The quantity 1 begins at No= 700 in yr. t=0 and grows at continuous annual rate of r=2% find Q1 in yr. t1=4?

Answer 1

Let the quantity at time (t) be N(t), where t>>years
we are given that
N(0) = 700
Increase in quantity after "t" years = N(t) - N(o)

% increase in quantity after "t" years
= 100* [N(t) - N(o)] / N(t)

Annual continuous rate of growth in quantity =
100* [N(t) - N(o)] / N(t) } * (1/ t) = 2 >>>> (given)
so rewriting
50 [N(t) - N(o)] = t* N(t)
N(t) {50 - t} = 50 N(o)

N(t) = 50 N(o) / {50 - t}
we have N(o) = 700, at t = 4 years, quantity N(4) will be

Q1 = N(4) = 50 *700 / {50 - 4}
= 760.8 >>> = 760 (for whole number)

Answer 2

If compounding is done continuously:

Q(1) = N(0)e^rt
Q(1) = 700e^(0.02)4
Q(1) = 700(1.08329)
Q(1) = 758.30

If compounding is done at same rate each year, but is not compounded continuously:

Q(1) = N(0)(1 + .02)^4
Q(1) = 700(1 + .02)^4
Q(1) = 700(1.08243)
Q(1) = 757.70