Given a varying force of interest where $\delta=kt$, where an unknown amount of money will double itself, find an expression for K

Question

The book says the answer is $\frac{ln(2)}{50}$ but I do not see how unless this a typo or I am missing something very fundamental

It is known that an amount of money will double itself in 10 years at a varying force of interest of $\delta=kt$. Find an expression for k.

Well the double of $A(t)=pe^{kt}$ the principals cancel each other out and since $e^{5k}$ is half of $e^{10k}$:

$2e^{5k}=e^{10k} \to ln(2)+5k=10k \to ln(2)=5k \to k=\frac{ln(2)}{5}$

The formula you wrote for $A(t)$ ($pe^{kt}$) is only true when the force of interest is constant. When the force of interest is variable, you need to integrate it. — Minus One-Twelfth, Jun 12 '20 at 21:58

score 2 · Answer 1 · answered Jun 12 '20 at 21:57

2

Hints: Actually the general formula is $$A(t)=A(0)e^{\int_0^t \delta(\tau)\, d\tau}.$$

In this case, $\delta(\tau)=k\tau$, so $\int_0^t \delta(\tau)\, d\tau = \frac{1}{2}kt^2$. Hence in this case, we have $$A(t)=A(0)e^{\frac{1}{2}kt^2}.$$ Now solve for $k$ if $A(10)=2A(0)$.

answered Jun 12 '20 at 21:57

Minus One-Twelfth

7,413

Okay: so I should get something like $e^{10k}=2e^{kt^2\frac{1}{2}}$ – K. Gibson Jun 12 '20 at 22:08
then plug in 10? – K. Gibson Jun 12 '20 at 22:08
Not quite. Remember, if $A(t)=A(0)e^{\frac{1}{2}kt^2}$, then $A(0)$ is just $A(0)$ (no $t$ involved), and $A(10) =A(0)e^{\frac{1}{2}k\times 10^2}$. – Minus One-Twelfth Jun 13 '20 at 08:17

Given a varying force of interest where $\delta=kt$, where an unknown amount of money will double itself, find an expression for K

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