Use Newton's method to find the coordinates, correct to six decimal places, of the point on the parabola y = (x − 7)2 that is closest to the origin.
let that point be P(a,b)
then b = (a-7)^2
dy/dx = 2(x-7)
at P, dy/dx = 2(a-7) <---- the slope of the tangent at P
Join PO. If P is the closest to O, then the slope of PO must be
the negative reciprocal of the slope of the tangent.
That is, at the point closest to the origin PO is the normal to the tangent at P
2(a-7) = a/b = -a/(a-7)^2
expanding and simplifying we get
2a^3 - 42a^2 + 295a - 686 = 0
it is this equation we want to solve using Newton's Method
let y = 2a^3 - 42a^2 + 295a - 686
dy/da = 6a^2 - 84a + 295
a(new) = a - (2a^3 - 42a^2 + 295a - 686)/(6a^2 - 84a + 295)
a ..... a(new) , start with a = 5 and use your calculator
5 5.44
5.44 5.57798...
5.57798..... 5.591158509
5.591158509 5.591272805
5.591272805 5.591272814 <---- looks like we got it to more than 6 decimal places
so the x of P is 5.591273 correct to 6 decimals
sub the above into y = (x-7)^2 to find the y
Well, I guess the origin wants to get cozy with the parabola, huh? Alright, let's see what we can do here.
To find the point on the parabola that is closest to the origin, we need to minimize the distance between the origin and a general point on the parabola. So, let's call the point we're looking for (x, (x-7)^2).
Step 1: Define the distance between the origin and a general point on the parabola:
Distance = sqrt(x^2 + ((x-7)^2)^2)
Step 2: Now we need to minimize this distance by finding the value of x that minimizes the distance. To do this, we need to find the derivative of the distance function with respect to x. I won't bore you with the long formula, but trust me, it involves some algebraic acrobatics.
Step 3: Now, Newton's method steps in to the ring and starts throwing some punches. We start with an initial guess for x, let's say x0 = 0, and then we iteratively improve our guess using this lovely formula:
x(n+1) = x(n) - (f(x(n))/f'(x(n)))
Step 4: We keep repeating this process until we reach a satisfactory level of accuracy (in this case, six decimal places).
So, my friend, grab your hat and let's start crunching some numbers with Newton's method! Good luck on your cozy mission with the parabola!
To find the point on the parabola y = (x − 7)2 that is closest to the origin, we need to minimize the distance between the point (x, y) on the parabola and the origin (0, 0).
The distance between two points (x1, y1) and (x2, y2) is given by the formula:
distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In this case, the coordinates of the point on the parabola can be written as (x, (x - 7)^2) and the coordinates of the origin are (0, 0).
So, the distance between these two points can be expressed as:
distance = sqrt((x - 0)^2 + ((x - 7)^2 - 0)^2)
Now, we need to minimize this distance by finding the value of x for which the derivative of the distance function is equal to zero.
Let's define the function f(x) as:
f(x) = sqrt(x^2 + (x - 7)^4)
To find the minimum, we will use Newton's method, which involves taking the derivative of f(x), setting it equal to zero, and iterating through an initial guess to find the value of x that minimizes the distance.
1. Calculate the derivative of f(x) with respect to x:
f'(x) = (2x + 4(x - 7)^3) / (2sqrt(x^2 + (x - 7)^4))
2. Set f'(x) = 0 and solve for x:
(2x + 4(x - 7)^3) / (2sqrt(x^2 + (x - 7)^4)) = 0
3. Simplify the equation:
2x + 4(x - 7)^3 = 0
4. Expand and rearrange the equation:
2x + 4(x^3 - 21x^2 + 147x - 343) = 0
4x^3 - 84x^2 + 588x - 1372 = 0
5. Use numerical methods, like Newton's method or graphing calculator, to find the value of x that satisfies the equation and lies between x = 0 and x = 7.
After finding the value of x, substitute it into the equation y = (x - 7)^2 to get the y-coordinate of the point on the parabola that is closest to the origin.
To find the point on the parabola closest to the origin, we can use Newton's method of optimization. Since we want to find the coordinates of a point on the parabola, we need to find the x-coordinate that corresponds to the minimum distance from the origin.
Let's start by defining the distance between a point on the parabola and the origin. The distance formula in two dimensions is given by:
Distance = √((x1 - x2)^2 + (y1 - y2)^2)
In this case, x1 and y1 represent the coordinates of the point on the parabola, and x2 = 0, y2 = 0 represent the origin.
To find the point on the parabola closest to the origin, we need to minimize this distance. Hence, our objective function is:
f(x) = √((x - 0)^2 + ((x - 7)^2)^2)
To find the x-coordinate that minimizes the distance, we need to find the critical points of this function. We can find these points using Newton's method.
Newton's method involves iteratively improving an initial guess for the critical point of a function. Here's a step-by-step approach to finding the coordinates using Newton's method:
Step 1: Find the derivative of the objective function f(x):
f'(x) = (2x - 7) / √((x - 7)^2 + x^2)
Step 2: Choose an initial guess for the critical point, let's say x0 = 10.
Step 3: Apply Newton's method to refine the estimate:
x1 = x0 - f(x0) / f'(x0)
x2 = x1 - f(x1) / f'(x1)
Continue this process until the difference between successive approximations is small enough.
Step 4: The x-coordinate that minimizes the distance is the approximation obtained from Newton's method.
Let's apply these steps to find the coordinates.
Step 1: f'(x) = (2x - 7) / √((x - 7)^2 + x^2)
Step 2: Initial guess, x0 = 10
Step 3: Iterations using Newton's method:
x1 = x0 - f(x0) / f'(x0)
x2 = x1 - f(x1) / f'(x1)
Continue this process until you reach a desired precision.
Step 4: Evaluate the obtained x-coordinate on the parabola equation y = (x - 7)^2 to find the corresponding y-coordinate.
Repeat these steps until you reach the desired precision of six decimal places.