# Thread: Calculus again!!!

1. ## Calculus again!!!

A ship is sailing due north at 12 kn when it observes another ship dead at a distance of 15 n. mi. The second ship is sailing due east at 9 kn. What is the closest distance of approach of the two ships?

2. Originally Posted by Raina
A ship is sailing due north at 12 kn when it observes another ship dead at a distance of 15 n. mi. The second ship is sailing due east at 9 kn. What is the closest distance of approach of the two ships?
immagine these two boats on a coordinate system. The first boat is at
(0,-15) and the is at the origin.

the coordinate of the 1st boat is given by the equation

$(0,12t-15)$

and the coordinate of the 2nd is

$(9t,0)$

We can find the distance between using the pythagorean theorem.

$d^2=(0-9t)^2+((12t-15)-0)^2$

$d=\sqrt{225t^2-360t+225}$

$d=3\sqrt{25t^2-40t+25}$

taking the derivative we get

$d'=\frac{3(25t-20)}{\sqrt{25t^2-40t+25}}$

setting the numerator equal to zero we get

$0=3(25t-20) \iff 0=25t-20 \iff 20 =25t \iff t=\frac{4}{5}$

$d=3\sqrt{25(\frac{4}{5})^2-40(\frac{4}{5})+25}$

$d=3\sqrt{16-32+25}=3\sqrt{9}=9$

Yeah!!

I hope this helps