Question

find the equation of the tangent line to the curve ( y = 4sec(x)-8cos(x) ) at the point ( (pi/3,4) ). write your answer in the form ( y = mx + b ) where ( m ) is the slope and ( b ) is the ( y - )intercept.

Explanation:

Step1: Find the derivative of the function

The derivative of \(y = 4\sec(x)-8\cos(x)\) is \(y^\prime=4\sec(x)\tan(x)+8\sin(x)\).

Step2: Evaluate the derivative at \(x = \frac{\pi}{3}\)

Substitute \(x=\frac{\pi}{3}\) into \(y^\prime\):
\(\sec(\frac{\pi}{3}) = 2\), \(\tan(\frac{\pi}{3})=\sqrt{3}\), \(\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}\)
\(y^\prime|_{x = \frac{\pi}{3}}=4\times2\times\sqrt{3}+8\times\frac{\sqrt{3}}{2}=8\sqrt{3}+4\sqrt{3}=12\sqrt{3}\)
So the slope \(m = 12\sqrt{3}\)

Step3: Use the point - slope form \(y - y_1=m(x - x_1)\)

We have \(x_1=\frac{\pi}{3}\), \(y_1 = 4\)
\(y-4=12\sqrt{3}(x-\frac{\pi}{3})\)
\(y=12\sqrt{3}x-4\pi\sqrt{3}+4\)

Answer:

\(y = 12\sqrt{3}x+(4 - 4\pi\sqrt{3})\)