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Use related rates to find the water level in a tank shaped like a right circular cone

by RoRi

Given a water tank in the shape of a right circular cone with radius of the base 4 feet and altitude of 10 feet. Water is added to the tank at a constant rate of 5 cubic feet per minute. Find the rate at which the water level is rising when the depth of the water is 5 feet when:

  1. the vertex of the cone is pointed up;
  2. the vertex of the cone is pointed down.
  1. First, we give the general setup for the problem. The following diagram may be helpful,

    Rendered by QuickLaTeX.com

    Next, we compute the radius of the waterline in terms of the height (or altitude) h,

        \[ r = -\frac{4}{10}h = -\frac{2}{5} h \quad \implies \quad \frac{dr}{dt} = -\frac{2}{5} \frac{dh}{dt}. \]

    Then, using the formula for the volume of a right circular cone we have and letting V be the volume of the water we have V is equal to the volume of the entire tank less the volume of the empty (smaller) cone above the water:

        \begin{align*} V &= \frac{1}{3} \pi (4^2) (10^2) - \frac{1}{3} \pi r^2 (10-h) \\ &= \frac{1600 \pi}{3} - \frac{10}{3} \pi r^2 + \frac{1}{3} \pi r^2 h \\ &= \frac{1600 \pi}{3} - r^2 \left( \frac{10 \pi}{3} - \frac{\pi}{3} h \right). \end{align*}

    This implies

        \[ \frac{dV}{dt} = -2r \left( \frac{10 \pi}{3} - \frac{\pi}{3} h \right) \frac{dr}{dt} - r^2 \left( - \frac{\pi}{3} \frac{dh}{dt} \right). \]

    (Here, we needed to use the product rule for derivatives since both r and h are functions of t, so when we differentiate their product we have be careful to use the product rule and get both terms shown above.)

    The problem gives us that the rate of change in the volume of water is 5 cubic feet per minute, so,

        \[ \frac{dV}{dt} = 5. \]

    When h = 5, we have r = 2. Substituting these values and \frac{dr}{dt} = -\frac{2}{5} \frac{dh}{dt} obtained above we have,

        \begin{align*} 5 &= -2 (2) \left( \frac{10 \pi}{3} - \frac{\pi}{3} (5) \right) \left( -\frac{2}{5} \frac{dh}{dt} \right) - (2^2) \left( -\frac{\pi}{3} \frac{dh}{dt} \right) \\[9pt] &= \frac{-20 \pi}{3} \frac{-2}{5} \frac{dh}{dt} + \frac{4 \pi}{3} \frac{dh}{dt} \\[9pt] &= \frac{60 \pi}{15} \frac{dh}{dt} \\[9pt] \implies \frac{dh}{dt} &= \frac{5}{4 \pi}. \end{align*}

  2. Part (b) will be almost the same as part (a) with a few minor changes. Here is the diagram we will work from,

    Rendered by QuickLaTeX.com

    Now, the radius of the waterline in terms of the height of the water h is

        \[ r = \frac{4}{10} h = \frac{2}{5} h \quad \implies \quad \frac{dr}{dt} = \frac{2}{5} \frac{dh}{dt}. \]

    Then, again using the volume of a right circular cone and letting V denote the volume of the water (things are a bit simpler this time since the water is a cone, and we don’t have to subtract anything) we have

        \[ V = \frac{1}{3} \pi r^2 h \quad \implies \quad \frac{dV}{dt} = \frac{2}{3} \pi r h \frac{dr}{dt} + \frac{1}{3} \pi r^2 \frac{dh}{dt}. \]

    (As in part (a), we had to be careful to use the product rule since both h and r are functions of t). The problem then gives us that the rate of change in the volume of water is 5 cubic feet per minute, so again,

        \[ \frac{dV}{dt} = 5. \]

    When h = 5 we still have r = 2. Substituting these values along with \frac{dr}{dt} = \frac{2}{5} \frac{dh}{dt} we have

        \begin{align*} 5 &= \frac{2}{3} \pi (2) (5) \left( \frac{2}{5} \frac{dh}{dt} \right) + \frac{1}{3} \pi (2^2) \frac{dh}{dt} \\[9pt] &= \frac{8 \pi}{3} \frac{dh}{dt} + \frac{4 \pi}{3} \frac{dh}{dt} \\[9pt] &= \frac{12 \pi}{3} \frac{dh}{dt} \\[9pt] \implies \frac{dh}{dt} &= \frac{5}{4 \pi}. \end{align*}

    The same answer as in part (a).

5 comments

  2. neta says:

    say r,h.v denote the radius, height and volume of the small cone then we can express r in terms of h thus we can express v in terms of h thus we can calculate dv/dh. since dv/dt is given we can find dh/dt from: (dv/dt)=(dv/dh) (dh/dt), for part b and for part a replace h by (10 – h),
    thist way was easier for me.

  5. Jonathan Andres Sanabria Osorio says:

    Why do you elevate to the square the height of the biggest cone? Isn’t the volume of a cone equal to one third of pi for the radius elevated to the square and the height? Therefore the biggest volume shouldn’t have 100 elevated to the square.

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