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?

Printable View

- Sep 8th 2008, 05:47 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, 05:59 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, 07:06 AMjvignacio
- Sep 8th 2008, 07:56 AMjvignacio

input x = into:

to get a gradient of

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

is that the correct path? - Sep 8th 2008, 08:49 AMKrizalid
Yes.

- Sep 8th 2008, 12:07 PMTKHunny
- Sep 9th 2008, 02:55 AMjvignacio
- Sep 9th 2008, 03:03 AM11rdc11
Plug in the x coordinate given into the function and see if it matches with the y coordinate given.

- Sep 9th 2008, 03:04 AMjvignacio
- Sep 9th 2008, 10:54 AMTKHunny
NOW we're done.

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