How do you find an equation of the tangent line to the curve at the given point #y = 2xcosx # and #(pi, -2pi)#?

Question

William Abdalla · Accepted Answer

#2x+y=0.# #y=2xcosx rArr dy/dx=2{x(cosx)'+(cosx)(x)'}# [Product Rule for Diff.] #:. dy/dx=2(-xsinx+cosx)# #:.[dy/dx]_(x=pi,y=-2pi)=2(-pisinpi+cospi)=-2.# This means that the slope m of tangent (tgt.) line to the given curve is #m=-2.# Tgt. line passes thro. the pt. #(pi,-2pi).# Therefore, its eqn. is #y-(-2pi)=(-2)(x-pi),# # i.e., y+2pi+2x-2pi=0,# or, #2x+y=0.#

Emma Aldridge · Answer

undefined To find the equation of the tangent line to the curve y = 2xcosx at the point (pi, -2pi), we need to find the slope of the tangent line and then use the point-slope form of a linear equation.

1. Find the derivative of the function y = 2xcosx using the product rule:
   dy/dx = 2cosx - 2xsinx

2. Evaluate the derivative at x = pi:
   dy/dx = 2cos(pi) - 2pi*sin(pi)
          = 2(-1) - 2pi(0)
          = -2

3. The slope of the tangent line is -2.

4. Use the point-slope form of a linear equation, y - y₁ = m(x - x₁), where (x₁, y₁) is the given point and m is the slope:
   y - (-2pi) = -2(x - pi)

5. Simplify the equation:
   y + 2pi = -2x + 2pi

6. Rearrange the equation to the standard form:
   2x + y = 0

Therefore, the equation of the tangent line to the curve y = 2xcosx at the point (pi, -2pi) is 2x + y = 0.