Science:Math Exam Resources/Courses/MATH103/April 2010/Question 06 (a)

MATH103 April 2010

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Question 06 (a)
The height of fluid, $h(t)$ in a cylindrical container is controlled by a pump so that it satisfies the differential equation ${\frac {dh}{dt}}=-kh^{1/3},\quad h(0)=h_{0}$ where $k>0$, $h_{0}>0$ are constants, and $h_{0}$ is the initial height of the fluid. (a) Solve this differential equation to determine the height $h(t)$ at any later time $t$.

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
Try separation of variables.

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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. Using separation of variables we can write ${\frac {dh}{dt}}=-kh^{1/3}$ as $h^{-1/3}\,dh=-k\,dt,\quad {\text{which we integrate to get}}\quad \int h^{-1/3}\,dh=\int -k\,dt.$ Both integrals are solved straight forward to yield ${\frac {3}{2}}h^{2/3}=-kT+C_{1}$ which we solve for h to obtain $h(t)=\left(C_{2}-{\frac {2}{3}}kt\right)^{3/2},$ where C2 is another constant. All that's left to do now is to use the initial condition h(0) = h0 to solve for C2: $h_{0}=h(0)=\left(C_{2}-{\frac {2}{3}}k\cdot 0\right)^{3/2},\quad {\text{which yields}}\quad C_{2}=h_{0}^{2/3}.$ Therefore the height h(t) at any time t is given by $h(t)=\left(h_{0}^{2/3}-{\frac {2}{3}}kt\right)^{3/2}.$ To double-check your answer you can take the derivative of h(t) and check that the given differential equation is indeed satisfied: ${\begin{aligned}{\frac {dh}{dt}}&={\frac {3}{2}}\left(h_{0}^{2/3}-{\frac {2}{3}}kt\right)^{1/2}\cdot \left(-{\frac {2}{3}}k\right)\\&=-k\underbrace {\left(h_{0}^{2/3}-{\frac {2}{3}}kt\right)^{1/2}} _{=h^{1/3}}\end{aligned}}$

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