What is the slope of the tangent line of #r=-8sin(theta/4)+4cos(theta/2)# at #theta=(2pi)/3#?

Question

Emma Aldridge · Accepted Answer

the tangent line is #y = -1.73205 - 0.57735 (x-1)#

Given #r(theta)# we have

#{ (x(theta)=r(theta)cos(theta)), (y(theta)=r(theta)sin(theta)) :}#

# { (dx/(d theta) = -r(theta)sin(theta)+cos(theta)((dr)/(d theta))), (dy/(d theta) = r(theta)cos(theta)+sin(theta)((dr)/(d theta))) :} #

but

#r(theta) = -8 Sin(theta/4)+ 4 Cos(theta/2)#

then

# { (dx/(d theta) = 3 Cos((3 theta)/4) - 5 Cos((5 theta)/4) - Sin(theta/2) - 3 Sin((3 theta)/2)), (dy/(d theta)=Cos(theta/2) + 3 (Cos((3 theta)/2) + Sin((3 theta)/4)) - 5 Sin((5 theta)/4)) :} #

but

#(dy)/(dx) = (((dy)/(d theta)))/(((dx)/(d theta)))#

at point #theta_0 = 2 pi/3# we have

#p_0 ={r(theta_0)cos(theta_0),r(theta_0)sin(theta_0)}={1.0,-1.73205}#

and

#m_0 = ((dy)/(dx) )_{theta_0} = -0.57735#

The tangent straight reads

#y = -1.73205 - 0.57735 (x-1)#

Christopher Agresta · Answer

undefined To find the slope of the tangent line at a specific point on a polar curve, you can use the derivative of the polar function with respect to θ, and then evaluate it at the given θ value.

Given the polar function r = -8sin(θ/4) + 4cos(θ/2), the derivative of r with respect to θ is:

dr/dθ = (-8/4)cos(θ/4) - (4/2)sin(θ/2)

Now, to find the slope of the tangent line at θ = (2π)/3, substitute this value into the derivative:

dr/dθ = (-8/4)cos((2π)/12) - (4/2)sin((2π)/6)

Evaluate the trigonometric functions:

dr/dθ = (-8/4)cos(π/6) - (4/2)sin(π/3)

Now calculate the values of cos(π/6) and sin(π/3):

cos(π/6) = √3/2
sin(π/3) = √3/2

Substitute these values into the expression:

dr/dθ = (-8/4)(√3/2) - (4/2)(√3/2)
       = -4√3/2 - 2√3/2
       = -6√3/2

So, the slope of the tangent line at θ = (2π)/3 is -6√3/2.