Use Newton's method to find the coordinates, correct to six decimal places, of the point on the parabola y = (x − 5)^2 that is closest to the origin.
or use one of the various online Newton method calculators to verify your work.
Actually the first one I tried by hand and then after that I tried plugging into the calculator but I think I had a bad guess cuz it hadn't gotten 6 decimal places the same. It kept changing
Just did this for you.
Look back to your earlier post of this
To find the coordinates of the point on the parabola that is closest to the origin, we can use Newton's method. Newton's method is an iterative process that can be used to approximate the root of a function. In this case, we want to find the x-coordinate of the point on the parabola that is closest to the origin, so we need to find the root of the derivative of the function.
Let's start by finding the derivative of the function y = (x - 5)^2. The derivative of y with respect to x is:
dy/dx = 2(x - 5)
Now, we can set up the Newton's method iteration formula:
x1 = x0 - f(x0)/f'(x0)
where x1 is the next approximation, x0 is the current approximation, f(x0) is the value of the function at x0, and f'(x0) is the value of the derivative at x0.
In this case, we want to find the root of the derivative function dy/dx = 2(x - 5). Since the formula for Newton's method uses an initial approximation, we need to choose a starting value for x, denoted as x0.
Let's choose x0 = 1 as the initial approximation. Now we can iterate using the Newton's method formula to find a more accurate approximation for the root:
x1 = x0 - f(x0)/f'(x0)
After performing the calculation, we repeat the process with the newly obtained value of x, plugging it into the formula again, and repeat until we reach the desired level of accuracy.
Here is a step-by-step breakdown of the Newton's method iterations:
1. Choose an initial approximation: x0 = 1
Calculate f(x0) and f'(x0):
f(x0) = (x0 - 5)^2 = (1 - 5)^2 = 16
f'(x0) = 2(x0 - 5) = 2(1 - 5) = -8
Calculate x1 using the Newton's method formula:
x1 = x0 - f(x0)/f'(x0) = 1 - 16/-8 = 3
2. Repeat the process:
Calculate f(x1) and f'(x1):
f(x1) = (x1 - 5)^2 = (3 - 5)^2 = 4
f'(x1) = 2(x1 - 5) = 2(3 - 5) = -4
Calculate x2 using the Newton's method formula:
x2 = x1 - f(x1)/f'(x1) = 3 - 4/-4 = 4.5
3. Repeat the process:
Calculate f(x2) and f'(x2):
f(x2) = (x2 - 5)^2 = (4.5 - 5)^2 = 0.25
f'(x2) = 2(x2 - 5) = 2(4.5 - 5) = -1
Calculate x3 using the Newton's method formula:
x3 = x2 - f(x2)/f'(x2) = 4.5 - 0.25/-1 = 4.375
4. Repeat the process until the desired level of accuracy is reached.
By continuing this iteration process, we can find an approximation for the x-coordinate of the point on the parabola that is closest to the origin.
Look, I took this subject back in 1956. I am not going to do half a million Newton's method approximations for you. I do not need the practice, you do.
Sketch it, draw tangents. Try, then if you have trouble post.