Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 11

How fast is the distance from the plane to the station is increasing when it is 2mi away from the station?
a.) State the quantities given in the problem.
b.) State the unknown.
c.) Illustrate a picture of the situation for any time $t$.
d.) Write an expression that relates the quantities.
e.) Solve the problem

a.) Given: $\displaystyle y = \text{altitude} = 1 mi$

$\qquad \displaystyle \frac{dx}{dt} = \text{speed} = 500 mi/h$

b.) Unknown: The rate of change of the distance of the plane to the station, $\displaystyle \frac{dz}{dt}$

c.)







d.) By using Pythagorean Theorem, we get $z^2 = 1^2 + x^2$. The altitude is constant because the plane only moves horizontally.

e.)


$
\begin{equation}
\begin{aligned}

& \cancel{2} z \frac{dz}{dt} = \cancel{2}x \frac{dx}{dt} ;
\\
\\
& \frac{dz}{dt} = \frac{x}{z} \left( \frac{dx}{dt} \right)

\end{aligned}
\end{equation}
$


To get the value of $x$, we substitute the $z$ to the equation


$
\begin{equation}
\begin{aligned}

z^2 =& 1^2 + x^2
\\
\\
(2)^2 =& 1^2 + x^2
\\
\\
x^2 =& 2^2 - 1^2
\\
\\
x =& \sqrt{3}

\end{aligned}
\end{equation}
$


Therefore,

$\displaystyle \frac{dz}{dt} = \frac{\sqrt{3}}{2} (500)$

$\fbox{$\large \frac{dz}{dt} = 250 \sqrt{3} mi/h $}$