The height of a closed cylinder is 5 times its base radius.

a) find the changes of its base radius if the surface area changes from 48πcm2 to 50πcm2.

b) if the radius increases at a constant rate of 0.5cms-1, find the rate of change of volume at the constant when the radius is 6cm.

1

Jawapan

2016-10-04T14:58:09+08:00
R = x                   surface area of cylinder = 2πrh + 2πr²
h = 5x                                                       = 48π
                                                                 = 50π

2π(x)(5x) + 2π(x)²
10πx² + 2πx²
12πx²

12πx² = 48π                                        change of the radius
       x² = 48π / 12π                               = 2.04 - 2
       x  = √4                                           = 0.04 cm
       x  =  2 cm

12πx² = 50π
      x²  = 50π / 12π
      x   = √(25 / 6)
      x   = 2.04 cm

dV/dt = ?                            dV/dt = (dV/dr) x (dr/dt)
V = πr²h                                      = [2π(6)(10)] [0.5]
dr/dt = 0.5                                   = 60π cms-1
dV/dr = 2πrh