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

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

Hence,

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

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