Question

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 !

Answer 1

16 goes in the blank

Explanation:

V(c) = 2*π*r*h

Differentiating boh sides

DV(c)/Dt = 2πr Dh/Dt now radius is 8 m

DV(c)/Dt = 8π Dh/Dt

That expression gives the relation of changes in V and h

DV(c)/Dt is the speed of growing of the volume

Dh/Dt is the speed of increase in height

so if the cylinder is filling at a rate of 2 m³/min the height will increase at a rate of 16π m/min