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?
if t = 0 now
y(0) = A 3^0 = A
so
A = 1000
and
y(t) = 1000 3^(t/a)
y(5) = 1000 3^(5/a)
100 = 3^t/a
log 100 = (t/a) log 3
2 = (t/a).477
t/a = 4.19
y(t) = 1000 3^(t/a)
let x = (t/a)
y = 1000 3^x
then
dy/dx = ln 3 * 3^x
dy/dt =dy/dx dx/dt
dy/dt = ln 3 * 3^(t/a) *(1/a)
dy/dt =(1/a)ln 3 * 3^(t/a)
put in t = 0 and t =10
To find the value of A, we can use the given information that in the year 2016, the number of people having the disease is 1000.
(i) We can substitute t = 0 and y(t) = 1000 into the equation to solve for A:
1000 = A * 3^(0/a)
Since any number raised to the power of 0 is equal to 1, we can simplify the equation:
1000 = A * 1
Therefore, A = 1000.
Now, let's address the remaining questions:
(ii) To find the number of people expected to have the disease in five years, we can substitute t = 5 into the equation:
y(5) = 1000 * 3^(5/a)
(iii) To find when 100,000 people are expected to have the disease, we can substitute y(t) = 100,000 into the equation:
100000 = 1000 * 3^(t/a)
To solve for t, we need to use logarithms:
log(3^(t/a)) = log(100000/1000)
Using the logarithmic property, we can bring the exponent down:
(t/a) * log(3) = log(100)
Now, solve for t by isolating it:
t = (a * log(100)) / log(3)
(iv) To determine how fast the number of people with the disease is expected to grow now and ten years from now, we can calculate the derivative with respect to t:
dy(t)/dt = (A * 3^(t/a)) * (log(3)/a)
Evaluate the derivative when t = 0 (now) and t = 10 (ten years from now) to find the rate of growth at those points in time.