# Related Rates: rate of change of the distance between two objects [closed]

A boat starts off $$172$$ miles directly west from the city. It travels due south at a speed of $$30$$ miles per hour. After travelling $$126$$ miles, how fast is the distance between the boat and city changing?

This is on my study guide for an exam and I'm confused on what equation to use?

## closed as off-topic by Eevee Trainer, max_zorn, Shailesh, John Omielan, mrtaurhoMar 5 at 6:08

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• Asking "how fast is the distance between the boat and city changing" means that it wants you to find the rate of change of the distance between the boat and the city. "The time spent for the 126 miles is 126/30. The expected duration is 172/(126/30)=172*30/126=40.96 hours." This calculation would measure how fast the boat is moving, or how fast the distance between the boat and its starting point is changing. This does not tell us how fast the distance between the boat and the city are changing. – Jake O Mar 5 at 14:35
• @JakeO, this makes sense. I did not claim that I understood the question! – NoChance Mar 5 at 19:31
• @NoChance, no problem! Just wanted to clarify what the question was asking :) – Jake O Mar 6 at 6:06

The distance between the city and the boat can be expressed by this function (it's nothing more than the Pythagorean theorem): $$D(t)=\sqrt{172^2+(30t)^2}\ mi$$

The boat is going to cover a distance of 126 miles at time: $$30t=126\implies t=4.2\ h$$

How fast the distance between the boat and the city is changing after the boat has traveled 126 miles south is the derivative of the distance function with respect to time evaluated at $$t=4.2$$ hours: $$D'(4.2)\ mph$$

Here's an illustration of how I found the distance function: • D(t) is not just the distance...It is the "shortest distance". – NoChance Mar 5 at 19:34

Call distance downward $$a(t)$$

Call distance to the city $$b(t)$$, both as functions of time.

So by Pythagoras theorem we get $$b(t)^2 - a(t)^2 = 172^2$$

If we differentiate both sides, by chain rule, $$2(b(t))(db/dt) - 2(a(t))(da/dt) = 0$$ $$2(213.21)(db/dt) - 2(126)(30) =0$$ $$db/dt = 7560/426.42$$ $$db/dt = 17.73 mph$$