Find an equation of the tangent line to the curve at the given point.
y = sec x, (π/3, 2)
y'(x) = secx tanx
y'(pi/3) = 2√3
(y-2) = 2√3 (x-π/3)
To find the equation of the tangent line to the curve at the given point, we need two pieces of information: the slope of the tangent line and a point on the line.
Step 1: Find the derivative of the function y = sec(x) to get the slope of the tangent line at any given point.
The derivative of y = sec(x) can be found using the chain rule. Recall that the derivative of sec(x) is sec(x) * tan(x):
dy/dx = sec(x) * tan(x)
Step 2: Evaluate the derivative at the given point (π/3) to find the slope of the tangent line at that point.
Let's substitute π/3 into the derivative equation:
dy/dx = sec(π/3) * tan(π/3)
To simplify, we need to evaluate sec(π/3) and tan(π/3):
sec(π/3) = 1/cos(π/3) = 2
tan(π/3) = √3
Therefore, the slope of the tangent line at the point (π/3, 2) is:
dy/dx = 2 * √3
Step 3: Use the point-slope form of a line to write the equation of the tangent line.
The point-slope form of a line is given by: y - y₁ = m(x - x₁), where (x₁, y₁) is the given point and m is the slope of the line.
Substituting the values into the equation:
y - 2 = 2 * √3 (x - π/3)
Now simplify the equation:
y - 2 = 2√3x - 2√3(π/3)
y - 2 = 2√3x - 2√3π/3
y = 2√3x - 2√3π/3 + 2
So, the equation of the tangent line to the curve y = sec(x) at the point (π/3, 2) is y = 2√3x - 2√3π/3 + 2.