Question

Draining Conical Reservoir Water is flowing at the rate of 50 \(\mathrm{m}^{3} / \mathrm{min}\) from a concrete conical reservoir (vertex down) of base radius 45 \(\mathrm{m}\) and height 6 \(\mathrm{m} .\) (a) How fast is the water level falling when the water is 5 \(\mathrm{m}\) deep? (b) How fast is the radius of the water's surface changing at that moment? Give your answer in \(\mathrm{cm} / \mathrm{min.}\)

Short answer

When the water is 5m deep, the water level is falling at a rate of 0.241m/min and the radius of the water's surface is decreasing at a rate of 18.1 cm/min.

Step-by-step solution

Step 1:Write down known values and sketch the situation

First, define the given values and visualize the problem in a sketch. Here, the volume of water leaving the reservoir is given as -50 m^3/min (negative because it's draining), the base radius of the reservoir is 45m, its height is 6m. Let's denote the radius and height of the water in the reservoir at any time t by r(t) and h(t) respectively.

Write down the volume of a cone

The volume \(V\) of a cone with base radius \(r\) and height \(h\) is given by the formula \(V = 1/3 \pi r^2 h\). The negative rate of change of volume is given, which is \(-dV/dt = 50\) m³/min.

Express r in terms of h

Since we know the reservoir is a cone, the radius and height are proportional. Therefore, r/h would be equivalent to the ratio of base radius over the height of the reservoir, \(r/h = 45/6 = 15/2\). Therefore, r = (15/2)h.

Find dh/dt

Substitute \(r = 15/2 h\) into the volume formula \(V = 1/3 \pi (15/2h)^2 h\) which gives \(V = 175/8 \pi h^3\). Differentiate both sides with respect to \(t\) which gives -50 = 525/8 \pi h^2 dh/dt. Solve for dh/dt when \(h =5\) (i.e., when the water level is 5m deep), which gives \(dh/dt = -0.241\) m/min.

Find dr/dt

Through the relation \(r= 15/2 h\), we find dr/dt = (15/2) dh/dt = -0.181} m/min. To convert it to cm/min, simply multiply it by 100, which gives dr/dt = -18.1 cm/min

Interpreting the result

The negative sign of dh/dt and dr/dt indicates that the water level and the radius of water are both decreasing. Hence, when the water is 5m deep, the water level is falling at a rate of 0.241m/min and the radius of the water surface is decreasing at a rate of 18.1 cm/min.