question two airplanes are flying in the air at the same height. airplane a is flying east at 400 mi/h and…

Explanation:

Step1: Define variables

Let $x$ be the distance of airplane $A$ from the airport, $y$ be the distance of airplane $B$ from the airport, and $z$ be the distance between the two airplanes. By the Pythagorean - theorem, $z^{2}=x^{2}+y^{2}$.

Step2: Differentiate with respect to time $t$

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

Step3: Find the initial values of $x$, $y$, and $z$

We know that airplane $A$ is flying east at a speed of $\frac{dx}{dt}=- 400$ mi/h (negative because $x$ is decreasing) and airplane $B$ is flying north at a speed of $\frac{dy}{dt}=-200$ mi/h (negative because $y$ is decreasing). Initially, $x = 60$ miles and $y = 80$ miles. Then, by the Pythagorean - theorem, $z=\sqrt{60^{2}+80^{2}}=\sqrt{3600 + 6400}=\sqrt{10000}=100$ miles.

Step4: Substitute the values into the differentiated equation

Substitute $x = 60$, $y = 80$, $z = 100$, $\frac{dx}{dt}=-400$, and $\frac{dy}{dt}=-200$ into $z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$. We get $100\frac{dz}{dt}=60\times(-400)+80\times(-200)$. First, calculate the right - hand side: $60\times(-400)+80\times(-200)=-24000-16000=-40000$. Then, solve for $\frac{dz}{dt}$: $\frac{dz}{dt}=\frac{-40000}{100}=-400$ mi/h.

Answer:

400