# Distribution function problem

## Homework Statement

Consider a population of individuals with a disease. Suppose that $$t$$ is the number of years since the onset of the disease. The death density function, $$f(t) = cte^{-kt}$$, approximates the fraction of the sick individuals who die in the time interval [t, t+Δt] as follows:
Fraction who die: $$f(t)\Delta t = cte^{-kt} \Delta t$$
where $$c$$ and $$k$$ are positive constants whose values depend on the particular disease.

(a) Find the value of c in terms of k.

(b) Express the cumulative death distribution function in the form below. Your answer will be in terms of k.

$$C(t)=\left\{\begin{array}{cc}A(t),& t < 0\\ B(t), & t \geq 0\end{array}\right$$

## Homework Equations

$$P(t) = \int_{-\inf}^t p(x) dx$$

## The Attempt at a Solution

To solve part a, I know that $$\lim_{t\rightarrow\infty} P(t) = 1$$. So c and k must have values so that it equals one. So I integrate $$f(t)$$ with the relevant equations to get:

$$\frac{-(kt+1) \cdot e^{-kt} \cdot c}{k^2}$$

from negative infinity to t.

The problem is that this diverges to negative infinity and doesn't give me a meaningful answer. So what do I do?

Also, how do I get the the bar with a superscript and subscript for 'from a to b' in tex?

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tiny-tim
Homework Helper
Hi altegron!
Consider a population of individuals with a disease. Suppose that $$t$$ is the number of years since the onset of the disease.

$$P(t) = \int_{-\infty}^t p(x) dx$$
No, nobody died before the onset of the disease!

(and that bar would be $$|_a^b\ \big|_a^b\ \Big|_a^b\ \bigg|_a^b\ \Bigg|_a^b$$)

Hi altegron!

No, nobody died before the onset of the disease!

(and that bar would be $$|_a^b\ \big|_a^b\ \Big|_a^b\ \bigg|_a^b\ \Bigg|_a^b$$)
Thanks for the response,

That makes sense -- I guess that means I should integrate from 0 to infinity then?

$$\lim_{x\rightarrow\infty} \int_0^x cte^{-kt} dt = 1$$

Which is easier written as

$$\int_0^{\infty} cte^{-kt} dt = 1$$

Which is:

$$\frac{-(kt+1) \cdot e^{-kt} \cdot c}{k^2} \bigg|_0^{\infty} = 1$$

The infinity term goes to zero and the zero term (negated for subtraction) is:

$$\frac{(k \cdot (0)+1) \cdot e^{-k \cdot (0)} \cdot c}{k^2} = 1$$

$$\frac{(0+1) \cdot e^{0} \cdot c}{k^2} = 1$$

$$\frac{c}{k^2} = 1$$

So, finally:

$$c = k^2$$

That seems plausible but I don't know how to check it.

Then for part b, I know $$A(t) = 0$$ because the disease hasn't started yet.

For $$B(t)$$ we use the integral we've already computed, from zero to t years. We know the value at zero is 1 (actually, -1, but it is being subtracted), so we have:

$$B(t) = 1 - \frac{(kt+1) \cdot e^{-kt} \cdot c}{k^2}$$

And that looks correct on my calculator when I graph it with k=1.

tiny-tim