calculus problem?

　
V = πr²h ..... where r, h are in meters, and V is in cubic meters
M = D * V .... where mass (M) is in kg, constant density (D) is in kg/m^3

M = Dπr²h

dM/dt = Dπ (2rh dr/dt + r² dh/dt)

h = 50 m
r = (4 m)/2 = 2 m
dh/dt = 3m/yr
dr/dt = (2 cm/yr)/2 = 1 cm/yr = 0.01 m/yr
D = 5000 kg/m^3

dM/dt = Dπ (2rh dr/dt + r² dh/dt)
dM/dt = (5000 kg/m^3) π (2 (2 m) (50 m) (0.01 m/yr) + (2 m)² (3m/yr))
dM/dt = (5000 kg/m^3) π (2 m^3/yr + 12 m^3/yr)
dM/dt = (5000 kg/m^3) π (14 m^3/yr)
dM/dt = 70000π kg/yr ≈ 219911 kg/yr

Mass is increasing at a rate of about 219911 kg/yr

m : mass of the tree in kg
V : volume of the tree in m³
h : height of the tree in m
r : radius of the tree in m
t : time taken in years

V = πr²h

m = density × V
m = 5000πr²h

dm/dt = 5000π[r²(dh/dt) + h(dr²/dt)]
dm/dt = 5000π[r²(dh/dt) + 2hr(dr/dt)]
dm/dt = 5000πr[r(dh/dt) + 2h(dr/dt)]

Take π = 3.14
When r = 2, h = 50, dh/dt = 3 and dr/dt = 0.02 :
dm/dt = 5000 × 3.14 × 2 × (2 × 3 + 2 × 50 × 0.02) kg/year
Increasing rate of mass, dm/dt ≈ 251000 kg/year