Question

How do you find the equation of the tangent line to the curve `y=4secx–8cosx` at the point `(pi/3,4)`?

Answer 1

`y = (12sqrt(3))x + 4(1-pisqrt(3))`

or equivalently:

`y = 20.78x - 17.77`

Explanation:

The equation of the tangent line to the curve at a point requires two steps:

Step 1: find the slope of the tangent line by finding the derivative of y
Step 2: use the given point and the slope to find the linear equation of the tangent line

Step 1: derivative

`y = 4secx - 8cosx`
`y' = 4secxtanx + 8 sinx`

We now substitute in the x-coord of the point given to compute the value of the derivative at that x.
`y'(pi/3) = 4sec(pi/3)tan(pi/3) + 8 sin(pi/3)`
`y'(pi/3) = 4(2)(sqrt(3)) + 8(sqrt(3)/2) = 8sqrt(3)+4sqrt(3) = 12sqrt(3)`

This is the slope of the tangent line.

Step 2: obtain equation of line

Point slope form: `(y-y_0) = m(x-x_0)`
Let `y_0 = 4` and `x_0 = pi/3`. `m= 12sqrt(3)`

`y-4 = 12sqrt(3) (x-pi/3)`

`y - 4 = (12sqrt(3))x-4pisqrt(3)`

`y = (12sqrt(3))x + 4(1-pisqrt(3))`

or equivalently:

`y = 20.78x - 17.77`

This is the equation of the tangent line to the function given at the point `(pi/3, 4)`