What is the slope of the tangent line of #r=12sin(theta/3)*cos(theta/2)# at #theta=(3pi)/2#?

1 Answer
Feb 13, 2016

#-3*sqrt(2)#

Explanation:

First, we need to find the derivative of the equation with respect to #theta#

#d/(d theta) r=12*d/(d theta)[sin(theta/3)*cos(theta/2)]#

where we have already factored out the constant. Using the chain rule we get two terms:

#d/(d theta) r=12*[d/(d theta)sin(theta/3)]*cos(theta/2)#

#+ 12* sin(theta/3)*[d/(d theta) cos(theta/2)]#

Each of the derivatives in the square brackets can be completed as follows:

#d/(d theta)sin(theta/3) = 1/3 cos(theta/3)#
and
#d/(d theta) cos(theta/2)=-1/2 sin(theta/2)#

plugging these back in above

#d/(d theta) r=4 cost(theta/3) cos(theta/2) - 6 sin(theta/3) sin(theta/2)#

next, we plug in the value of #theta for which we want the slope and simplify

#d/(d theta) r|_(theta = 3pi/2)= 4cos(pi/2)cos((3pi)/4) - 6sin(pi/2)sin((3pi)/4)#

#d/(d theta) r|_(theta = 3pi/2)= 0-6*sqrt(2)/2 = -3*sqrt(2)#