Question

1. air traffic control an airplane is flying at an altitude of 7 mi and passes directly over a radar antenna as shown in the figure. when the plane is 10 mi from the antenna (s = 10), the radar detects that the distance s is changing at the rate of 300 mph. what is the speed of the airplane at that moment? the plane flies at a fixed altitude of 7 miles above the ground. given: find:

Explanation:

Step1: Establish the relationship

By the Pythagorean theorem, $s^{2}=x^{2}+7^{2}=x^{2} + 49$.

Step2: Differentiate both sides with respect to time $t$

Using the chain - rule, $2s\frac{ds}{dt}=2x\frac{dx}{dt}$. Then $\frac{dx}{dt}=\frac{s}{x}\cdot\frac{ds}{dt}$.

Step3: Find the value of $x$ when $s = 10$

When $s = 10$, from $s^{2}=x^{2}+49$, we have $10^{2}=x^{2}+49$, so $x^{2}=100 - 49=51$ and $x=\sqrt{51}$.

Step4: Substitute the known values

We know that $s = 10$, $x=\sqrt{51}$, and $\frac{ds}{dt}=300$. Then $\frac{dx}{dt}=\frac{10}{\sqrt{51}}\times300=\frac{3000}{\sqrt{51}}\approx421.4$ mph.

Answer:

$\frac{3000}{\sqrt{51}}\approx421.4$ mph