Question

A ship leaves port traveling southwest at a rate of 12 mi/hr. At noon, the ship reaches its closest approach to a radar station, which is on the shore 1.5 mi from the port. If the ship maintains its speed and course, what is the rate of change of the tracking angle θ between the radar station and the ship at 1:30 P.M. (see figure)? (Hint: Use the Law of Sines.)

Accepted Answer

First, understand the setup: The ship is traveling southwest, forming a right triangle with the radar station and the path of the ship. The radar station is 1.5 miles from the port, which is the closest point of approach. Define the variables: Let x be the distance the ship has traveled southwest from the closest point of approach. Let θ be the angle between the line from the radar station to the ship and the line from the radar station to the closest point of approach. Use the Law of Sines: In the triangle formed, the Law of Sines states that $$ \frac{\sin(	heta)}{x} = \frac{\sin(90^\circ)}{1.5} . $$Since $$ \sin(90^\circ) = 1 $$, this simplifies to $$ \sin(	heta) = \frac{x}{1.5} . $$Differentiate with respect to time: To find the rate of change of θ, differentiate both sides of the equation $$ \sin(	heta) = \frac{x}{1.5} $$ with respect to time t. Use the chain rule: $$ \frac{d}{dt}[\sin(	heta)] = \cos(	heta) \cdot \frac{d	heta}{dt} $$ and $$ \frac{d}{dt}[\frac{x}{1.5}] = \frac{1}{1.5} \cdot \frac{dx}{dt} . $$Solve for $$ \frac{d	heta}{dt} $$: Set $$ \cos(	heta) \cdot \frac{d	heta}{dt} = \frac{1}{1.5} \cdot \frac{dx}{dt} . $$Given that the ship travels at 12 mi/hr, $$ \frac{dx}{dt} = 12 . $$Substitute this value into the equation and solve for $$ \frac{d	heta}{dt} .$$

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Key Concepts

Law of Sines

Rate of Change

Trigonometric Functions