The discussion revolves around finding the gradient of stationary points for the function y = 1 - 2sin(x) within the domain 0 < x < 2π. Participants are exploring the implications of differentiating the function and determining where it crosses the x-axis.
The conversation is active, with participants questioning each other's interpretations and clarifying the problem statement. Some guidance has been offered regarding the differentiation process and the conditions for finding x-axis crossings, but no consensus has been reached on the final values or methods.
There is a noted change in the problem statement regarding the domain of the function, which has led to further exploration of the implications for finding gradients at specific points. Participants are also working with the constraints of not having calculators available for exact calculations.
rock.freak667 said:Gradient of the stationary point? Do you need to find that or the gradient? Because when you put dy/dx=0 it sort of implies what the gradient will be.
pip_beard said:Sorry. the qu reads 'the curve y=1-2sinx has domain 0<x<pi. find the gradients of the curve at the points where the curve crosses the x-axis
pip_beard said:so do i differentiate it and make it = 0?
so i get -2cosx=0
pip_beard said:so do i differentiate it and make it = 0?
so i get -2cosx=0
pip_beard said:how to i rearrange this to work it out?
pip_beard said:-2sinx=1
sinx=-1/2
pip_beard said:-2sinx=1
sinx=-1/2
pip_beard said:principle value of -0.523 (but out of range) the pi-(-0.523) = 3.6651
therfore value = 3.6651 and 6.2832?
pip_beard said:ohhh... did sin^-1... so ...
x principle = 0.479425
so pi - 0.4879425 = 2.66216