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Kavina's question at Yahoo! Answers regarding an initial value problem

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MarkFL

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Feb 24, 2012
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Here is the question:

Calculus question on differential equations?

Water in a barrel (upright cylinder) is leaking out at a rate proportional to the square root of the depth of the water. If the water level was 29 cm 3 minutes ago, and is at 25 cm now, how many more minutes (from now) will it take for the barrel to be empty? Please help :)
Here is a link to the question:

Calculus question on differential equations? - Yahoo! Answers

I have posted a link there to this topic so the OP may find my response.
 
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MarkFL

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Feb 24, 2012
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Hello Kavina,

Let $V(t)$ be the volume in $\text{cm}^3$ of water in the cylindrical barrel at time $t$ in minutes, and $h$ be the depth in $\text{cm}$ of the water in the barrel.

We are given:

$\displaystyle \frac{dV}{dt}=-k\sqrt{h}$ where $0<k$ in the constant of proportionality.

The formula for the volume of a cylinder is:

$\displaystyle V=\pi r^2h$

Differentiating this with respect to time (and observing that $\pi r^2$ will remain constant) we find:

$\displaystyle \frac{dV}{dt}=\pi r^2\frac{dh}{dt}$

Now, equating the two expressions for $\displaystyle \frac{dV}{dt}$ we have:

$\displaystyle \pi r^2\frac{dh}{dt}=-k\sqrt{h}$

$\displaystyle \frac{dh}{dt}=-\frac{k}{\pi r^2}\sqrt{h}$

Now, we may redefine the constant of proportionality and write the IVP:

$\displaystyle \frac{dh}{dt}=-k\sqrt{h}$ where $h(0)=29,\,h(3)=25$.

The ODE associated with the IVP is separable:

$\displaystyle h^{-\frac{1}{2}}\,dh=-k\,dt$

Integrate:

$\displaystyle \int h^{-\frac{1}{2}}\,dh=-k\int\,dt$

$\displaystyle 2h^{\frac{1}{2}}=-kt+C$

Use initial conditions to find parameter $C$

$\displaystyle 2(29)^{\frac{1}{2}}=-k(0)+C$

$\displaystyle C=2\sqrt{29}$

and so we have:

$\displaystyle 2h^{\frac{1}{2}}=-kt+2\sqrt{29}$

Now, use other given point to determine the constant of proportionality $k$:

$\displaystyle 2(25)^{\frac{1}{2}}=-k(3)+2\sqrt{29}$

$\displaystyle k=\frac{2(\sqrt{29}-5)}{3}$

Hence, we may write:

$\displaystyle t=\frac{3(\sqrt{29}-\sqrt{h})}{\sqrt{29}-5}$

Now, to find when the barrel will be empty, we may let $h=0$ and we find:

$\displaystyle t(0)=\frac{3\sqrt{29}}{\sqrt{29}-5}$

Since "now" is given to be $t=3$ we must subtract 3 from this to find the answer to the question:

$\displaystyle \frac{3\sqrt{29}}{\sqrt{29}-5}-3=\frac{3\sqrt{29}-3(\sqrt{29}-5)}{\sqrt{29}-5}=\frac{15}{\sqrt{29}-5}$
 
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