Question

find an equation of the tangent line to the curve at the given point. y=secx (pi/3,2)

Answer 1

\[y=\sec x\]

Answer 2

with points (\[(\pi/3,2)\]

Answer 3

I'll lay out the procedure for solving this type of problem without giving the solution. If you still can't solve it, post again indicating where you got stuck.

The first step is to find the derivative of your function:\[\frac{ d }{ dx } \sec x= ?\]If you don't have that derivative handy, you can calculate it by using the fact that\[\sec x=\frac{ 1 }{ \cos x }\]The result here will be a general formula for the derivative of sec x.

The second step is to evaluate this formula where\[x=\pi/3\]This tells you the slope of the tangent line at the point you're interested in.

Finally, you plug this slope into the point-slope form for the equation of a line, along with the coordinates given for this point. The result is the equation of the tangent line at that point.

Answer 4

derivative = secxtanx

Answer 5

???

Answer 6

so the slope would be 3.46
rounding of course

Answer 7

Or, to be more precise,\[2\sqrt{3}\]We get this result by noting that sec x * tan x = (sin x)/cos^2 x), where sin(pi/3) = (sqrt3)/2 and cos(pi/3) = 1/2. If this is for a class, you may be expected to produce the precise result based on your knowledge of the sine and cosine of this special angle rather than an approximation from a calculator.

Okay, so you have the slope and you also have the coordinates of a point through which the line passes. All you need to do now is plug those items into the point-slope equation for a line to produce an equation for the line you need, completing the problem.