Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 20

At what rate is the boat approachng the dock when it is 8m from the dock?
Illustration:







Required: $\displaystyle \frac{dx}{dt}$ when $x = 8m$

Solution:

By using Pythagorean Theorem we have,

$z^2 = 1^2 + x^2 \qquad$ Equation 1

taking the derivative with respect to time


$
\begin{equation}
\begin{aligned}

\cancel{2} z \frac{dz}{dt} =& \cancel{2}x \frac{dx}{dt}
\\
\\
\frac{dx}{dt} =& \frac{z}{x} \frac{dz}{dt} \qquad \text{Equation 2}

\end{aligned}
\end{equation}
$


We can get the value of $z$ by substituting $x = 8$ in equation 1


$
\begin{equation}
\begin{aligned}

z^2 =& 1^2 +x^2
\\
\\
z^2 =& 1^2 + 8^2
\\
\\
z =& \sqrt{65} m

\end{aligned}
\end{equation}
$


Now, using equation 2 to solve for the unknown


$
\begin{equation}
\begin{aligned}

\frac{dx}{dt} =& \frac{\sqrt{65}}{8} (1)
\\
\\
\frac{dx}{dt} =& \frac{\sqrt{65}}{8} m/s

\end{aligned}
\end{equation}
$


This means that the boat is approaching the dock at a rate of $\displaystyle \frac{\sqrt{65}}{8} m/s$ when the boat is $8 m$ from the dock.