Science:Math Exam Resources/Courses/MATH102/December 2012/Question B 01

MATH102 December 2012

  •  QA 1  •  QA 2  •  QA 3  •  QB 1  •  QB 2  •  QB 3  •  QB 4  •  QB 5  •  QB 6  •  QC 1  •  QC 2  •  QC 3(a)  •  QC 3(b)  •  QC 3(c)  •  QC 4(a)  •  QC 4(b)  •  QC 4(c)  •  QC 51  •  QC 52  •  

   Other MATH102 Exams

  •  December 2014  •  December 2011  •  December 2012  •  December 2013  •  December 2016  •  December 2015  •  

Question B 01

For the following questions, refer to the slope fields in the figure below. Match the plots below with the differential equations by filling in the box with A, B, C, D, or NOT (for "None Of These"). (a) The equation ${\frac {dx}{dt}}=1-t^{2}$ matches the slope field ______. (b) The equation ${\frac {dx}{dt}}=t(t-1)(t+1)$ matches the slope field ______. (c) The equation ${\frac {dx}{dt}}=x^{2}-1$ matches the slope field ______. (d) The equation ${\frac {dx}{dt}}=1-x^{2}$ matches the slope field ______.

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
To read slope field graphs, one chooses a t and x value and goes to that point of the graph. The slope of the line that appears there is the value of dx/dt at those particular x and t values (for example, flat horizontal lines have slope 0 and so dx/dt=0 there).

Hint 2

Secondly, check, if the slope field is independent of $x$ or $t$. Therefore, restrict the slope field to vertical or horizontal lines (maybe using a ruler), and verify if the restrictions look equal in any of these two directions. The ODE is independent of $t$ (only depending on $x$), if the vertical restrictions look equal. The ODE is independent of $x$ (only depending on $t$), if the horizontal restrictions look equal.

Hint 3
Thirdly, check at which $x$ (for $t$-independent slope fields) the slopes become horizontal. At these $x$, the appropriate ODE has an equilibrium, is equal to zero. Or, check at which $t$ (for $x$-independent slope fields) the slopes become horizontal. At these $t$, the appropriate ODE is equal to zero.

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. Field A: The ODE must be independent of $t$, because restricted to every vertical line (restricted to any $t$), the slope field looks equal. We have equilibriums at $x=-1$ and $x=1$, because the slopes get more and more horizontal, the closer they are at $x=\pm 1$. For $x\in (-1,1)$ the slope is positive. This fits exactly to the ODE ${\frac {dx}{dt}}=1-x^{2}$ (d). Field B: The ODE must be independent of $t$, because restricted to every vertical line (restricted to any $t$), the slope field looks equal. We have equilibriums at $x=-1$ and $x=1$, because the slopes get more and more horizontal, the closer they are at $x=\pm 1$. For $x\in (-1,1)$ the slope is negative. This fits exactly to the ODE ${\frac {dx}{dt}}=x^{2}-1$ (c). Field C: The ODE must be independent of $x$, because restricted to every horizontal line (restricted to any $x$), the slope field looks equal. We have equilibriums at $t=-1$ and $t=1$, because the slopes get more and more horizontal, the closer they are at $t=\pm 1$. For $t\in (-1,1)$ the slope is positive, for $t<-1$ and $t>1$ the slope is negative. This fits exactly to the ODE ${\frac {dx}{dt}}=1-t^{2}$ (a). ODE (b) and Field D: This ODE causes a slope field where the slopes are horizontal for every $x$ and $t=0$. This is not the case for the Field D. Hence, ODE (b) and Field D cannot fit.

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Antiderivative sketching, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag