At a given moment, a plane passes directly above a radar station at an altitude of 6 km6 km and the plane's speed is 800 km/h.800 km/h. Let thetaθ be the angle that the line through the radar station and the plane makes with the horizontal. How fast is thetaθ changing 24 min24 min after the plane passes over the radar station? (Use decimal notation. Give your answer to three decimal places.)

the plane is moving with constant speed therefore x(t) = 222.222*t => no forces are interacting horizontally with the plane therefore acceleration is 0, then v is constant and x(t) is a linear function which coefficient is v.

now we have a triangle with an angle theta, one side is x(t), and the other is 6000m. we can get theta by tan(theta) = 6000/(222.222*t). 24 minutes are 1440 seconds so if we replace such value, we get the theta angle by solving for theta => theta = arctan(6000/(222.222*1440)) = 0.019 radians or 1.074 degrees. Now if you want to know the exchange rate of theta we have to differentiate the expression with respect to t:

[tex]\frac{d}{dt} \theta = \frac{d}{dt}(arctan(\frac{6000}{222.222*t} )) = \frac{1}{1+\frac{6000}{222.222*t} } * -\frac{27}{t^2} = -\frac{27}{t^2+729.001}[/tex]

then replace t with 1440 and you will get that theta is changing by -0.000013 (1.3E-5) radians or -7.458E-4 degrees every second which has a lot of sense since the plane is getting out of your line of sight due to the earth's curvature