Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 42

At what rate is the distance between the people changing 15 minutes later.
Given: Speed of the person who walks east = $\displaystyle \frac{3\text{mi}}{h}$
Speed of the person who walks northeast = $\displaystyle \frac{3\text{mi}}{h}$

Required: rate of change of distance between the two people after 15 minutes
We use cosine law to relate all the given parameters

$a^2 = b^2 + c^2 - 2 (b)(c) \cos \theta \qquad \Longleftarrow \text{ Equation 1}$





$
\begin{equation}
\begin{aligned}
\text{where } a &= \text{ distance between the two people}\\
b &= \text{ distance walked by the person due northeast}\\
c &= \text{ distance walked by the person due east}\\
\theta &= \text{ angle between the two people}
\end{aligned}
\end{equation}
$


Also, by inspection, we found out that the angle between the two people is fixed and is equal to $45^\circ$



$
\begin{equation}
\begin{aligned}
a^2 &= b^2 + c^2 - 2 (b) (c) \cos (45)\\
a^2 &= b^2 + c^2 - 2 (b) (c) \left( \frac{\sqrt{2}}{2}\right)\\
a^2 &= b^2 + c^2 - \sqrt{2} (b) (c)

\end{aligned}
\end{equation}
$


Taking the derivative with respect to time
$\displaystyle 2a \frac{da}{dt} = 2b \frac{db}{dt} + 2c \frac{dc}{dt} - \sqrt{2} \left[ b \frac{db}{dt} + c \frac{db}{dt}\right] \qquad \Longleftarrow \text{ Equation 2}$

To get the value of the distance transversed $b$ and $c$ after 15 minutes, we will use their corresponding speed.


$
\begin{equation}
\begin{aligned}
b & = 2 \frac{\text{mi}}{h} (15 \text{ minutes)} \left( \frac{1h}{60 \text{ minutes}}\right) 0.5\text{mi} && \Longleftarrow \text{ make sure to be consistent with the units}\\
\\
c & = 3 \frac{\text{mi}}{h} (15 \text{ minutes)} \left( \frac{1h}{60 \text{ minutes}}\right) 0.75\text{mi}
\end{aligned}
\end{equation}
$

To solve the value of $a$, we will use the Equation 1

$
\begin{equation}
\begin{aligned}
a^2 &= b^2 + c^2 - 2 (b)(c) \cos \theta\\
\\
a &= \sqrt{(0.5)^2 + (0.75)^2 - 2 ().5)(0.75) \cos (45)}\\
\\
a &= 0.5312 \text{mi}
\end{aligned}
\end{equation}
$


Solving for the required, we use Equation 2, plugging in the given and completed value we get...
$\displaystyle 2a \frac{da}{dt} = 2b \frac{db}{dt} + 2c \frac{dc}{dt} - \sqrt{2} \left[ b \frac{dc}{dt} + c \frac{db}{dt} \right]$

Solving for $\displaystyle \frac{da}{dt}$
$\displaystyle \frac{da}{dt} = \frac{2(0.5)(2) + 2(0.75)(3) - \sqrt{2} \left[ (0.5)(3)+(0.75)(2) \right]}{2(0.5312)}$

$\boxed{\displaystyle \frac{da}{dt} = 2.1248 \frac{\text{mi}}{h}}$