Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 44

Suppose that a boat leaves a dock at 2:00PM and travels due south at a speed of $\displaystyle 20 \frac{\text{km}}{h}$. Another boat has been heading due east at $\displaystyle 15 \frac{\text{km}}{h}$ and reaches the same dock at 3:00PM. At what time were the two boats closes together.




Let $t$ be the time since 2 PM.
By using Pythagorean Theorem.

$
\begin{equation}
\begin{aligned}
z^2 &= x^2 + y^2 = [15(t-1)]^2 + (20 t)^2\\
\\
z^2 &= \sqrt{[15(t-1)]^2 + (20t)^2}
\end{aligned}
\end{equation}
$

Taking the derivative of $z$ with respect to time...
$\displaystyle z' = \frac{2[15(t-1)](15)+2(20t)(20)}{2\sqrt{[15(t-1)]^2 + (20t)^2}}$

when $z' = 0$,

$
\begin{equation}
\begin{aligned}
0 &= 450(t-1) + 800t\\
\\
0 &= 450t - 450 + 800t\\
\\
t &= \frac{450}{1250} = 0.36 \text{ hour}\\
\\
t &= 0.36 \text{ hour} \left( \frac{60 \text{minutes}}{1 \text{hour}} \right) = 21.6 \text{minutes}\\
\\
t &= 21 \text{minutes } + 0.6 \text{minutes } \left( \frac{60s}{1 \text{minute}} \right) \\
\\
t &= 21 \text{minutes } + 36 \text{seconds}
\end{aligned}
\end{equation}
$


Therefore, we can say that the two boats are closest together at $2:21:36\text{PM}$