a pond that lies in a crater is being drained att he rate of 1000m3/min.If the crater has the shape of an inverted cone of radius 200(at he original water level) and original depth is 50 m (at the center) find that rate at which the water level begins to fall.How fast is it falling when the water at he center is only 20 m deep?
did that
Can you please gimme the whole procedure ?
To solve this problem, we can use the concept of related rates. The problem gives us information about the rate at which the pond is being drained, and asks us to find the rate at which the water level is falling and how fast it is falling when the depth is 20 m.
Let's denote the radius of the cone at a given depth h as r(h), where r(0) = 200 (since the original water level is at radius 200) and h(0) = 50 (since the original depth is 50 m). We need to find dh/dt, the rate at which the depth is changing, when h = 20 m.
First, let's establish a relationship between the height h, the radius r, and the volume V of the cone. The formula for the volume of a cone is V = (1/3)πr²h.
Given that the pond is being drained at a rate of 1000 m³/min, the rate of change of volume with respect to time is dV/dt = -1000 m³/min (negative because the volume is decreasing).
To find dh/dt, we differentiate the volume equation with respect to time:
dV/dt = (1/3)π(2rhr)dr/dt + (1/3)πr²dh/dt
Simplifying, we can solve for dh/dt:
-1000 = (2/3)πr²dr/dt + (1/3)πr²dh/dt
Since we are interested in finding dh/dt when h = 20 m, we substitute r and h into the equation:
-1000 = (2/3)π(200²)(dr/dt) + (1/3)π(200²)(dh/dt)
Now, we need to find dr/dt, the rate at which the radius is changing when h = 20 m. To do this, we differentiate the equation for the radius of the cone with respect to time:
r² = (200/50)h
Differentiating both sides with respect to time:
2r(dr/dt) = (200/50)(dh/dt)
Simplifying, we solve for dr/dt:
dr/dt = (200/50)(dh/dt) / (2r)
Now we can substitute this expression for dr/dt back into the previous equation:
-1000 = (2/3)π(200²)[(200/50)(dh/dt) / (2r)] + (1/3)π(200²)(dh/dt)
Simplifying further, we can solve for dh/dt:
-1000 = (2/3)π(200/50)(dh/dt) + (1/3)π(200/50)(dh/dt)
-1000 = (4/3)π(200/50)(dh/dt)
dh/dt = -1000 / [(4/3)π(200/50)]
Now we can compute the value of dh/dt:
dh/dt = -1000 / [(4/3) × 3.14 × (200/50)]
Finally, we can substitute the value of dh/dt into the expression when h = 20 m to find the rate at which the water level is falling:
dh/dt = -1000 / [(4/3) × 3.14 × (200/50)]
dh/dt = -1000 / [1.57 × 4 × 4]
dh/dt ≈ -39.97 m/min
Therefore, the rate at which the water level is falling when the water at the center is only 20 m deep is approximately 39.97 m/min.