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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

Hello,

I got problem with A homework

"find an equation of the tangent line to curve at the given point.

$y=sec(x)$. $(pi/3,2)$

progress:

$y'=sec(x)tan(x)$. So basicly that sec(x) dont say me much so i rewrite it as $1/cos(x)$

$y'=1/cos(x)•tan(x)$ now i can put $pi/3$ on the function to calculate the slope.

i get that the slope is $m=2•sqrt(3)$ and now we use the tangent equation $y-y1=m(x-x1)$

So we got $y-2=2sqrt(3)(x-pi/3)$ and i basicly answer $y=2sqrt(3)(x-pi/3)+2$

Is this correct? I am sure i am thinking correct but not 100%

I got problem with A homework

"find an equation of the tangent line to curve at the given point.

$y=sec(x)$. $(pi/3,2)$

progress:

$y'=sec(x)tan(x)$. So basicly that sec(x) dont say me much so i rewrite it as $1/cos(x)$

$y'=1/cos(x)•tan(x)$ now i can put $pi/3$ on the function to calculate the slope.

i get that the slope is $m=2•sqrt(3)$ and now we use the tangent equation $y-y1=m(x-x1)$

So we got $y-2=2sqrt(3)(x-pi/3)$ and i basicly answer $y=2sqrt(3)(x-pi/3)+2$

Is this correct? I am sure i am thinking correct but not 100%

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