Science:Math Exam Resources/Courses/MATH100/December 2018/Question 06

MATH100 December 2018

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Question 06
Particle $A$ travels with a constant speed of 2 units per minute on the $x$ -axis starting at the point (4,0) and moving away from the origin, while particle $B$ travels with a constant speed of 1 unit per minute on the $y$ -axis starting at the point (0,8) and moving towards the origin. Find the rate of change of the distance between the two particles when the distance between the two particles is exactly 10 units.

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?

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Hint
Use Pythagoras' theorem to relate the positions of the two particles.

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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. Let $x$ be the distance from the origin to the particle $A$ and let $y$ be the distance from the origin to the particle $B$ . Also Let $a$ be the distance between $A$ and $B$ . Notice that $x,y$ and $a$ are functions of time and we shall thus write them as $x(t),y(t),a(t)$ respectively. By the given information, we have ${\frac {dx}{dt}}=2$ units/min and ${\frac {dy}{dt}}=-1$ units/min. Using the initial positions, then we see that $x(t)=4+2t$ and $y(t)=8-t$ . We can relate these two functions by Pythagoras' theorem: we have $x(t)^{2}+y(t)^{2}=a(t)^{2}$ , and therefore $a(t)^{2}=x(t)^{2}+y(t)^{2}=(4+2t)^{2}+(8-t)^{2}=16+16t+4t^{2}+64-16t+t^{2}=80+5t^{2}.$ Now, we will first determine the time $t$ at which the distance between the particles is $10$ units. We do this by setting $a(t)=10$ and solving for $t$ . This gives $100=80+5t^{2}$ and therefore $t=2$ (we can ignore the negative root since time cannot be negative). Next, we want to find $a'(t)$ when $a(t)=10$ . By differentiating both sides of the equation $a(t)^{2}=80+5t^{2}$ , we get $2a(t)a'(t)=10t$ . Now we can plug in $a(t)=10$ and $t=2$ , which gives $\color {blue}a'(2)=1{\text{ unit per min}}$ .