Home > Calculus > Determining the Slope and Tangent Lines for a Polar Curve

What is the slope of the tangent line of #r=theta-3cos((2theta)/3+(pi)/2)# at #theta=(-5pi)/6#?

Answer 1

Aurora Alvarado

The slope #m~~ -1.34#

The reference Tangents with Polar Coordinates gives us the equation:

#dy/dx = ((dr)/(d theta)sin(theta)+rcos(theta))/((dr)/(d theta)cos(theta)-rsin(theta))" [1]"#

Compute #(dr)/(d theta)#

#(dr)/(d theta) = 1+2sin((2theta)/3+pi/2)#

#(dr)/(d theta) = 1+2cos((2theta)/3)#

The substitute the equation for r and #(dr)/(d theta)# into equation [1]:

#dy/dx = ((1+2cos((2theta)/3))sin(theta)+(theta-3cos((2theta)/3+(pi)/2))cos(theta))/((1+2cos((2theta)/3))cos(theta)-(theta-3cos((2theta)/3+(pi)/2))sin(theta))#

The slope, m, is the above equation evaluated at #theta=(-5pi)/6#

#m = ((1+2cos((2((-5pi)/6))/3))sin((-5pi)/6)+((-5pi)/6-3cos((2((-5pi)/6))/3+(pi)/2))cos((-5pi)/6))/((1+2cos((2((-5pi)/6))/3))cos((-5pi)/6)-((-5pi)/6-3cos((2((-5pi)/6))/3+(pi)/2))sin((-5pi)/6))#

#m~~ -1.34#

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Answer from HIX Tutor

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