# Related rates problem: Why is the first boat $15$ miles north of the crossing point?

I am trying to solve a related rates question, but I can't solve it myself.

The question is:

Two boats are traveling at $$30$$ miles/hr, the first going north and the second going east. The second crosses the path of the first $$10$$ minutes after the first one was there. At what rate is their distance increasing when the second has gone $$10$$ miles beyond the crossing point?

I try to use Pythagorean theorem solve it. The distance between the two boats is $$\sqrt{x^2+y^2}$$, where $$x$$ is how far the second boat beyond the crossing point, $$y$$ is how far the first boat beyond the crossing point. My problem is I don't know how far is the first boat beyond the point when the second boat has gone $$10$$ miles beyond the point?

I have looked at the solution; it said the distance is $$15$$ miles, but I don't know how to get this result.

• Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. Dec 5 '19 at 2:01

Ten minutes is one sixth of an hour. Since the first boat is traveling at $$30~\text{mi}/\text{h}$$, it travels $$\left(30~\frac{\text{mi}}{\text{h}}\right)\left(\frac{1}{6}\right) = 5~\text{mi}$$ in ten minutes. Therefore, the first boat is five miles north of the second boat when the second boat crosses the path of the first boat.
The two boats are traveling at the same speed. Therefore, when the second boat has traveled ten miles east of the crossing point, the first boat has traveled an additional ten miles north of the crossing point. Thus, the first boat is $$5~\text{mi} + 10~\text{mi} = 15~\text{mi}$$ north of the crossing point when the second boat is ten miles east of the crossing point.