Difficult related rates word problem

[Math_Junkie]

A circular cylindrical oil drum of length 1m is lying on its side. It contains diesel fuel but has a small leak at the base so that fuel is leaking at a rate proportional to the depth of fuel remaining in the drum (i.e. dV/dt = - kah, where V is volume of fuel in m[sup:llcx27b0]3[/sup:llcx27b0], h is the depth of fluid (in meters) in the drum, a is the drum radius (in
meters), and k is a positive constant with units of m/s). Using the notation in the diagram:

a) Obtain expressions for the depth h and volume V of the fuel remaining in the drum in terms of the angle ?
and the radius a. These should be valid for 0 ? ? ? ?.
b) Determine the rate at which h is changing (i.e. dh/dt) as a function of a, k and ?.
c) Rearrange to give dh/dt as a function of a, k and h.

Related rates problem? I've tried to make sense of what this question is asking me and I have no clue how to put the pieces together to answer part a).

 

[soroban]

Hello, Math_Junkie!

I'll get you started . . .

A circular cylindrical oil drum of length 1m is lying on its side.
It contains diesel fuel but has a small leak at the base so that fuel is leaking
at a rate proportional to the depth of fuel remaining in the drum

\(\displaystyle \text{That is: }\:\frac{dV}{dt} \,=\, - kah\)
\(\displaystyle \text{where }V\text{ is volume of fuel in }m^3,\,h\text{ is the depth of fluid }(m),\)
\(\displaystyle a\text{ is the drum radius }(m)\text{, and }k\text{ is a positive constant }(m/s).\)

              * * *
          *     C     *
       Ao - - - o - - - oB
       *  *     |     *  *
            *   |   * a
      *       * |@*       *
      *         o         *
      *        O|         *
                |
       *        |a       *
        *       |       *
          *     |     *
              * o *
                D

\(\displaystyle \text{a) Obtain expressions for the depth }h\text{ and volume }V\text{ of the fuel in the drum}\)
. . \(\displaystyle \text{ in terms of angle }\theta \text{ and radius }a.\;\text{ These should be valid for }\,0 \leq \theta \leq \pi.\)

\(\displaystyle \text{In right triangle }OCB\!:\:\cos\theta \,=\,\frac{CO}{a} \quad\Rightarrow\quad CO \,=\,a\cos\theta\)

\(\displaystyle \text{The height of the fuel is: }\:h \;=\;CD \;=\;CO + OD \;=\;a\cos\theta + a\)

\(\displaystyle \text{Therefore: }\:\boxed{h \:=\:a(1 + \cos\theta)}\,\text{ meters}\)


\(\displaystyle \angle AOC = \theta \quad\Rightarrow\quad \angle AOD \,=\,\pi - \theta\)

\(\displaystyle \text{Area of sector }AOD \:=\:\tfrac{1}{2}a^2(\pi - \theta) \:=\:\text{Area of sector }BOD\)

\(\displaystyle \text{Area of }\Delta AOB\:=\:\tfrac{1}{2}a^2\sin2\theta\)

\(\displaystyle \text{Hence, Area of fuel} \:=\:a^2(\pi - \theta) + \tfrac{1}{2}a^2\sin2\theta\)


\(\displaystyle \text{Since the length of the tank is 1 meter,}\)

. . \(\displaystyle \boxed{V \;=\;a^2\bigg[(\pi-\theta) + \tfrac{1}{2}\sin2\theta\bigg]}\,\text{ cubic meters}\)