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.
http://wordplay.blogs.nytimes.com/2011/03/14/numberplay-pi-in-the-sky/
To find the equation for the lines that are tangent to the curve y = cot^2(x) at the point (π/4, 1), we can follow these steps:
Step 1: Find the derivative of y = cot^2(x) using the chain rule.
The derivative of cot^2(x) can be found using the chain rule for derivatives.
dy/dx = d/dx[cot^2(x)] = d/dx[(cot(x))^2]
To find the derivative of cot(x), we can use the identity: d/dx[cot(x)] = -csc^2(x).
Therefore, applying the chain rule, we have:
dy/dx = 2cot(x)*(-csc^2(x)) = -2cot(x)csc^2(x)
Step 2: Determine the slope of the tangent line at the point (π/4, 1).
To find the slope of the tangent line, substitute π/4 into the derivative.
dy/dx = -2cot(π/4)csc^2(π/4)
Since cot(π/4) = 1 and csc(π/4) = √2, we can substitute these values:
dy/dx = -2(1)(√2)^2 = -4
Therefore, the slope of the tangent line at (π/4, 1) is -4.
Step 3: Use the point-slope form of the equation to find the equation of the tangent line.
The point-slope form of a linear equation is: y - y1 = m(x - x1), where (x1, y1) is a point on the line, and m is the slope.
Using the point (π/4, 1) and the slope -4, we have:
y - 1 = -4(x - π/4)
Simplifying the equation:
y - 1 = -4x + π
y = -4x + π + 1
Therefore, the equation of the tangent line to the curve y = cot^2(x) at the point (π/4, 1) is y = -4x + π + 1.