a plane flying horizontally at an altitude of 2 mi and a speed of 480 mi/h passes directly over a radar…

Explanation:

Step1: Establish a right - triangle relationship

Let $y = 2$ (constant altitude), $x$ be the horizontal distance of the plane from the point directly above the radar station, and $z$ be the distance from the plane to the radar station. By the Pythagorean theorem, $x^{2}+y^{2}=z^{2}$. Since $y = 2$, we have $x^{2}+4=z^{2}$.

Step2: Differentiate with respect to time $t$

Differentiating both sides of $x^{2}+4=z^{2}$ with respect to $t$ gives $2x\frac{dx}{dt}=2z\frac{dz}{dt}$, which simplifies to $x\frac{dx}{dt}=z\frac{dz}{dt}$.

Step3: Find $x$ when $z = 5$

When $z = 5$, using $x^{2}+4=z^{2}$, we substitute $z = 5$ into it: $x^{2}+4 = 25$, so $x=\sqrt{25 - 4}=\sqrt{21}$.

Step4: Substitute known values and solve for $\frac{dz}{dt}$

We know that $\frac{dx}{dt}=480$ (speed of the plane), $x=\sqrt{21}$, and $z = 5$. Substituting into $x\frac{dx}{dt}=z\frac{dz}{dt}$, we get $\sqrt{21}\times480 = 5\times\frac{dz}{dt}$. Then $\frac{dz}{dt}=\frac{480\sqrt{21}}{5}=96\sqrt{21}\approx440$.

Answer:

440