How do you find the equation of the line tangent to #y=4secx–8cosx# at the point (pi/3,4)?

Question

Christopher Agresta · Accepted Answer

I found: #y=12sqrt(3)x-4(pisqrt(3)-1)#

We know that the equation of the tangent can be evaluated as: #(y-y_0)=m(x-x_0)# where: #m=#slope of the tangent; #x_0,y_0# are the coordinates of the tangence point. Now, to find the slope #m# we evaluate the DERIVATIVE of our function and evaluate it at #x=pi/3# (given); so: #y'=(4sin(x))/(cos^2(x))+8sin(x)# now we substitute #x=pi/3# #y'(pi/3)=m=(4sin(pi/3))/(cos^2(pi/3))+8sin(pi/3)=12sqrt(3)# and for our equation: #(y-4)=12sqrt(3)(x-pi/3)# #y=12sqrt(3)x-12pisqrt(3)/3+4# #y=12sqrt(3)x-4(pisqrt(3)-1)#

hope it helps!

Paisley Johnson · Answer

undefined To find the equation of the line tangent to the curve y = 4sec(x) - 8cos(x) at the point (π/3, 4), we need to find the derivative of the function and evaluate it at the given point.

The derivative of y = 4sec(x) - 8cos(x) can be found using the chain rule and the derivative rules for sec(x) and cos(x).

dy/dx = 4sec(x)tan(x) + 8sin(x)

Now, substitute π/3 into the derivative expression to find the slope of the tangent line at that point.

dy/dx = 4sec(π/3)tan(π/3) + 8sin(π/3)

Simplifying, we get:

dy/dx = 4(2/√3)(√3/3) + 8(√3/2)

dy/dx = 8/√3 + 4√3

dy/dx = (8 + 4√3√3)/√3

dy/dx = (8 + 12√3)/√3

Now, we have the slope of the tangent line at the point (π/3, 4).

Using the point-slope form of a line, we can write the equation of the tangent line:

y - 4 = [(8 + 12√3)/√3](x - π/3)

Simplifying further, we get:

y - 4 = (8 + 12√3)(x - π/3)/√3

This is the equation of the line tangent to y = 4sec(x) - 8cos(x) at the point (π/3, 4).