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

Question

Elijah Abram · Accepted Answer

#y = 2sqrt3(x-pi/3)+2 color(white)"XXX"#

The equation for the tangent line at the point #(x_1,f(x_1))# is:

#y = f'(x_1)(x-x_1)+f(x_1)#

(This is really just the point-slope form of a line in disguise!)

The problem gives us that #x_1=pi/3# and #f(x_1)=2#. All we need to do is find #f'(x_1)# and then plug all of our values into the point-slope equation.

#f'(x)=d/dxsecx=secxtanx#

#therefore f'(pi/3)=sec(pi/3)tan(pi/3)=2*sqrt3#

Now, we need to just plug in all of these values to give us the tangent line equation:

#y = f'(x_1)(x-x_1)+f(x_1)#

#y = 2sqrt3(x-pi/3)+2#

Or if the problem prefers slope intercept form:

#y=2sqrt3x-(2pisqrt3)/3+2#

Charles Arocha · Answer

undefined To find the equation of the tangent line to the curve y=secx at (pi/3,2), we need to find the derivative of the function y=secx and evaluate it at x=pi/3. The derivative of y=secx is dy/dx = secx * tanx. Evaluating this at x=pi/3, we get dy/dx = sec(pi/3) * tan(pi/3). Simplifying this, we have dy/dx = 2 * sqrt(3).

Now, we have the slope of the tangent line, which is dy/dx = 2 * sqrt(3). To find the equation of the tangent line, we use 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. Plugging in the values, we have y - 2 = (2 * sqrt(3))(x - pi/3). Simplifying this equation gives the equation of the tangent line to the curve y=secx at (pi/3,2).