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.
Nardos, did you post to the wrong question by accident. Because what you posted does not even relate to my question
Two cars A and B are moving along a straight road in the same direction withe velocity of 25km/h and 40km/h, respectively. Find the velocity of car B relatively to car A?
let the point of contact be (a,b), then
b = (a-5)^2 = a^2 - 10a + 25
sketch the tangent and the normal
y' = 2(x-5) = 2x - 10
so the slope of the tangent at (a,b)
= 2a - 10
and the slope of the NORMAL at (a,b)
= -1/(2a - 10) or 1/(10-2a)
but using the old grade 9 way of doing slope ...
slope of normal = (b-0)/(a-0) = b/a
thus:
1/(10-2a) = b/a
a = 10b - 2ab
subbing in b
a = 10(a^2 - 10a + 25) - 2a(a^2 - 10a + 25)
a = 10a^2 -100a + 250 - 2a^3 + 20a^2 - 50a
2a^3 - 30a^2 + 151a - 250 = 0
using x instead of a,
f(x) = 2x^3 - 30x^2 + 151x - 250
f ' (x) = 6x^2 - 60x + 151
by Newtons' Method, the simplified expression is
newx = (4x^3 - 30x^2 + 250)/(6x^2 - 60x + 151)
according to my sketch x = 3.5 might be a good starting point
newx = 3.765227...
next newx = 3.765227..
I think we got it after two iterations
carry on, you have the x, now find the y
Reiny,
Aren't we supposed to use distance formula for this though?
To find the coordinates of the point on the parabola that is closest to the origin using Newton's method, we need to minimize the distance between the origin and a point on the parabola.
Let's denote the coordinates of the point on the parabola as (x, (x-5)^2). The distance between this point and the origin (0, 0) is given by the distance formula:
distance = sqrt(x^2 + (x-5)^4)
To minimize this distance, we can minimize the square of the distance since the square root function is increasing. So, we consider minimizing the function:
f(x) = x^2 + (x-5)^4
To find the coordinates of the point closest to the origin, we need to find the value of x that minimizes this function. We can use Newton's method to find this value.
Here's the step-by-step process to apply Newton's method:
1. Start with an initial guess for x, denoted as x0.
2. Calculate the derivative of the function f(x) with respect to x, denoted as f'(x).
3. Determine the next guess for x using the formula:
x1 = x0 - f(x0)/f'(x0)
4. Repeat steps 2 and 3 using the updated guess x1, until you reach the desired level of accuracy.
Let's use the above steps to find the coordinates of the point closest to the origin:
Step 1: Select an initial guess for x (x0). Let's start with x0 = 1.
Step 2: Calculate the derivative of the function f(x) = x^2 + (x-5)^4. The derivative f'(x) is:
f'(x) = 2x + 4(x-5)^3
Step 3: Use the formula x1 = x0 - f(x0)/f'(x0) to find the next guess for x:
x1 = x0 - (x0^2 + (x0-5)^4) / (2x0 + 4(x0-5)^3)
Repeat steps 2 and 3 until you reach the desired level of accuracy.
Continue iterating until the difference between consecutive approximations is less than the desired level of accuracy. In this case, let's aim for an accuracy of six decimal places.
By following these steps, you can repeatedly apply Newton's method to find the coordinates of the point on the parabola that is closest to the origin, correct to six decimal places.