Questions: A plane flying horizontally at an altitude of 3 miles and a speed of 540 mi / h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it has a total distance of 4 miles away from the station. (Round your answer to the nearest whole number.) mi / h

Solution Steps

We need to find the rate at which the distance from the plane to the radar station is increasing when the plane is 4 miles away from the station. The plane is flying horizontally at an altitude of 3 miles and a speed of 540 miles per hour.

Let \( x \) be the horizontal distance from the plane to the radar station, and \( y \) be the altitude of the plane. The distance \( s \) from the plane to the radar station can be found using the Pythagorean theorem: \[ s^2 = x^2 + y^2 \] Given:

• \( y = 3 \) miles (constant)
• \( \frac{dx}{dt} = 540 \) miles per hour (horizontal speed of the plane)
• \( s = 4 \) miles (total distance from the plane to the station)

Differentiate both sides of the equation \( s^2 = x^2 + y^2 \) with respect to time \( t \): \[ 2s \frac{ds}{dt} = 2x \frac{dx}{dt} \] Simplify: \[ s \frac{ds}{dt} = x \frac{dx}{dt} \] Solve for \( \frac{ds}{dt} \): \[ \frac{ds}{dt} = \frac{x}{s} \frac{dx}{dt} \]

Using the Pythagorean theorem: \[ 4^2 = x^2 + 3^2 \] \[ 16 = x^2 + 9 \] \[ x^2 = 7 \] \[ x = \sqrt{7} \]

Substitute \( x = \sqrt{7} \), \( s = 4 \), and \( \frac{dx}{dt} = 540 \) into the equation: \[ \frac{ds}{dt} = \frac{\sqrt{7}}{4} \cdot 540 \] \[ \frac{ds}{dt} = \frac{540 \sqrt{7}}{4} \] \[ \frac{ds}{dt} = 135 \sqrt{7} \]

Calculate \( 135 \sqrt{7} \): \[ 135 \sqrt{7} \approx 135 \cdot 2.6458 \approx 357.183 \] Round to the nearest whole number: \[ \frac{ds}{dt} \approx 357 \text{ miles per hour} \]

Final Answer