Science:Math Exam Resources/Courses/MATH100/December 2013/Question 07/Solution 1

If the distance between the spectator and the launchpad is $x$ and the distance between the launchpad and the rocket is $y$, the distance $z$ between the rocket and the spectator is $\displaystyle z={\sqrt {x^{2}+y^{2}}}$. We are given that $x$ = 4 km and $y$ = 3 km. Thus $z$ = 5 km.

Note that since the distance between the spectator and the launchpad is not changing, ${\frac {dx}{dt}}=0\,\mathrm {km/s}$. Differentiating both sides implicitly with respect to time $t$:

${\begin{aligned}z^{2}&=x^{2}+y^{2}\\2z\cdot {\frac {dz}{dt}}&=2x\cdot {\frac {dx}{dt}}+2y\cdot {\frac {dy}{dt}}\\{\frac {dz}{dt}}&={\frac {1}{z}}\left(x\cdot {\frac {dx}{dt}}+y\cdot {\frac {dy}{dt}}\right)\\&={\frac {1}{5\,\mathrm {km} }}\left(4\,\mathrm {km} \cdot 0\,\mathrm {km/s} +3\,\mathrm {km} \cdot 0.7\,\mathrm {km/s} \right)\\&={\color {blue}0.42\,\mathrm {km/s} }\end{aligned}}$