Science:Math Exam Resources/Courses/MATH110/April 2012/Question 08/Solution 1

Let $x$ be the distance of the ship from the drop point, $y$ be the distance of the probe from the drop point, and $r$ be the distance between the ship and the probe. Drawing a labelled picture looks like this:

We want to find $\frac{dr}{dt}$, at time 10s after the probe is dropped. We know that $\frac{dx}{dt}=5m/s$ and $\frac{dy}{dt}=4m/s$.

By the Pythagorean Theorem,

$x^2+y^2=r^2.$

In order to find the relationship between the time rates of change of these variables, we differentiate both sides of this equation with respect to time, in order to obtain

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=2r\frac{dr}{dt}$

Dividing both sides by $2r$ in order to isolate $\frac{dr}{dt}$, we find

$\frac{dr}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{r}.$

We are interested in the value of $\frac{dr}{dt}$, at time $10s$ after the probe is dropped. At this point in time,

$x=5m/s\cdot 10s=50m,$

$y=4m/s\cdot 10s=40m,$

and

$\begin{array}{rl}r& =\sqrt{x^2+y^2}=\sqrt{(50m{)}^2+(40m{)}^2}\\ & =10\sqrt{5^2+4^2}m=10\sqrt{41}m.\end{array}$

Hence,

$\begin{array}{rl}\frac{dr}{dt}& =\frac{(50m)(5m/s)+(40m)(4m/s)}{10\sqrt{41}m}\\ & =\frac{25m/s+16m/s}{\sqrt{41}}\\ & =\sqrt{41}m/s.\end{array}$

That is, $10s$ after the probe is dropped, the distance between the probe and ship is increasing at a rate of $\sqrt{41}m/s$.