Marcelo Mcclain

2023-03-08

Water is leaking out of an inverted conical tank at a rate of 10,000 cm3/min at the same time water is being pumped into the tank at a constant rate If the tank has a height of 6m and the diameter at the top is 4 m and if the water level is rising at a rate of 20 cm/min when the height of the water is 2m, how do you find the rate at which the water is being pumped into the tank?

repicotit48p

Beginner2023-03-09Added 3 answers

Let V be the volume of water in the tank, in $c{m}^{3}$; let h be the depth/height of the water, in cm; and let r be the radius of the surface of the water (on top), in cm. Since the tank is an inverted cone, so is the mass of water. Since the tank has a height of 6 m and a radius at the top of 2 m, similar triangles implies that $\frac{h}{r}=\frac{6}{2}=3$ so that $h=3r$

The volume of the inverted cone of water is then $V=\frac{1}{3}\pi {r}^{2}h=\pi {r}^{3}$

Now differentiate both sides with respect to time $t$ (in minutes) to get $\frac{dV}{dt}=3\pi {r}^{2}\cdot \frac{dr}{dt}$ (the Chain Rule is used in this step).

If $V}_{i$ is the volume of water that has been pumped in, then $\frac{dV}{dt}=\frac{d{V}_{i}}{dt}-10000=3\pi \cdot {\left(\frac{200}{3}\right)}^{2}\cdot 20$ (when the height/depth of water is 2 meters, the radius of the water is $\frac{200}{3}$ cm).

Hence $\frac{d{V}_{i}}{dt}=\frac{800000\pi}{3}+10000\approx 847758\frac{{\text{cm}}^{3}}{min}$

The volume of the inverted cone of water is then $V=\frac{1}{3}\pi {r}^{2}h=\pi {r}^{3}$

Now differentiate both sides with respect to time $t$ (in minutes) to get $\frac{dV}{dt}=3\pi {r}^{2}\cdot \frac{dr}{dt}$ (the Chain Rule is used in this step).

If $V}_{i$ is the volume of water that has been pumped in, then $\frac{dV}{dt}=\frac{d{V}_{i}}{dt}-10000=3\pi \cdot {\left(\frac{200}{3}\right)}^{2}\cdot 20$ (when the height/depth of water is 2 meters, the radius of the water is $\frac{200}{3}$ cm).

Hence $\frac{d{V}_{i}}{dt}=\frac{800000\pi}{3}+10000\approx 847758\frac{{\text{cm}}^{3}}{min}$