Science:Math Exam Resources/Courses/MATH102/December 2016/Question A 06

MATH102 December 2016

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Question A 06
The graphs below represent the position, velocity and acceleration of a child swinging on a swing. Identify the correct relationships between these functions. (i) $B''(t)=C'(t)=A(t)$ (ii) $B''(t)=A'(t)=C(t)$ (iii) $A''(t)=C'(t)=B(t)$ (iv) $A''(t)=B'(t)=C(t)$ (v) $C''(t)=A'(t)=B(t)$ (vi) $C''(t)=B'(t)=A(t)$

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Start with the fact that where a differentiable function $x(t)$ has a (local) max/min (zero slope) its derivative must become zero (intersects x-axis). The same reasoning will help you determine which one is the second derivative as $x''(t)$ is the derivative of the derivative.

Checking a solution serves two purposes: helping you if, after having used the hint, 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. At the points where $x(t)$ has a min/max (slope of tangent line =0), $v(t)=x'(t)$ must be equal to zero, i.e. intersects x-axis. This fact eliminates the option of $x(t)=B(t)$, because if so, we then see that at its maximum point neither $A(t)$ nor $C(t)$ vanishes. Now we have two choices, for each of which we check whether the graphs match: If $x(t)=C(t)$, we see that at $C$'s max and min, $B(t)$ intersects x-axis, this means that $v(t)=x'(t)=B(t)$ so we must have $(x')'(t)=B'(t)=A(t)$ which implies that where $B(t)$ has a max or min $A(t)$ must become zero, however, we've already seen that at $B$'s maximum $A(t)$ is NOT zero. $\Rightarrow {\text{NOT a choice}}$ . If $x(t)=A(t)$, we see that at $A$'s max and min, $C(t)$ intersects x-axis, this means that $v(t)=x'(t)=C(t)$ so we must have $(x')'(t)=C'(t)=B(t)$ which implies that where $C(t)$ has a max or min $B(t)$ must become zero, which we see that it is in fact true. Therefore, the correct choice is $x(t)=A(t)$, $x'(t)=A'(t)=C(t)$, and $x''(t)=\color {blue}A''(t)=C'(t)=B(t)$. Answer: $\color {blue}(iii)$