Question

A jet flying at 450 mi/hr and traveling in a straight line at a constant elevation of 500 ft passes directly over a spectator at an air show. How quickly is the angle of elevation (between the ground and the line from the spectator to the jet) changing 2 seconds later?

Accepted Answer

First, understand the scenario: The jet is flying horizontally at a constant speed of 450 mi/hr, which needs to be converted to feet per second for consistency with the elevation given in feet. Use the conversion factor: 1 mile = 5280 feet. Next, set up the relationship between the angle of elevation $$ 	heta $$, the horizontal distance $$ x $$ from the spectator to the jet, and the constant elevation $$ h = 500  ft. $$The tangent of the angle $$ 	heta  is $$given by $$ 	an(	heta) = \frac{h}{x} . $$Differentiate the equation $$ 	an(	heta) = \frac{h}{x} $$ with respect to time $$ t  to $$find $$ \frac{d	heta}{dt} . $$Use implicit differentiation: $$ \sec^2(	heta) \cdot \frac{d	heta}{dt} = -\frac{h}{x^2} \cdot \frac{dx}{dt} . $$Determine $$ \frac{dx}{dt} $$, the rate at which the horizontal distance $$ x  is $$changing. Since the jet is moving at 450 mi/hr, convert this speed to feet per second: $$ \frac{450 	imes 5280}{3600}  ft/s. $$Finally, substitute the known values into the differentiated equation. At $$ t = 2 $$ seconds, calculate $$ x $$ using $$ x = 	ext{speed} 	imes t . $$Then, solve for $$ \frac{d	heta}{dt} $$ using the values of $$ x $$, $$ h $$, and $$ \frac{dx}{dt} .$$

A jet flying at 450 mi/hr and traveling in a straight line at a constant elevation of 500 ft passes directly over a spectator at an air show. How quickly is the angle of elevation (between the ground and the line from the spectator to the jet) changing 2 seconds later?

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Key Concepts

Related Rates

Trigonometric Functions

Differentiation