A water tank has the shape of an inverted circular cone with base radius 2m and height 4m. If water is being pumped into the tank at a rate of 2m3 /min, find the rate at which the water level is rising when the water is 3m deep
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Answer:
Step-by-step explanation:
Delhi (1962) · Author has 3.5K answers and 515.1K answer views
Determine Water Line
Draw an isosceles triangle on the page. This is the inverted circular cone’s cross-section. The cone’s height H=4 meters and its radius R=2 meters is one-half the isosceles triangle’s base.
Draw a horizontal line through the isosceles triangle on the page. This water line shows two similar triangles, the whole isosceles triangle and a smaller triangle above the water line with height h meters and radius r meters.
HR=hr⇒r=RHh similar triangles
Determine Water Volume
Imagine this water line is perpendicular to the water depth w=H−
Answer:
The volume of the tank is a circular cone, therefore it is given by:
V = πr²h/3
V = volume, r = base radius, h = height or water level
Differentiate both sides with respect to time t. The base radius r doesn't change over time so treat it as a constant:
dV/dt = πr²(dh/dt)/3
Given values:
dV/dt = 2m³/min
r = 2m
Plug in and solve for dh/dt:
2 = π(2)²(dh/dt)/3
dh/dt = 0.48m/min
The water level is increasing at a rate of 0.48m/min