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
$\displaystyle (0,12t-15)$
and the coordinate of the 2nd is
$\displaystyle (9t,0)$
We can find the distance between using the pythagorean theorem.
$\displaystyle d^2=(0-9t)^2+((12t-15)-0)^2$
$\displaystyle d=\sqrt{225t^2-360t+225}$
$\displaystyle d=3\sqrt{25t^2-40t+25}$
taking the derivative we get
$\displaystyle d'=\frac{3(25t-20)}{\sqrt{25t^2-40t+25}}$
setting the numerator equal to zero we get
$\displaystyle 0=3(25t-20) \iff 0=25t-20 \iff 20 =25t \iff t=\frac{4}{5}$
$\displaystyle d=3\sqrt{25(\frac{4}{5})^2-40(\frac{4}{5})+25}$
$\displaystyle d=3\sqrt{16-32+25}=3\sqrt{9}=9$
Yeah!!
I hope this helps