Question

The number of people expected to have a disease in t years is given by
y(t) = A.3^(t/a) (i) If now (year 2016) the number of people having the disease is 1000, find the value of A.
(ii) How many people are expected to have the disease in five years?
(iii) When are 100,000 people expected to have the disease?
(iv) How fast is the number of people with the disease expected to grow now and ten years from now?

Answer 1

If we let a = 9,
y(t) = A x 3^(t/9)

(i) If y(t) = 1000 a t = 0, then
1000 = A (3)^(0/a)
A = 1000

(ii) At t= 5
y(5) = 1000 (3)^(5/9)
y(5) = 1841

(iii) At y(t) = 100,000
100,000 = 1000 (3)^(t/9)
t = 37.7 years

(iv) Taking the derivative and substituting, t = 0 , and t = 10
for t = 0
dy/dt = 122 people/year

for t = 10
dy/dt = 414 people/year