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 ? !

16 goes in the blank

Step-by-step 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  

i need points

step-by-step explanation:

sorry bubs