What is the slope of the tangent line of #r=2theta-3sin((13theta)/8-(5pi)/3)# at #theta=(7pi)/6#?

Question

Ezekiel Alexander · Accepted Answer

#color(blue)(dy/dx=([(7pi)/3-3 sin((11pi)/48)] cos ((7pi)/6)+[2-(39/8)cos ((11pi)/48)]*sin ((7pi)/6))/(-[(7pi)/3-3 sin((11pi)/48)] sin ((7pi)/6)+[2-(39/8)cos ((11pi)/48)] cos ((7pi)/6)))#
SLOPE #color(blue)(m=dy/dx=-0.92335731861741)#

The solution:

The given
#r=2theta-3 sin((13theta)/8-(5 pi)/3)# at #theta=(7pi)/6#

#dy/dx=(r cos theta+r' sin theta)/(-r sin theta+r' cos theta)#

#dy/dx=([2theta-3 sin((13theta)/8-(5 pi)/3)] cos theta+[2-3(13/8)cos ((13theta)/8-(5 pi)/3)]*sin theta)/(-[2theta-3 sin((13theta)/8-(5 pi)/3)] sin theta+[2-3(13/8)cos ((13theta)/8-(5 pi)/3)] cos theta)#

Evaluating #dy/dx# at #theta=(7pi)/6#

#dy/dx=([2((7pi)/6)-3 sin((13((7pi)/6))/8-(5 pi)/3)] cos ((7pi)/6)+[2-3(13/8)cos ((13((7pi)/6))/8-(5 pi)/3)]*sin ((7pi)/6))/(-[2((7pi)/6)-3 sin((13((7pi)/6))/8-(5 pi)/3)] sin ((7pi)/6)+[2-3(13/8)cos ((13((7pi)/6))/8-(5 pi)/3)] cos ((7pi)/6))#

#dy/dx=([(7pi)/3-3 sin((91pi)/48-(5 pi)/3)] cos ((7pi)/6)+[2-(39/8)cos ((91pi)/48-(5 pi)/3)]*sin ((7pi)/6))/(-[(7pi)/3-3 sin((91pi)/48-(5 pi)/3)] sin ((7pi)/6)+[2-(39/8)cos ((91pi)/48-(5 pi)/3)] cos ((7pi)/6))#
#color(blue)(dy/dx=([(7pi)/3-3 sin((11pi)/48)] cos ((7pi)/6)+[2-(39/8)cos ((11pi)/48)]*sin ((7pi)/6))/(-[(7pi)/3-3 sin((11pi)/48)] sin ((7pi)/6)+[2-(39/8)cos ((11pi)/48)] cos ((7pi)/6)))#

#color(blue)(dy/dx=-0.92335731861741)#

#x=r cos theta=(2theta-3 sin((13theta)/8-(5 pi)/3))*cos theta#
#x=[(7pi)/3-3 sin((91pi)/48-(5 pi)/3)] cos ((7pi)/6)#
#x=[(7pi)/3-3 sin((11pi)/48)] cos ((7pi)/6)#
#x=-4.6352670975528#

#y=r sin theta=(2theta-3 sin((13theta)/8-(5 pi)/3))*sin theta#
#y=[(7pi)/3-3 sin((91pi)/48-(5 pi)/3)] sin ((7pi)/6)#
#y=[(7pi)/3-3 sin((11pi)/48)] sin ((7pi)/6)#
#y=-2.6761727065385#

Using Point-Slope Form:
Equation of the tangent line is

#y-y_1=m(x-x_1)#
#y--2.6761727065385=-0.92335731861741(x--4.6352670975528)#

Check the graph:

God bless....I hope the explanation is useful.

Adam Adkins · Answer

undefined To find the slope of the tangent line of $ r = 2\theta - 3\sin\left(\frac{13\theta}{8} - \frac{5\pi}{3}\right) $ at $ \theta = \frac{7\pi}{6} $, you first need to differentiate the polar equation with respect to $ \theta $ to find $ \frac{dr}{d\theta} $. Then evaluate $ \frac{dr}{d\theta} $ at $ \theta = \frac{7\pi}{6} $ to get the slope of the tangent line at that point.