Thread: Compute tangent line to curve... help :/

A) Computer the tangent line to the curve x^4-yx^2+y^4 =1 at the point (-1,1)

Ok so I did the derivative =
4x^3-2x y' + 4x^3 y' = 0
Do I just take the point x (-1) put it in the formula for X to find y' and that's it? :S

B) Iterate Newton's method 2 times to produce a rational number approximating 4^-3. Use 2 as your initial guess.

Xn+1 = Xn - f(Xn)/(f'(Xn))

so... I'm lost about there :/

A mistake:

Ok so I did the derivative =
4x^3-2x y' + 4x^3 y' = 0

The derivative is:

Thanks, can't believe I missed that.
But what do I do with it now?

Originally Posted by roushrsh

Thanks, can't believe I missed that.
But what do I do with it now?

sub -1 for x , 1 for y and calculate the value of the derivative (slope) at (-1,1)

use the point-slope form of a linear equation to get the tangent line equation.

Ah ok, makes sense

So 3y' = 2
y'=2/3

y-y1 = m(x-x1)
y-1 = 2/3(x+1)
y=2/3x + 5/3

For the next one do I start with 3 = 2 - (4^-3)/(4^-4) ? Then repeat? :S

Originally Posted by roushrsh

B) Iterate Newton's method 2 times to produce a rational number approximating 4^-3. Use 2 as your initial guess.

Xn+1 = Xn - f(Xn)/(f'(Xn))

please check this problem ...

why would one wish to use Newton's method to produce a rational number approximating , when is a rational number to start with?

did you mean to type ?

Yes, sorry, I meant the latter with the sqrt
the
^3sqrt4

Xn+1 = Xn - f(Xn)/(f'(Xn))

use with

Why would I use x^3 -4 ? :S (=3x^2 for the f'(x))
Thanks
EDIT: nvm, I found that out, thanks. I'll try it

So would it be
xn+1 = xn-f(xn)/f'(xn)
f(x) = x^3-4 x0 = 2
f'(x) = 3x^2
x1 = 2-4/12 = 2-1/3 = (1+2/3)
x2 = (1+2/3) - ((1+2/3)^3-4)/(3(1+2/3)^2) = 1.561111...
Makes sense