Related Rates Question Water is flowing at the rate of 10 m^3/min from a concrete conical reservoir (vertex down) of base radius 45 m and height 6 m How fast is the water level falling when the water is 5 m deep and how fast is the radius of the water's surface changing at that moment?
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You need to remember the volume of cone formula such that:
`V = (1/3)*pi*r^2*h` ( `pi*r^2*h = ` area of base multiplied by height)
You need to write radius in terms of height, hence, using similar triangles yields
`45/6 = r/h =gt r = 45h/ 6 =gt r = 15h/2`
Substituting `15h/2` for r in equation of volume yields:
`V = (1/3)*pi*(15h/2)^2*h`
`V = 75pi*h^3/4`
You need to find how fast the water level falls at height h=5 m, hence you need to differentiate the volume with respect to t such that:
`(dV)/(dt) = 225pi*h^2/4*(dh)/(dt)`
You need to find `(dh)/(dt) ` such that:
`(dh)/(dt) = (4*(dV)/(dt))/(225*pi*h^2)`
You need to substitute -10 for `(dV)/(dt)` and 5 for h such that:
`(dh)/(dt) = (4*(-10))/(225*pi*5^2)`
`(dh)/(dt) = (4*(-2))/(225*pi*5) = -8/(225*pi)`
`(dh)/(dt)~~ (-8)/(3532.5)`
`(dh)/(dt) ~~ -0.0022 m/min`
You need to find how fast the radius of water's surface is changing at the moment, hence you need to differentiate `r = 15h/2` with respect to t such that:
`(dr)/(dt) = (15/2)(dh)/(dt)`
You need to find `(dr)/(dt)` at `h=5 m` such that:
`(dr)/(dt) = (15/2)(dh)/(dt) |_h=5m`
You need to substitute `-8/(225*pi)` for `((dh)/(dt)) in (dh)/(dt) = ((dr)/(dt))/(15/2) |_h=5m ` such that:
`(dr)/(dt) = (-8/(225*pi))*(15/2) |_h=5m`
`(dr)/(dt) = -4/(15*pi)`
`(dr)/(dt) ~~ -0.08492 m/sec`
Hence, evaluating how fast water level falls at height h=5 m yields `-0.0022` m/min and evaluating how fast the radius of water's surface is changing at h=5m yields `-0.08492 ` m/sec.
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