Find an equation of the tangent line to the curve at the given point.
((pi/6),(2 square root of (3)/3))
y = sec (x)
y = secx
y' = secx tanx
at x = pi/6,
y' = 2/√3 * 1/√3 = 2/3
so, now we have a point and a slope, so the line is
y - 2/√3 = 2/3 (x-pi/6)
http://www.wolframalpha.com/input/?i=plot+y%3Dsec%28x%29+and+y+%3D+2%2F3+%28x-pi%2F6%29+%2B+++2%2F%E2%88%9A3
To find the equation of the tangent line to the curve at a given point, you need to find the derivative of the curve and then substitute the coordinates of the given point into the derivative equation.
Given that the curve is y = sec(x), we need to find the derivative of this function. The derivative of sec(x) can be found using the chain rule and is equal to sec(x)tan(x).
Now, substitute the x-coordinate ((pi/6)) of the given point into the derivative equation sec(x)tan(x):
dy/dx = sec(x)tan(x)
dy/dx = sec(pi/6)tan(pi/6)
To find sec(pi/6), we would first find the value of cos(pi/6), since sec(x) is the reciprocal of cos(x). The value of cos(pi/6) is equal to sqrt(3)/2. Taking the reciprocal of sqrt(3)/2 gives 2/sqrt(3), which simplifies to (2√3)/3.
Similarly, to find tan(pi/6), we need to find the value of sin(pi/6)/cos(pi/6). The value of sin(pi/6) is equal to 1/2, and cos(pi/6) is equal to sqrt(3)/2. Therefore, tan(pi/6) = (1/2) / (sqrt(3)/2) = 1/sqrt(3), which simplifies to √3/3.
Substituting the values, we get:
dy/dx = (2√3)/3 * (√3)/3
dy/dx = 2/3
So, the slope of the tangent line is 2/3.
Now, to find the equation of the tangent line, we need to use the point-slope form of a line. We have the coordinates of the given point as ((pi/6), (2√3)/3).
Using the point-slope form, we can write the equation of the tangent line as:
y - y1 = m(x - x1)
Substituting the values, we get:
y - (2√3)/3 = (2/3)(x - pi/6)
Now, simplify and rearrange the equation to get the final equation of the tangent line:
y - (2√3)/3 = (2/3)x - (2/3)(pi/6)
y - (2√3)/3 = (2/3)x - (1/3)pi
This is the equation of the tangent line to the curve y = sec(x) at the point ((pi/6), (2√3)/3).