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

MATH100 December 2013

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Question 07

Full-Solution Problem. Justify your answer and show your work. Full simplification of numerical answers is required unless explicitly stated otherwise. At a distance of 4 kilometres from the launch site, a spectator is observing a rocket being launched. If the rocket lifts off vertically and is rising at a speed of 0.7 kilometres/second when it is at an altitude of 3 kilometres, how fast is the distance between the rocket and the spectator changing at that instant?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
What is the distance between the rocket and the spectator at that instant? If the distance between the spectator and the launchpad is $x$ and the distance between the launchpad and the rocket is $y$, what is the distance $z$ between the rocket and the spectator?

Hint 2
From the hint above, the distance between the rocket and the spectator is $\displaystyle z={\sqrt {x^{2}+y^{2}}}$. How can we manipulate this equation to relate the rate of change in $z$ to the rates of change of $x$ and $y$?

Hint 3
Differentiate both sides of the equation $\displaystyle z^{2}=x^{2}+y^{2}$ with respect to time $t$ and solve for ${\frac {dz}{dt}}$.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $t$: ${\begin{aligned}z^{2}&=x^{2}+y^{2}\\2z\cdot {\frac {dz}{dt}}&=2x\cdot {\frac {dx}{dt}}+2y\cdot {\frac {dy}{dt}}\\{\frac {dz}{dt}}&={\frac {1}{z}}\left(x\cdot {\frac {dx}{dt}}+y\cdot {\frac {dy}{dt}}\right)\\&={\frac {1}{5\,\mathrm {km} }}\left(4\,\mathrm {km} \cdot 0\,\mathrm {km/s} +3\,\mathrm {km} \cdot 0.7\,\mathrm {km/s} \right)\\&={\color {blue}0.42\,\mathrm {km/s} }\end{aligned}}$

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