The rate of change of money in a continuously compounded account is given by A'(t) = 50e^.05t

find...
a. the function A(t)
b. the total amount accumulated in the account 10 years after the original deposit
c. the number of years that it would take for the account to triple it original deposit

thank you, 10 points for best correct answer

a.............................
A(t) = ∫ A'(t) dt
A(t) = 50 ∫ e^(.05t) dt
A(t) = 1000 e^(.05t) + c

b.................................
A(10) - A(0) = 1000 [√(e) - 1] ≈ $648.72

c.................................
A(t)/A(0) = 3
1000e^(.05t)/1000 = 3
e^(.05t) = 3
.05t = ln(3)
t = 20 ln(3) ≈ 21.97 years