How do you find the equation of the line tangent to #y=secx#, at (pi/3,2)?

Question

Ezra Adams · Accepted Answer

#y=2sqrt3x-(2pisqrt3)/3+2# Name the point #" "(pi/3,2)# by A. #" "# The line passes through the point #A(pi/3,2)" "#. #" "# The slope of this tangentbis determined by performing #y'# at #" "# #x=pi/3# #" "# Differentiating #y# #" "# #y' = (secx)'=tanxsecx# #" "# The slope of tangent line at #x=pi/3# is: #" "# #y'_(x=pi/3) =tan(pi/3)sec (pi/3)=sqrt3 xx 2=color (brown)(2sqrt3)# #" "# The equation of the tangent line with slope#" "color (brown)(2sqrt3)# #" "# passing through#" "A (pi/3,2)# is: #" "# #y-y_A= y_(x_A)(x-x_A)# #" "# #y-2=2sqrt3 (x-pi/3)# #" "# #y=2sqrt3x-(2pisqrt3)/3+2#

Aurora Alvarado · Answer

undefined To find the equation of the line tangent to y=secx at (pi/3,2), we need to find the derivative of y=secx and evaluate it at x=pi/3.

The derivative of y=secx is dy/dx = secx * tanx.

Evaluating the derivative at x=pi/3, we have dy/dx = sec(pi/3) * tan(pi/3).

Using the trigonometric values, sec(pi/3) = 2 and tan(pi/3) = sqrt(3).

Therefore, dy/dx = 2 * sqrt(3).

Now, we have the slope of the tangent line.

Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point (pi/3, 2) and m is the slope (2 * sqrt(3)), we can substitute the values to find the equation of the tangent line.

Thus, the equation of the line tangent to y=secx at (pi/3,2) is y - 2 = 2 * sqrt(3)(x - pi/3).