Hey can you guys walk me through this problem please::
Water is flowing at the rate of 6 meters cubed/min from a resovoir shaped like a hemispherical sphere with a radius of 13 meters. Answer the following questions using the volume formula V=(pi/3)ysquared(3R-y) when the water is y units deep....at what rate is the water level changing when the water is 8 m deep?
2006-12-11 15:49:29 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
your a genius wow! can you explain it a bit though, sorry im a hs student with a terrible teacher
2006-12-11 15:59:55 · update #1
how did you get that volume formual is essentialy what im asking...
2006-12-11 16:08:51 · update #2
V = ⅓πy²(3R - y)
= πy²R - ⅓πy³
Now dV/dt = dV/dy * dy/dt (using the chain rule)
dV/dt = 6 m³/min
So 6 = (2πyR - πy²) * dy/dt
= (2π*8*13 - π*8²)* dy/dt when y = 8 m
= (208 - 64)π * dy/dt
= 144π
Thus dy/dt = 6/(144π)
= (1/24π) m/min
≈ 0.0133 m/min
= 1.33 cm/min
2006-12-11 16:23:29 · answer #1 · answered by Wal C 6 · 0⤊ 1⤋
[a] floor section = circumference * top = 2? r * h top = 10cm + 0.1cm * t h = 10 + 0.1t A = 10? * (10 + 0.1t) = 100? + ? t dA/dt = price of replace of section with relation to time dA/dt = (100? + ? t)' = ? cm^2/sec [b] quantity = portion of base * top = ? r^2 * (10 + 0.1t) V = 25? (10 + 0.1t) V = 250? + 2.5? t dV/dt = 2.5? cm^3/sec wish this helps!
2016-11-25 22:06:50 · answer #2 · answered by ? 4 · 0⤊ 0⤋
V = pi(y^2)R - (pi/3)y^3
dV/dt = pi[2Ry(dy/dt) - y^2(dy/dt)] = 6
6/[pi(2Ry - y^2)] = dy/dt = falling at 0.013 m/min
2006-12-11 15:56:39 · answer #3 · answered by jacinablackbox 4 · 0⤊ 0⤋
dV/dt = (dV/dy)(dy/dt)
V = (π/3)(3R - y)y^2
V = (π/3)(3Ry^2 - y^3)
dV/dy = (π/3)(6Ry - 3y^2)
dy/dt = (dV/dt)/(dV/dy)
dy/dt = 6/[(π/3)(6*13*8 - 3*8^2)]
dy/dt = 18/[(π)(624 - 192)]
dy/dt = 1/(24π)
dy/dt = 0.0133 m/min
2006-12-11 16:19:47 · answer #4 · answered by Helmut 7 · 0⤊ 0⤋