One train travels west toward Denver (the origin) at $2 \mathrm{mi} / \mathrm{min}(120 \mathrm{mph})$, while another train travels north away from Denver at $1.5 \mathrm{mi} / \mathrm{min}(90 \mathrm{mph}$ ). At time $t=0$ (here $t$ is in minutes) the first train is 20 mi east of Denver and the second train is 10 mi north of Denver. Show how to use the derivative to find the time when the trains are closest together and find the distance between the two trains at that time.

Step 1 ：Given two trains, one moving west towards Denver at a speed of 2 mi/min and another moving north away from Denver at a speed of 1.5 mi/min. At time t=0, the first train is 20 mi east of Denver and the second train is 10 mi north of Denver.

Step 2 ：We can use the Pythagorean theorem to find the distance between the two trains at any given time. The distance is a function of time, given by \(d = \sqrt{(20 - 2t)^2 + (1.5t + 10)^2}\).

Step 3 ：To find the time when the trains are closest together, we take the derivative of the distance function and set it equal to zero. This gives us \(\frac{d}{dt} = \frac{6.25t - 25.0}{\sqrt{(20 - 2t)^2 + (1.5t + 10)^2}} = 0\).

Step 4 ：Solving this equation gives us the time at which the trains are closest together, \(t = \boxed{4}\) minutes.

Step 5 ：Substituting this time back into the distance function gives us the minimum distance between the two trains, \(d = \sqrt{(20 - 2*4)^2 + (1.5*4 + 10)^2} = \boxed{20}\) miles.