Science:Math Exam Resources/Courses/MATH102/December 2016/Question B 03 (c)

MATH102 December 2016

  •  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  •  

   Other MATH102 Exams

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

Question B 03 (c)

The velocity of a protein inside a cell starting at the top and falling under the force of gravity is given by the differential equation ${\frac {dv}{dt}}=g-kv$ where $g\approx 10^{7}\mu m/s^{2}$ is the gravitational acceleration, $kv$ is the acceleration due to drag and $k=10^{13}s^{-1}$. (Note that this model ignores diffusion of the protein which makes it a really bad model but we’re exploring what gravity alone would do to a protein.) (c) If the protein is initially at rest at the top of the cell, how long does it take for the protein to reach a velocity of $(1-e^{-2})v_{ss}$ (roughly 85% of the way to steady state)? You do not have to solve the equation but instead simply state the solution.

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
Write the equation in a form that is ready to be solved.

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. We can rewrite the equation in the form ${\frac {dv}{dt}}=g-kv=-k(v-{\frac {g}{k}})=-k(v-v_{ss})$ where $v_{ss}={\frac {g}{k}}$ (recall part (a)). This means that if we denote $z=v-v_{ss}$ then since $v_{ss}$ is a constant we have ${\frac {dz}{dt}}={\frac {dv}{dt}}$ so ${\frac {dz}{dt}}=-kz,z_{0}=v(0)-v_{ss}=0-10^{-6}=-10^{-6}$, which has a solution $z(t)=-10^{-6}e^{-kt}\ {\text{or}}\ v-10^{-6}=-10^{-6}e^{-kt}$ . Now, the time to go from $v(0)=0$ to $v(t)=(1-e^{-2})v_{ss}=(1-e^{-2})10^{-6}$ is equivalent to solving the following equation for t: $(1-e^{-2})10^{-6}-10^{-6}=-10^{-6}e^{-kt}$ i.e. $e^{-kt}={\frac {(1-e^{-2})10^{-6}-10^{-6}}{-10^{-6}}}=e^{-2}$. Indeed $\color {blue}t={\frac {2}{k}}$.