An upright cylindrical tank with radius 8 m is being filled with water at a rate of 2 m3/min. How fast is the height of the water increasing? Part 1 of 3. If h is the water's height, the volume of the water is V = πr2h. We must find dV/dt. Differentiating both sides of the equation gives Dv/Dt= πr2 Dh/Dt Subsituting for r , this becomes Dv/Dt ____________ π Dh/Dt What goes in the blank ? Thanks !

Accepted Solution

Answer:16 goes in the blankStep-by-step explanation:V(c) = 2*π*r*hDifferentiating boh sidesDV(c)/Dt   =  2πr Dh/Dt    now radius is 8 mDV(c)/Dt   =  8π Dh/DtThat expression gives the relation of changes in V and h DV(c)/Dt  is the speed of growing of the volumeDh/Dt  is the speed of increase in heightso if the cylinder is filling at a rate of  2 m³/min   the height will increase at a rate of 16π m/min