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.
To find the coordinates of the point on the parabola that is closest to the origin, we can use Newton's method. Here are the steps:
Step 1: Define the function representing the distance from the origin to a point on the parabola.
Let's call this function f(x) = sqrt(x^2 + (x-5)^4).
We are looking for the value of x that minimizes this function.
Step 2: Take the derivative of f(x) with respect to x.
f'(x) = (2x + 4(x-5)^3) / (2 * sqrt(x^2 + (x-5)^4))
Step 3: Set up the Newton's method iteration.
Start with an initial guess, let's say x0 = 1.
The Newton's method iteration formula is: x1 = x0 - f(x0)/f'(x0)
Step 4: Calculate x1 using the iteration formula.
Plug in x0 = 1 into f(x) to find f(1).
Then plug in x0 = 1 into f'(x) to find f'(1).
Use the values to calculate x1 = x0 - f(x0)/f'(x0).
Step 5: Continue the iteration until convergence.
Repeat Step 4 with x1 as the new x value until the difference between consecutive x values is negligible, say less than 0.000001.
By following these steps, we can find the coordinates of the point on the parabola that is closest to the origin.
To find the point on the parabola that is closest to the origin, we need to minimize the distance between the origin (0,0) and a general point on the parabola (x, (x-5)^2).
We can express the distance between the two points as a function and then minimize that function using Newton's method. Let's call the distance d:
d = √((x - 0)^2 + ((x - 5)^2 - 0)^2)
= √(x^2 + (x-5)^4)
Now, let's minimize d by finding the critical points, where the derivative of d with respect to x equals zero. We will use Newton's method to solve this:
1. Take the first derivative of d with respect to x:
d' = (1/2) * (x^2 + (x-5)^4)^(-1/2) * (2x + 4(x-5)^3)
2. Set the derivative equal to zero and solve for x:
(1/2) * (x^2 + (x-5)^4)^(-1/2) * (2x + 4(x-5)^3) = 0
3. Simplify the equation and solve for x. This involves squaring both sides of the equation and collecting similar terms:
x^2 + (x-5)^4 = 0
4. Since we cannot solve this equation algebraically, we can use numerical methods to approximate the solution. One way to do this is by using an iterative method like Newton's method.
Initialize an approximate value of x, let's call it x0. Choose an initial guess value close to the expected solution. In this case, since we want the point closest to the origin, we can choose x0 = 0.
Repeat the following steps until the desired accuracy is achieved:
- Compute the next approximation of x, x1, using the formula:
x1 = x0 - f(x0)/f'(x0)
- Update x0 with the value of x1.
- Repeat until the desired accuracy is reached.
5. Once we have a sufficiently accurate approximation of x, we can find the corresponding y-coordinate by evaluating y = (x-5)^2.
Following the steps above, you can use Newton's method to approximate the coordinates of the point on the parabola that is closest to the origin, correct to six decimal places.
y' = 2(x-5)
Let x0=1
x1 = 1 - y/y' = 1 - 16/-8 = 3
x2 = 3 - 4/-4 = 4
x3 = 4 - 1/-2 = 6
x4 = 6 - 1/2 = 5.5
and so on for six decimal places.