Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 36

At what rate is cart $B$ moving toward $Q$ at the moment when cart $A$ is 5ft from $Q$?



We know that $L_A + L_B = 39$, if we take the derivative with respect to time
we will get $\displaystyle \frac{dL_A}{dt} + \frac{dL_B}{dt} = 0$ so,
$\displaystyle \frac{dL_A}{dt} = - \frac{dL_B}{dt} \qquad \Longleftarrow \text{ Equation 1}$
Also, by using Pythagorean Theorem in both triangle we obtain $L_A^2 = 12^2 + X_A^2$ and $L_B^2 = 12^2 + X_B^2$

Taking the derivative with respect to time,

$
\begin{equation}
\begin{aligned}
\cancel{2} L_A \frac{dL_A}{dt} &= \cancel{2}X_A \frac{dX_A}{dt} & \text{and} && \cancel{2}L_B \frac{dL_B}{dt} &= \cancel{2}X_B\frac{dX_B}{dt}\\
\\
\frac{dL_A}{dt} &= \frac{X_A}{L_A} \frac{dX_A}{dt} &&& \frac{dL_B}{dt} &= \frac{X_B}{L_B} \frac{dX_B}{dt}
\end{aligned}
\end{equation}
$


Substituting these values in Equation 1
$\displaystyle \frac{X_A}{L_A} \frac{dX_A}{dt} = - \left( \frac{X_B}{L_B} \frac{dX_B}{dt} \right)$

We have,
$\displaystyle \frac{dX_B}{dt} = - \frac{L_B X_A}{L_A X_B} \frac{dX_A}{dt}$; but $L_B = 39 - L_A$
$\displaystyle \frac{dX_B}{dt} = - \frac{(39-L_A) x_A}{L_A X_B}\frac{dX_A}{dt} \qquad \Longleftarrow \text{ Equation 2}$

Using the equations we in obtain Pythagorean Theorem, we know that when $X_A = 5$,
$L_A = \sqrt{5^2 + 12^2} = 13$
Hence,
$X_B = \sqrt{(39-L_A)^2-12^2} = \sqrt{(39-13)^2 - 12^2} = \sqrt{532}$

Now, plugging all values in Equation 2 we have,

$
\begin{equation}
\begin{aligned}
\frac{dX_B}{dt} &= \frac{-(39,13)(5)}{13(\sqrt{532})} (2)\\
\\
\frac{dX_B}{dt} &= -0.8671 \frac{\text{ft}}{s}
\end{aligned}
\end{equation}
$

This means that as cart $A$ is being pulled away from $Q$. the distance of cart $B$ to $Q$ is decreasing at a rate $\displaystyle 0.8671 \frac{\text{ft}}{s}$