Science:Math Exam Resources/Courses/MATH100/December 2014/Question 07 (b)

MATH100 December 2014

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Question 07 (b)
Particle $1$ moves on the $y$-axis starting at the point $(0,6)$ and travels towards the origin with constant speed of $2$ units per second. Particle $2$ moves on the $x$-axis starting at the origin and travels in a positive $x$-direction with constant speed of $1$ unit per second.what moment in the first $3$ seconds is the distance minimised?

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 possible values of $t\in [0,3]$ could give the minimum distance? Check these.

Hint 2
It may be easier to minimise the distance squared, which is equivalent to minimising the distance.

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 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. So we need to minimise the distance $d$ for $t\in [0,3]$. We instead minimise $d^{2}$, which is equivalent, using the closed interval method: ${\begin{aligned}d(0)^{2}=36&&d(3)^{2}=45-72+36=9\end{aligned}}$ The critical numbers are then given by ${\frac {d}{dt}}(d^{2})=10t-24=0$ So the only critical number is $t=12/5$, for which holds that $12/5\in [0,3]$. Then ${\begin{aligned}d(12/5)^{2}&={\frac {144}{5}}-24\cdot {\frac {12}{5}}+36\\&={\frac {144-2\times 144+180}{5}}\\&={\frac {36}{5}}<9\end{aligned}}$ By the closed interval method, the minimum distance is reached when $t={\frac {12}{5}}$ seconds.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. So we need to minimise the distance $d$ for $t\in [0,3]$. We do this using the closed interval method: ${\begin{aligned}d(0)={\sqrt {36}}=6&&d(3)={\sqrt {45-72+36}}={\sqrt {9}}=3\end{aligned}}$ The critical numbers are then given by ${\frac {d}{dt}}(d)={\frac {1}{2}}(5t^{2}-24t+36)^{-1/2}(10t-24)={\frac {10t-24}{2{\sqrt {5t^{2}-24t+36}}}}=0$ So the only critical number is when $10t-24=0$. This is at $t=12/5$. Then ${\begin{aligned}d(12/5)&={\sqrt {{\frac {144}{5}}-24\cdot {\frac {12}{5}}+36}}\\&={\sqrt {\frac {144-2\times 144+180}{5}}}\\&={\sqrt {\frac {36}{5}}}<{\sqrt {\frac {36}{4}}}=3\end{aligned}}$ By the closed interval method the minimum distance is reached when $t={\frac {12}{5}}$ seconds.

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