simplify: \[ \frac{5}{4} \cdot \frac{d\theta}{dt} = \frac{1000}{40000} \]

\[ \frac{5}{4} \cdot \frac{d\theta}{dt} = \frac{1}{40} \]

Solve for \(\frac{d\theta}{dt}\): \[ \frac{d\theta}{dt} = \frac{1}{40} \cdot \frac{4}{5} = \frac{1}{50} \]

Final Answer