Science:Math Exam Resources/Courses/MATH102/December 2016/Question B 04 (b)
• QA 1 • QA 2 • QA 3 • QA 4 • QA 5 • QA 6 • QA 7 • QA 8 • QB 1 • QB 2 • QB 3(a) • QB 3(b) • QB 3(c) • QB 4(a) • QB 4(b) • QB 4(c) • QB 5(a) • QB 5(b) • QB 6(a) • QB 6(b) • QB 7(a) • QB 7(b) • QB 7(c) • QB 8 •
Question B 04 (b) 

The velocity of a particle is determined by its position so that the position function satisfies the equation where . The particle is at initially. (b) Sketch the phase line for the equation and determine what happens to the position of the particle as ? 
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 

Draw the graph of and find interval of and . Based on this, sketch the phase line. 
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.

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. The below is the phase line for the given equation. As you see, based on the graph of , we can easily get for and , while for and . Therefore, on , is increasing (i.e, move to the right on the graph), while on , is decreasing (i.e., move to the left). This gives the phase line as you see in the picture above. In particular, if we start at the point , since it lies on it moves to the left and approaches to as . Answer: 