Z(t) represents the water height at time t; Z = 1m, then what is dz(t)/dt for 1m. Given: Radius (flask) : r(t) = 3.2/√(z+1) Bottom of flask, where z = 0, then r = 3.2 Time to escape shackles: 7 minutes Rate: 0.6π m3 Magician’s height: 1.82 meters Volume of magician’s body: 0.53 m3 Height of cube: 0.5 m 0.6π m3 / minute x 7 minutes = 13.19 m3 Magician wants to be able to get out once the water reaches his head = Magician’s height + height of the block Help from the teacher: volume of the cone V = (1/3)(pi)(r^2)h = (1/3)(pi)(z)[3.2/sqrt(z+1)]^2 = 3.41(pi)(z)/(z+1) volume of the cube Vc = 0.125 m^3 volume of the magician: Vm = 0.52 m^3 volume of the water: Vw = V - (Vc + Vm) the volume of the water in the flask can be written as a function of the height of the water above ground. Z(t) represents the height of water above the ground level at time t. A formula for the rate of change of the height is a function of time or dz(t)/dt. Z dz(t)/dt 1m
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The problem provides you a function of volume of water such that:
`V_w= V - (V_c + V_m)`
`V_w = 3.41 (pi*z)/(z+1) - 0.125 - 0.52`
`V_w = 3.41 (pi*z)/(z+1) - 0.645`
You need to differentiate the function of V_w with respect to t such that:
`(dV)/(dt) = (d(3.41 (pi*z)/(z+1) - 0.645))/(dz)*(dz)/(dt)`
`(dV)/(dt) = (3.41 (pi)(z+1) - 3.41 (pi*z))/((z+1)^2)*(dz)/(dt)`
`(dV)/(dt) = (3.41 pi*z + 3.41- 3.41 pi*z)/((z+1)^2)*(dz)/(dt)`
`(dV)/(dt) = (3.41)/((z+1)^2)*(dz)/(dt)`
The problem provides that `(dV)/(dt) = 0.6 pi ` hence, substituting 0.6 pi for (dV)/(dt) such that:
`0.6 pi = (3.41)/((z+1)^2)*(dz)/(dt)`
`(dz)/(dt) = (0.6 pi ((z+1)^2))/3.41`
You need to substitute 1 for z in equation `(dz)/(dt)` such that:
`(dz)/(dt)= (0.6 pi ((1+1)^2))/3.41 =gt (dz)/(dt)~~3.141`
Hence, evaluating `(dz)/(dt)` for z =1m yields `(dz)/(dt)~~3.141.`
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