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Thread: equation of tangent line

  1. #1
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    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.
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    Super Member bigwave's Avatar
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    sorry misread post
    Last edited by bigwave; Feb 12th 2010 at 08:02 PM. Reason: misread post
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  3. #3
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    Quote Originally Posted by bigwave View Post
    $\displaystyle y=6sec(x)$
    $\displaystyle x= \frac{\pi}{3}$
    $\displaystyle y'=6\cos{\frac{\pi}{3}}=3$

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

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

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

    Are you asking how to calculate $\displaystyle \sec \left(\frac{\pi}{3}\right)$ and $\displaystyle \tan \left(\frac{\pi}{3}\right)$ ?
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  5. #5
    Super Member bigwave's Avatar
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    ok sorry misread your post


    $\displaystyle y' = \frac{6\sin{x}}{\cos^2{x}}= 12\sqrt{3}$
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    Quote Originally Posted by drumist View Post
    You're correct. The methodology is the same though.

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

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

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

    Then plug it in to slope intercept form.
    $\displaystyle y=mx+b$
    $\displaystyle 12=(12\sqrt{3})(\frac{\pi}{3})+b$
    $\displaystyle 12=4\pi \sqrt{3}+b$
    $\displaystyle b=12-4\pi\sqrt{3}$
    $\displaystyle y=12x\sqrt{3}+12-4\pi\sqrt{3}$
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  7. #7
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    Quote Originally Posted by drumist View Post
    You're correct. The methodology is the same though.

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

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

    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
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  8. #8
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    Quote Originally Posted by bigwave View Post
    ok sorry misread your post


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

    how do you get the radical three?
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  9. #9
    Super Member bigwave's Avatar
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    basically you don't want a decimal in your answer but rather an exact one
    especially since we have trig functions in the works.

    $\displaystyle
    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 $\displaystyle y'$ mentioned above is accually easier to deal with
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  10. #10
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    Quote Originally Posted by bigwave View Post
    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?
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  11. #11
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    Quote Originally Posted by rhcp1231 View Post
    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
    $\displaystyle \pi/3$ 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 $\displaystyle \pi/3$ is $\displaystyle (1/2,\sqrt{3}/2)$. That means $\displaystyle \cos(\pi/3)=1/2$ and $\displaystyle \sin(\pi/3)=\sqrt{3}/2$. We notice that $\displaystyle \sec x = \frac{1}{\cos x}$ and $\displaystyle \tan x = \frac{\sin x}{\cos x}$ and we can find each of those values in exact form. (You can also now see where the $\displaystyle \sqrt{3}$ term comes from.)
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  12. #12
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    Quote Originally Posted by rhcp1231 View Post
    how do you get the radical three?
    $\displaystyle \sin {\frac {\pi}{3}}=\frac {\sqrt {3}}{2}$
    $\displaystyle \cos^{2}\frac{\pi}{3}=\frac{1}{2}^2=\frac{1}{4}$

    $\displaystyle \frac{6\sin{\frac{\pi}{3}}}{\cos^{2}\frac{\pi}{3}} =6\cdot \frac{\sqrt 3}{2}\cdot 4=24\cdot \frac{\sqrt 3}{2}=12\sqrt{3}$
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  13. #13
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    Quote Originally Posted by drumist View Post
    $\displaystyle \pi/3$ 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 $\displaystyle \pi/3$ is $\displaystyle (1/2,\sqrt{3}/2)$. That means $\displaystyle \cos(\pi/3)=1/2$ and $\displaystyle \sin(\pi/3)=\sqrt{3}/2$. We notice that $\displaystyle \sec x = \frac{1}{\cos x}$ and $\displaystyle \tan x = \frac{\sin x}{\cos x}$ and we can find each of those values in exact form. (You can also now see where the $\displaystyle \sqrt{3}$ term comes from.)

    so when i look at the unit circle the x-value is the cos and the y-value is the sin?
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  14. #14
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    Quote Originally Posted by rhcp1231 View Post
    so when i look at the unit circle the x-value is the cos and the y-value is the sin?
    Yes, exactly.
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  15. #15
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    Quote Originally Posted by drumist View Post
    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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