SOLUTION: A plane flies horizontally at an altitude of 3 km and passes directly over a tracking telescope on the ground. When the angle of elevation is π/4, this angle is decreasing at

Algebra ->  Equations -> SOLUTION: A plane flies horizontally at an altitude of 3 km and passes directly over a tracking telescope on the ground. When the angle of elevation is π/4, this angle is decreasing at       Log On


   

Question 1127029: A plane flies horizontally at an altitude of 3 km and passes directly over a tracking telescope on the ground. When the angle of elevation is π/4, this angle is decreasing at a rate of π/4 rad/min. How fast is the plane traveling at that time?
Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


We need to find the relationship between the change in the angle of elevation and the speed of the plane -- i.e., the change in the plane's distance from the tracking station.

Let x be the (horizontal) distance of the plane from the ground station; let y be the angle of elevation. Then, since the plane is flying at an altitude of 3km, x and y are related by

tan%28y%29+=+3%2Fx

Find the derivative with respect to time and solve for the plane's speed, dx/dt.

sec%5E2%28y%29%28dy%2Fdt%29+=+%28-3%2Fx%5E2%29%28dx%2Fdt%29

dx%2Fdt+=+%28-x%5E2%2F3%29sec%5E2%28y%29%28dy%2Fdt%29

The given information is that when the angle of elevation is pi/4, the rate of change in the angle of elevation is -pi/4 rad/min. At an angle of pi/4, x is 3, cos(x) is sqrt(2)/2, cos^2(x) = 1/2, sec^2(x) = 2. So

dx%2Fdt+=+%28-9%2F3%29%282%29%28-pi%2F4%29+=+3pi%2F2

The plane's speed is 3pi/2 km/min, or about 283 km/hr, about 177 mph.