A life insurance company invests $ 5000 in a bank account in...

A life insurance company invests $\$ 5000$ in a bank account in order to fund a death benefit of $\$ 20,000 .$ Growth in the investment over time can be modeled by the differential equation $$\frac{d A}{d t}=A i$$ where $i$ is the interest rate and $A(t)$ is the amount invested at time $t$ (in years). Calculate the interest rate that the investment must earn in order for the company to fund the death benefit in 24 years. Choose one of the following. Source: Society of Actuaries. a. $\frac{-\ln 2}{12}$ b. $\frac{-\ln 2}{24}$ c. $\frac{\ln 2}{24}$ d. $\frac{\ln 2}{12}$ e. $\frac{\ln 2}{6}$

Transcript

00:01 Hi there, so for this problem, we are told that a life insurance company invents 5 ,000 in a bank account in order to fund a death benefit of 20 ,000.

00:20 So we are given the following expression that will be the rate of change of a with respect to time is equal to a times i.

00:29 Now, this is the growth in the investment over a time interval, given by this differential equation, where i is the interest rate.

00:39 A is the amount invested at a time still, okay? so we need to calculate the interest rate that the investment must earn in order for the company to fund the death benefit in 24 years.

00:53 So what we are given for this are the following conditions, that the amount, at zero is the initial investment that will be 5 ,000, and then the amount after 24 years is equal to 20 ,000.

01:12 That's what those are the two values given for this.

01:16 So first of all, we need to solve the differential equation that we are given.

01:20 And for that, we separate the variables.

01:22 So we will have the differential in a divided by a is equal to the rate i times the differential in time.

01:29 Now, we integrate both sides of this.

01:33 And now we will have that.

01:35 For the right side of this, for the left side of this, we obtained an apparel logarithm of a.

01:43 And for the left side of this, we obtain the interest times the time plus a constant of integration c.

01:50 If we apply the exponential function in both sides of this, we obtained that the amount at any given time is equal to c times the exponential of the rate times the time.

01:58 Now, if we evaluate this up zero, we will obtain that this is just c, and then we know that this at zero is 5 ,000.

02:06 So with that, we can write now the amount at any given time is equal to 5 ,000, and then this times the exponential of minus the rate times the time.

02:17 Now, once we have this, we equal this to the other condition at 24...

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