hey i have im practising some problems and i dont know where to start on this question.

find an equation for the line tangent to the graph of y = secx at (pie/3,2)

any suggestions?

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- Sep 8th 2008, 04:47 AMjvignacioequation for the line tangent to graph
hey i have im practising some problems and i dont know where to start on this question.

find an equation for the line tangent to the graph of y = secx at (pie/3,2)

any suggestions? - Sep 8th 2008, 04:59 AMTKHunny
1) Verify that (pi/3,2) is ON the graph. This is a point on the line.

2) Find the derivative of the function, y' = ???

3) Evaluate the derivaive at x = pi/3. This is the slope of the line.

4) Remember your algebra and use the Point-Slope form of a line. - Sep 8th 2008, 06:06 AMjvignacio
- Sep 8th 2008, 06:56 AMjvignacio
$\displaystyle

y = secx

$

$\displaystyle

y' = tanx \cdot secx

$

input x = $\displaystyle \frac{\pi}{3}$ into:

$\displaystyle

y' = tan(\frac{\pi}{3}) \cdot sec(\frac{\pi}{3})

$

to get a gradient of

$\displaystyle

m = 2 \cdot \sqrt{3}

$

now y - y1 = m(x - x1) formula to get the equation

$\displaystyle

y - 2 = 2 \cdot \sqrt{3}(x - \frac{\pi}{3})

$

is that the correct path? - Sep 8th 2008, 07:49 AMKrizalid
Yes.

- Sep 8th 2008, 11:07 AMTKHunny
- Sep 9th 2008, 01:55 AMjvignacio
- Sep 9th 2008, 02:03 AM11rdc11
Plug in the x coordinate given into the function and see if it matches with the y coordinate given.

- Sep 9th 2008, 02:04 AMjvignacio
- Sep 9th 2008, 09:54 AMTKHunny
NOW we're done.

When the sneaky one shows up, you'll be the only one to get it right.