Rate of Growth?

A sum of $1000 at 5% interest compounded continuously will grow to V(t) = 1000e^0.05t dollars in t years.

Find the rate of growth after 10 years.

Rate of Growth?

Firstly, your equation does not correspond to what you said:

%26quot;5% compound interest%26quot; over t years

=%26gt; te principal becomes P(t) = 1000 (1.05)^t

and the growth is I(t) = 1000 [(1.05)^t - 1]

=%26gt; P(t) = 1000 e^(t ln(1.05))

That%26#039;s the formula you really wanted to set up.

Secondly, the question is ambiguous:

%26quot;Find the rate of growth after 10 years.%26quot;

- to mathematicians that means dV(t)/dt | t=10

- but in general you might colloquially mean

%26quot;the total amount of accumulated compound interest over 10 years%26quot;: i.e. I(t) | t =10

I(t) | t =10 is easy to evaluate using above:

1000 [(1.05)^10 - 1] = 62.9% of $1000 = $629

If you really mean %26quot;rate of growth%26quot; then:

dV(t)/dt = ln(1.05)*1000 e^(t ln(1.05))

Thus:

dV(t)/dt | t=10

= ln(1.05)*1000 e^(10 ln(1.05))

= 0.08 i.e. 8% (relative to the original principal P(0)=$1000)

Rate of Growth?

First find derivative f function and then substitute 10

dV/dt = 50 * e^0.05t

at t = 10,

dV/dt = 82.436 dollars per annum

Rate of Growth?

64.8%