Science:Math Exam Resources/Courses/MATH110/April 2014/Question 06/Solution 1

This is a related rates problem. So we go about it using the standard related rates approach.

1. Draw a diagram
2. Variables: Define the distance between the two ships to be $z$. We want to find $z'(t)$. Define the distance between the ships and the origin to be $a$ and $b$ respectively. Set the original location of ship B to be at the origin. After $4$ hrs, ship A is $50$ km east of the origin and ship B is $120$ km north of the origin.
3. Equations: The Pythagorean theorem will be useful here. The distance between them is $z={\sqrt {50^{2}+120^{2}}}=130km$. Since A is going away from the origin and B is also going away from the origin, we have $a'=20$ and $b'=30$.
4. Differentiating the distance formula: $z^{2}=a^{2}+b^{2}$, we have: $zz'=aa'+bb'$, substitute to get: $130z'=50\cdot 20+120\cdot 30$.
5. Rearrange to get $z'={\frac {460}{13}}\approx 35.38km/hr$. The distance between the ships are increasing at ${\frac {460}{13}}\approx 35.38km/hr$.