Answer:
a) 53.26 $/year
b) 81.07 $/year
Step-by-step explanation:
Data provided in the question:
Amount invested = $500
Interest rate = 7%
Future value, S(t) = [tex]500e^{0.07t}[/tex]
Now,
rate of growth of money = S'(t) = [tex]\frac{d(500e^{0.07t})}{dt}[/tex]
or
S'(t) = [tex]0.07\times500e^{0.07t}[/tex]
or
S'(t) = [tex]35e^{0.07t}[/tex]
a) at t = 6
S'(t) = [tex]35e^{0.07(6)}[/tex]
or
S'(t) = [tex]35e^{0.42}[/tex]
or
S'(t) = 53.26 $/year
b) at t = 12
S'(t) = [tex]35e^{0.07(12)}[/tex]
or
S'(t) = [tex]35e^{0.84}[/tex]
or
S'(t) = 81.07 $/year
Answer:
(a) $53.27 per year
(b) $81.07 per year
Step-by-step explanation:
The derivative of the function is ...
S'(t) = 0.07·500e^(0.07t) = 35e^(0.07t)
(a) The derivative evaluated at t=6 is ...
S'(6) = 35·e^0.42 ≈ 53.27
At t=6, the account is growing at the rate of $53.27 per year.
__
(b) The derivative evaluated at t=12 is ...
S'(12) = 35·e^0.84 ≈ 81.07
At t=12, the account is growing at the rate of $81.07 per year.
