equation for the line tangent to graph

• Sep 8th 2008, 04:47 AM
jvignacio
equation 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 AM
TKHunny
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 AM
jvignacio
Quote:

Originally Posted by TKHunny
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.

thank u ! ill give it a go and post it
• Sep 8th 2008, 06:56 AM
jvignacio
$
y = secx
$

$
y' = tanx \cdot secx
$

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

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

$
m = 2 \cdot \sqrt{3}
$

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

$

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

$

is that the correct path?
• Sep 8th 2008, 07:49 AM
Krizalid
Yes.
• Sep 8th 2008, 11:07 AM
TKHunny
Quote:

Originally Posted by jvignacio
$
y = secx
$

$
y' = tanx \cdot secx
$

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

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

$
m = 2 \cdot \sqrt{3}
$

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

$

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

$

is that the correct path?

Did you do my Step #1? You can be assured I would put one on the exam that was NOT actually on the curve.

Other than that...AWESOME!!
• Sep 9th 2008, 01:55 AM
jvignacio
Quote:

Originally Posted by TKHunny
Did you do my Step #1? You can be assured I would put one on the exam that was NOT actually on the curve.

Other than that...AWESOME!!

thakns mann!!
how do i check if its on the graph?
• Sep 9th 2008, 02:03 AM
11rdc11
Plug in the x coordinate given into the function and see if it matches with the y coordinate given.
• Sep 9th 2008, 02:04 AM
jvignacio
Quote:

Originally Posted by 11rdc11
Plug in the x coordinate given into the function and see if it matches with the y coordinate given.

yes it does! both 2 :)
• Sep 9th 2008, 09:54 AM
TKHunny
NOW we're done.

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