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

Question

Isabella Ackerman · Accepted Answer

#-3*sqrt(2)#

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)#

Christopher Agresta · Answer

undefined To find the slope of the tangent line of $ r = 12\sin(\frac{\theta}{3})\cos(\frac{\theta}{2}) $ at $ \theta = \frac{3\pi}{2} $, we first express the equation in terms of $ x $ and $ y $ using the polar coordinate conversion formulas. Then, we differentiate with respect to $ \theta $ and evaluate it at $ \theta = \frac{3\pi}{2} $. Finally, we find the tangent slope using the formula $ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} $ and substitute the given value of $ \theta $. The slope is $ \frac{1}{2} $.