Posted by hiii on .
All right, sorta hard to explain but I will try my best. I am suppose to find the equation for the that lines that are tangent to the curve y=cot^2(x) at the point (pie/4, 1)....(the x is not to the power of 2, a variable right next to the cot^2)
I am not sure how to do this and if you can help, that would be awesome. It's the cot that's really tripping me up :/
<rant> AARGH! Who are these people that don't understand that π is PI, not PIE? </rant>
y = cot^2(x)
y' = 2cot(x)(-csc^2(x)) = -2cot(x)csc^2(x)
at x=π/4, the slope is -2(1)(2) = -4
the line through (π/4,1) with slope -4 is
(y-1) = -4(x-π/4)
massage that equation into the form you most like.
how did u get -2(1)(2)? did u just plug the PI/4 into the formula?
yeah, that's the way one computes the value of f(x) -- plug in the value of x.
- calculus - Count Iblis,