posted by TayB .
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.
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?
Nardos, did you post to the wrong question by accident. Because what you posted does not even relate to my question
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
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
Aren't we supposed to use distance formula for this though?