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

1 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)

Explanation:

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:
Desmos.comDesmos.com

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