1. equation of tangent line

Find an equation of the tangent line to the curve at the given point. y = 6 sec x

P = (π/3, 12)

I know that the first step is taking the derivative of 6 sec x, but i dont quit know how to deal with the pi over 3 when i use it in the derivative to find the slope.

3. Originally Posted by bigwave
$y=6sec(x)$
$x= \frac{\pi}{3}$
$y'=6\cos{\frac{\pi}{3}}=3$

$y-12=(3)(x-\frac{\pi}{3})$
$
y=3x-\pi+12$

the derivative of sec(x) is sec(x)tan(x) though

4. Originally Posted by rhcp1231
the derivative of sec(x) is sec(x)tan(x) though
You're correct. The methodology is the same though.

$y' = 6 \sec x \tan x = 6 \sec \left(\frac{\pi}{3}\right) \tan \left(\frac{\pi}{3}\right)$

Are you asking how to calculate $\sec \left(\frac{\pi}{3}\right)$ and $\tan \left(\frac{\pi}{3}\right)$ ?

$y' = \frac{6\sin{x}}{\cos^2{x}}= 12\sqrt{3}$

6. Originally Posted by drumist
You're correct. The methodology is the same though.

$y' = 6 \sec x \tan x = 6 \sec \left(\frac{\pi}{3}\right) \tan \left(\frac{\pi}{3}\right)$

Are you asking how to calculate $\sec \left(\frac{\pi}{3}\right)$ and $\tan \left(\frac{\pi}{3}\right)$ ?
So, after you've differentiated to $\frac{\mathrm{d} y}{\mathrm{d} x}=6\sec x\tan x$

Then plug in your x value and you'll get:
$6\cdot \sec \frac{\pi}{3}\cdot \tan \frac{\pi}{3}=6\cdot 2\cdot \sqrt 3=12\sqrt 3=m$

Then plug it in to slope intercept form.
$y=mx+b$
$12=(12\sqrt{3})(\frac{\pi}{3})+b$
$12=4\pi \sqrt{3}+b$
$b=12-4\pi\sqrt{3}$
$y=12x\sqrt{3}+12-4\pi\sqrt{3}$

7. Originally Posted by drumist
You're correct. The methodology is the same though.

$y' = 6 \sec x \tan x = 6 \sec \left(\frac{\pi}{3}\right) \tan \left(\frac{\pi}{3}\right)$

Are you asking how to calculate $\sec \left(\frac{\pi}{3}\right)$ and $\tan \left(\frac{\pi}{3}\right)$ ?

yes i am asking how i keep that pi over three because if i simplify it into a decimal number it is wrong, so i guess this is almost just a basic arithmetic question lol

8. Originally Posted by bigwave

$y' = \frac{6\sin{x}}{\cos^2{x}}= 12\sqrt{3}$

how do you get the radical three?

9. basically you don't want a decimal in your answer but rather an exact one
especially since we have trig functions in the works.

$
y' = \frac{6\sin{(\frac{\pi}{3})}}{\cos^2{(\frac{\pi}{3 })}}= \frac{3\sqrt{3}}{\frac{1}{4}}= 12\sqrt{3}
$

the expression of $y'$ mentioned above is accually easier to deal with

10. Originally Posted by bigwave
basically you don't want a decimal in your answer but rather an exact one
especially since we have trig functions in the works.

how do i do that?

11. Originally Posted by rhcp1231
yes i am asking how i keep that pi over three because if i simplify it into a decimal number it is wrong, so i guess this is almost just a basic arithmetic question lol
$\pi/3$ is a common angle that we consider on the unit circle. Take a look at this:

File:Unit circle angles.svg - Wikipedia, the free encyclopedia

You probably have studied a version of this in a trig course.

The coordinate corresponding to $\pi/3$ is $(1/2,\sqrt{3}/2)$. That means $\cos(\pi/3)=1/2$ and $\sin(\pi/3)=\sqrt{3}/2$. We notice that $\sec x = \frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$ and we can find each of those values in exact form. (You can also now see where the $\sqrt{3}$ term comes from.)

12. Originally Posted by rhcp1231
how do you get the radical three?
$\sin {\frac {\pi}{3}}=\frac {\sqrt {3}}{2}$
$\cos^{2}\frac{\pi}{3}=\frac{1}{2}^2=\frac{1}{4}$

$\frac{6\sin{\frac{\pi}{3}}}{\cos^{2}\frac{\pi}{3}} =6\cdot \frac{\sqrt 3}{2}\cdot 4=24\cdot \frac{\sqrt 3}{2}=12\sqrt{3}$

13. Originally Posted by drumist
$\pi/3$ is a common angle that we consider on the unit circle. Take a look at this:

File:Unit circle angles.svg - Wikipedia, the free encyclopedia

You probably have studied a version of this in a trig course.

The coordinate corresponding to $\pi/3$ is $(1/2,\sqrt{3}/2)$. That means $\cos(\pi/3)=1/2$ and $\sin(\pi/3)=\sqrt{3}/2$. We notice that $\sec x = \frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$ and we can find each of those values in exact form. (You can also now see where the $\sqrt{3}$ term comes from.)

so when i look at the unit circle the x-value is the cos and the y-value is the sin?

14. Originally Posted by rhcp1231
so when i look at the unit circle the x-value is the cos and the y-value is the sin?
Yes, exactly.

15. Originally Posted by drumist
Yes, exactly.

ok ok, lol i forgot all of this stuff.

tan is just the x over the y and to find the sec, csc, i put the number under 1 right?

like to find the numerical value of the csc of pi over three i put radical three over two and place it as a fraction under 1, is this correct?

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