An aircraft is flying horizontally at a constant height of 400 ft above a fixed observational point..
At a certain instant the speed of the aircraft is 300 mph and the angle of elevation (fata) is 30 degrees and decreasing. How fast is (fata) decreasing at this instant?
The 400ft is the vertical distance (y) between the observer and the aircraft. As the aircraft fly away from the observer, the horizontal distance (x) between the observer and the aircraft increases. As x increases, the angle of elevation (`theta` ) decreases.
The above conditions can be described using a triangle at a certain time. Please see figure below.
When `theta` is 30, the horizontal distance can be determined using the tangent function.
`tan theta = y / x`
Substitute value of `theta` and y.
`tan 30 = 400/x`
So the horizontal distance is `400sqrt3` ft.
To determine the rate of `theta` at the time when theta is `30^o` , take the derivative of tangent function with respect to time (t). In taking the derivative, note that y is a constant.
`d/(dt) (tan theta)=d/dt(y/x)`
`d/(dt) (tan theta)=yd/dt(x^(-1))`
`sec^2theta *(d theta)/(dt)=y(-x^-2)*(dx)/(dt)`
`sec^2theta *(d theta)/(dt)=-y/x^2 *(dx)/(dt)`
Then, isolate `(d theta)/(dt)` .
`(d theta)/(dt)= -y/(x^2sec^2theta)*(dx)/(dt)`
`(d theta)/(dt)=-(ycos^2theta)/x^2 *(dx)/(dt)`
Note that dx/dt is the instantaneous speed of the aircraft when theta=30. So `(dx)/(dt) = 300` mi/hr.
Then, convert the unit of the speed to ft/hr in order to be consistent with the unit of x and y.
`(dx)/(dt) = 300 (mi)/(hr) * (5280 ft)/(1mi) = 15840000 (ft)/(hr)`
And then, substitute the values of x,y `theta` , and `(dx)/(dt)` to determine the rate of the angle of elevation.
`(d theta)/(dt)= -(400cos^2 30)/((400sqrt3)^2)*15840000`
Hence, when `theta=30` , the angle of elevation is decreasing at 9900 degree per hour.