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?
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.
$\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?