Question

PLS TRY TO ANSWER ASAP.

Answer 1

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:

       $V=\dfrac{2.08}{3}\pi x^3\\\\\\8\pi=\dfrac{2.08}{3}\pi x^3\\\\\\ 11.53846=x^3\\ \\ \\ x=2.25969\\ \\ \\ x^2=5.1062$m²

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.