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.