Find the equation of the tangent line to the curve $f(x)=2\sec(x) - 4$ at the point $[(\pi/6),f(\pi/6)]$

Question

I have tried to directly substitute $\pi/6$ into the equation and solve for the $y$ value, but it isn't turning out. I believe the slope is $4/3$ (I found this by deriving the equation). It is just the $y$ value I am having difficulty with.

Answer 1

If you have the slope, plug $\frac \pi 6$ into the equation to get a point on the curve, which is also on the tangent line. Now use the point-slope form for a line.

Answer 2

$2\sec(\frac{\pi}{6})\tan(\frac{\pi}{6}) = \frac{4}{3}$

So the slope is $\frac{4}{3}$, therefore:

$y = \frac{4}{3}x + b$

$y = 2\sec(\frac{\pi}{6})-4 = -1.6906$

$-1.6906 = \frac{4}{3}\cdot \frac{\pi}{6} + b$

$b = -1.6906 - \frac{4}{3}\cdot \frac{\pi}{6}$

$b = -2.3887$

$y = \frac{4}{3}x - 2.3887$

Answer 3

That's right. You first find the slope of the tangent line through differentiation. Then, you find the y-value for the given point, followed by using point-slope form.

$\frac{d}{dx}2\sec(x)-4$ = $2\frac{d}{dx}\sec (x)$ = $2\tan (x)\sec (2)$.
From here, $x = \frac{\pi}{6}$, which means...
$y_1 = 2\sec ({\frac{\pi}{6}})-4$ = $2\frac{2\sqrt{3}}{3}-4$ = $\frac{4\sqrt{3}}{3}-4$ = $\frac{4\sqrt{3}-12}{3}$.
$y-y_1 = m(x-x_1)$.
$y-\frac{4\sqrt{3}-12}{3}$ = $\frac{4}{3}(x-\frac{\pi}{6})$
$y$ = $\frac{4}{3}(x-\frac{\pi}{6})+\frac{4\sqrt{3}-12}{3}$
$y$ = $\frac{4}{3}x-\frac{4\pi}{18}+\frac{4\sqrt{3}-12}{3}$
$y$ = $\frac{4}{3}x+\frac{12\sqrt{3}-36-2\pi}{9}\approx \frac{4}{3}x-2.38873$