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.

Answer 1

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?

Answer 2

Nardos, did you post to the wrong question by accident. Because what you posted does not even relate to my question

Answer 3

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

Answer 4

Reiny,
Aren't we supposed to use distance formula for this though?