These two q's for my homework I am very confused on how to do:
First Question:
Using Newton’s method, approximate the value of √5 up to 2 decimal points
starting with x1 = 3.
2nd Question:
Thomas Malthus was an economist that predicted that the population grows exponentially while the food supply grows linearly. If p(t) gives the population and f(t) gives the number of people that can be supported by the amount of food produced at t years, we would be concerned about when p(t) = f(t). After this point, the pop. would be beyond capacity that the environment can support. Suppose p(t)= 1000e^(0.02t) and f(t)= 30t + 2000. Using Newton’s method with x1 = 60, find x2 to approximate when the functions intersect.
let x = √5
then x^2 - 5
let y = x^2 - 5
y' = 2x
Newton said:
newx = x - y/y'
= x - (x^2 - 5)/(2x)
= (2x^2 - x^2 + 5)/2x
= (x^2 + 5)/(2x)
x newx
3 2.3333....
2.333... 2.23809...
2.23809... 2.23606... could have stopped here
2.23606.. 2.236067878
2.236067978.. 2.236067977
2. let y = 1000 e^(.02t) - 30t - 2000
find y' and follow the same steps as I did in #1
Since you only have to find x2, (only one step)
I would not bother simplifying the expression
and just do one step of
newx = x = y/y'
For the first question:
To approximate the value of √5 using Newton's method, we need to find the root of the function f(x) = x^2 - 5, which represents the equation √5 = x.
Newton's method involves iteratively improving an initial guess to find a more accurate approximation. We start with an initial guess, which in this case is x1 = 3.
The iterative formula for Newton's method is:
x_(n+1) = x_n - f(x_n)/f'(x_n)
To apply this formula, we need to find the derivative of f(x). Taking the derivative of f(x) = x^2 - 5, we get f'(x) = 2x.
Now we can start the iterations:
Step 1:
x1 = 3
f(x1) = (3)^2 - 5 = 9 - 5 = 4
f'(x1) = 2(3) = 6
Using the iterative formula:
x2 = x1 - f(x1)/f'(x1) = 3 - 4/6 = 3 - 0.67 = 2.33
Step 2:
x2 = 2.33
f(x2) = (2.33)^2 - 5 ≈ 0.1089
f'(x2) = 2(2.33) ≈ 4.66
Using the iterative formula:
x3 = x2 - f(x2)/f'(x2) = 2.33 - 0.1089/4.66 ≈ 2.33 - 0.0234 ≈ 2.3066
Since we want to approximate up to two decimal places, we can continue these iterations until we reach the desired accuracy.
Continuing this process, you can keep iterating until you reach the desired level of accuracy. The iterations will converge towards the actual value of √5, which is approximately 2.236.
For the second question:
To find the intersection point of the functions p(t) and f(t), we need to find the value of t where p(t) - f(t) = 0.
First, you need to define the function g(t) = p(t) - f(t).
In this case, p(t) = 1000e^(0.02t) and f(t) = 30t + 2000.
Substituting these values, we have g(t) = 1000e^(0.02t) - (30t + 2000).
We want to find the value of t where g(t) = 0. In other words, we need to find the root of the function g(t).
Similar to the first question, we can use Newton's method to approximate the value of t where g(t) = 0.
Start with an initial guess, x1 = 60.
The iterative formula for Newton's method in this case will be:
x_(n+1) = x_n - g(x_n)/g'(x_n)
To apply this formula, we need to find the derivative of g(t).
Taking the derivative of g(t), we get g'(t) = 20e^(0.02t) - 30.
Now we can start the iterations:
Step 1:
x1 = 60
g(x1) = 1000e^(0.02(60)) - (30(60) + 2000) ≈ -614.05
g'(x1) = 20e^(0.02(60)) - 30 ≈ -9.68
Using the iterative formula:
x2 = x1 - g(x1)/g'(x1) ≈ 60 - (-614.05)/(-9.68) ≈ 60 + 63.46 ≈ 123.46
By repeating this process, you can continue iterating until you reach an approximate value for x where g(x) = 0.