# Derivatives in the real world

Two row boats start at the same location, and start traveling apart along straight lines which meet at an angle of $\pi/3$. Boat A is traveling at a rate of $10$ miles per hour directly east, and boat $B$ is traveling at a rate of $16$ miles per hour going both north and east. How fast is the distance between the rowboats increasing $2$ hours into the journey?

I cannot figure out why the equation between boats and the distance in this case is $r^2= x^2+y^2-2xy\cos(\pi/3)$. This equation was given in the solution to the problem.

• Try using the law of cosines. – Sak Jul 4 '14 at 1:29
• The equation you're having trouble with is called the law of cosines. Its like the pythagorean theorem but more general. – Spencer Jul 4 '14 at 1:30
• Do you mean it applies to triangles that are non right angled? – Minu Jul 4 '14 at 2:06
• Yes. When you apply it to right angled triangles at the right angle, the law of cosines is simply the Pythagorean Theorem. – Sak Jul 4 '14 at 2:11
• Can you help me with the derivative of the equation? Please explain each step. – Minu Jul 4 '14 at 2:24

Velocity of the first boat $\hat V_1= 10\hat i$

Velocity of the second boat $\hat V_2 = 16cos(\frac{\pi}{3})\hat i + 16sin(\frac{\pi}{3})\hat j$

After a time "t"

Displacement by boat 1 $\hat D_1 = 10t\hat i$

Displacement by boat 2 $\hat D_2 = 16tcos(\frac{\pi}{3})\hat i + 16tsin(\frac{\pi}{3})\hat j$

Using parallelogram law to find the Displacement between Boat 1 and Boat 2

$\hat D = \hat D_1 - \hat D_2$

$\hat D = (10-16cos(\frac{\pi}{3}))t\hat i - 16tsin(\frac{\pi}{3})\hat j$

Magnitude of the displacement (r) =$\sqrt{\left({\left(10t-16tcos(\frac{\pi}{3})\right)}^2 + {\left(16tsin(\frac{\pi}{3})\right)}^2\right)}$

If you simplify, then you get $r = \sqrt{\left(100t^2 + 256t^{2}{(cos(\frac{\pi}{3}))}^2 - 2.10.16t^{2}cos(\frac{\pi}{3}) + 256t^{2}{(-sin(\frac{\pi}{3}))}^2\right)}$

$r = \sqrt{\left(100t^2 + 256t^{2} - 2.10.16t^{2}cos(\frac{\pi}{3}) \right)}$

This is where you are getting the forumula that you mentioned in the book.

$r = \sqrt{\left(356t^2 -2.10.16.t^{2}\frac{1}{2}\right)}$

$r = \sqrt{\left(356t^2 - 160t^{2}\right)}$

$r = \sqrt{196t^2} = 14t$

After two hours into the jounrney, the distance would be $=14\times2 = 28$

The rate at which it is increasing would be

$dr/dt = 14$ miles per hour.

you get the "Rate at which the distance is increasing when boats are 2 hours into the journey" = $14$ miles.

This is the "Physics" way of solving the problem.

Thanks

Satish