Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 15

How fast is the distance between the cars increasing after two hours?

Given:

$\qquad$ speed of car A = $60 mi/h$

$\qquad$ speed of car B = $25 mi/h$

Required: Distance between the cars 2 hours later

Solution:







By using Pythagorean Theorem,


$\begin{equation} \begin{aligned} z^2 = x^2 + y^2 \qquad \text{Equation 1} \end{aligned} \end{equation}$


We multiply the instance $t = 4$ to the corresponding speed of the cars to get the distance at the moment 2 hours later.


$\begin{equation} \begin{aligned} x =& 25 mi/h (2h) = 50 mi \\ \\ y =& 60 mi/h(2h) = 120 mi \end{aligned} \end{equation}$


From equation 1, we get the value of $z$

$z = \sqrt{50^2 + 120^2} = 130 mi$

To get the unknown, we derive equation 1 with respect to time


$\begin{equation} \begin{aligned} z^2 =& x^2 + y^2 \\ \\ \cancel{2} z \frac{dz}{dt} =& \cancel{2}x \frac{dx}{dt} + \cancel{2} y \frac{dy}{dt} \\ \\ \frac{dz}{dt} =& \frac{\displaystyle x \frac{dx}{dt} + y \frac{dy}{dt} }{z} \\ \\ \frac{dz}{dt} =& \frac{50(25) + 120 (60)}{130} \\ \\ &\boxed{\displaystyle \frac{dz}{dt} = 65 mi/h} \end{aligned} \end{equation}$