Answer:
[tex]y'\approx 522 \ mi[/tex]
Step-by-step explanation:
Rate of Change
The instantaneous rate of change is computed as the derivative of a function with respect to an independent variable, usually time.
We know a plane is flying horizontally at an altitude h=2 miles when it passes directly over a radar station. Let's call x the horizontal distance the plane travels.
Since the change of the distance x over time is the speed, then x'=570 mi/h. The distance x, the height and the distance between the plane and the radar station (y) form a right triangle. By the Pythagora's theorem
[tex]y^2=x^2+h^2[/tex]
We need to compute the rate at which y is changing when y=5 mi.
Differentiating with respect to time:
[tex]2yy'=2xx'+2hh'[/tex]
Please note that the altitude is constant, thus h'=0 and:
[tex]2yy'=2xx'[/tex]
Solving for y'
[tex]\displaystyle y'=\frac{xx'}{y}[/tex]
We need to compute x when y=5 and h=2
[tex]x^2=y^2-h^2=25-4=21[/tex]
Thus we have
[tex]x=\sqrt{21}[/tex]
Fiinally
[tex]\displaystyle y'=\frac{570\sqrt{21}}{5}[/tex]
[tex]\boxed{y'\approx 522 \ mi}[/tex]
