PLS TRY TO ANSWER ASAP. A water tank at a filtration plant is built in the shape of an inverted cone with height 5.2 m and diameter 5 m

Question

PLS TRY TO ANSWER ASAP.
A water tank at a filtration plant is built in the shape of an inverted cone with height 5.2 m and diameter 5 m at the top. Water is being pumped into the tank at a rate of 1.2 m³/min. Find the rate at which the water level is rising when there is 8π m³ of water

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Clara 2 weeks 2021-09-15T09:04:10+00:00 1 Answer 0

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    2021-09-15T09:05:35+00:00

    Answer:

    • 0.075 m/min

    Explanation:

    You need to use derivatives which is an advanced concept used in calculus.

    1. Write the equation for the volume of the cone:

          V=\dfrac{1}{3}\pi r^2h

    2. Find the relation between the radius and the height:

    • r = diameter/2 = 5m/2 = 2.5m
    • h = 5.2m
    • h/r =5.2 / 2.5 = 2.08

    3. Filling the tank:

    Call y the height of water and x the horizontal distance from the axis of symmetry of the cone to the wall for the surface of water, when the cone is being filled.

    The ratio x/y is the same r/h

    • x/y=r/h
    • y = x . h / r

    The volume of water inside the cone is:

            V=\dfrac{1}{3}\pi x^2y

            V=\dfrac{1}{3}\pi x^2(2.08)\cdot x\\\\\\V=\dfrac{2.08}{3}\pi x^3

    4. Find the derivative of the volume of water with respect to time:

                \dfrac{dV}{dt}=2.08\pi x^2\dfrac{dx}{dt}

    5. Find x² when the volume of water is 8π m³:

           V=\dfrac{2.08}{3}\pi x^3\\\\\\8\pi=\dfrac{2.08}{3}\pi x^3\\\\\\  11.53846=x^3\\ \\ \\ x=2.25969\\ \\ \\ x^2=5.1062

    6. Solve for dx/dt:

          1.2m^3/min=2.08\pi(5.1062m^2)\dfrac{dx}{dt}

          \dfrac{dx}{dt}=0.03596m/min

    7. Find dh/dt:

    From y/x = h/r = 2.08:

            y=2.08x\\\\\\\dfrac{dy}{dx}=2.08\dfrac{dx}{dt}\\\\\\\dfrac{dy}{dt}=2.08(0.035964m/min)=0.0748m/min\approx0.075m/min

    That is the rate at which the water level is rising when there is 8π m³ of water.

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