Question

At noon, ship $A$ is $170km$ west of ship $B$. Ship $A$ is sailing east at $40km/h$ and ship $B$ is sailing north at $25km/h$. How fast is the distance between the ships changing at $4:00PM$?

Answer 1

Let the ships $A$ and $B$ be at the points $Q$ and $P$ at $12$ noon (when t=0).

$PQ=170km$

At time t,let the positions of the ships be $Q^{\prime }and{P}^{\prime }.$

Therefore $PQ\text{'}=40t-100$and$PP\text{'}=25t$

The distance between the two ships at time t,is$P\text{'}Q\text{'}=s$

$s^2={(40t-100)}^2+{\left(25t\right)}^2$

$={(40t-100)}^2+625t^2-----\left(i\right)$

differentiating with respect to t,

$2s\frac{ds}{dt}=2(40t-100)\ast 40+1250t$

$2s\frac{ds}{dt}=80(40t-100)+1250t$

The rate of seperation at time t is

$\frac{ds}{dt}=\frac{1}{2s}\left[80\right(40t-100)+1250]$

$=\frac{1}{s}\left[40\right(40t-100)+625t]-----\left(ii\right)$

When the time is 4.00pm,$t=4$

$s=\sqrt{{(40\times 4-100)}^2+{(25\times 4)}^2}$

$=\sqrt{3600+10000}$

$=\sqrt{13600}$\\ $=40\sqrt{49}$

Therefore their rate of separation at this time(from(ii))

$\frac{ds}{dt}=\frac{1}{40\sqrt{49}}\left(35\right(140-100)+2500)$

$\frac{ds}{dt}=\frac{1}{40\sqrt{49}}(1400+2500)$

$\frac{ds}{dt}=\frac{3900}{40\sqrt{49}}$

$\frac{ds}{dt}=\frac{195}{\sqrt{49}}\frac{km}{hr}$