Polar Differentiation

Let r = f( θ ) and ( x,y) is the rectangular representation of the point having the polar representation ( r, θ ) Then x = f( θ ) cos θ and y = f( θ ) sin θ &

Example 1 Let r = sin3 θ Find dy/dx Solution: x= sin3 θ cosθ & y= sin3 θ sinθ dx/dθ = - sin3 θ sinθ + 3cos3 θ cosθ dy/dθ = sin3 θ cosθ + 3cos3 θ sinθ

Example 2 Let r = 1 + cosθ Find 1. dy/dx 2. The slope of the tangent to the graph at (r,θ )= ( 1, π/2 ) 3. The equation of tangent to the graph in Cartesian coordinates

1.1. Solution: x= (1 + cosθ) cosθ = cosθ+ cos 2 θ y= (1 + cosθ) sinθ = sinθ+ sinθ cosθ= sinθ + (1/2) sin2θ dx/dθ = - sinθ - 2 cosθ sinθ dy/dθ = cosθ + cos2θ

2.2. The slope of the tangent at the point ( 1, π/2) is equal to:

3.3. The Cartesian representation of the point having the polar representation ( 1, π/2) is: (1.cos π/2, 1.sin π/2 ) = ( 0, 1 ) Thus the equation in of the tangent to the curve at the given point is the equation of the straight line through the Point (0, 1 ) and having the slope, which is : y – 1 = 1 ( x – 0 ) Or x – y + 1 = 0

Horizontal & vertical Tangents Example Let r = 4 - 4cos θ Find the points at which the curve of r has horizontal or vertical tangents.

Solution We have, x = cosθ ( 4 – 4cosθ ) y = sinθ ( 4 – 4cosθ ) dx/dθ = cosθ (4sinθ) + ( 4 – 4cosθ)(- sinθ) = 4[ 2cosθ sinθ – sinθ ] = 4 sinθ [ 2cosθ – 1 ] dy/dθ = sinθ (4sinθ) + ( 4 – 4cosθ)( cosθ) = 4[ sin 2 θ + cosθ – cos 2 θ] = 4[ 1 – cos 2 θ + cosθ – cos 2 θ = - 4[ 2cos 2 θ - cosθ +1] = 4 ( cosθ – 1 ) (2cosθ +1)

Horizontal tangents The curve has horizontal tangent for all θ in the set: { θ: dy/dθ = 0 and dx/dθ ≠ 0 } = { θ: " cosθ = 1 or cosθ = - 1/2 " and " sinθ ≠ 0, cosθ ≠ 1/2 " } = { 0, 2π, π – π/3, π +π/3 } - { 0, π, π, 2π, π/3, 2π - π/3 } = { π – π/3, π +π/3 } = = { 2 π/3, 4π/3 } Thus, the points, at which, the curve has a horizontal tangent are: ( 4 - 4cos (2 π/3), 2 π/3 ) = ( ( - 1/2), 2 π/3 ) = ( 6, 2 π/3 ) ( 4 - 4cos (4 π/3), 2 π/3 ) = ( ( - 1/2), 2 π/3 ) = ( 6, 4 π/3 )

Vertical Tangents The curve has vertical tangent for all θ in the set: { θ: dx/dθ = 0 and dy/dθ ≠ 0 } = { θ: " sinθ = 0 or cosθ = 1/2 " & " cosθ = 1, cosθ = - 1/2 " } = { 0, π, π, 2π, π/3, 2π - π/3 } - { 0, 2π, π – π/3, π +π/3 } = { π, π/3, 2π - π/3 } = = { π, π/3, 5π/3 } Thus, the points, at which, the curve has a vertical tangent are: (4 - 4cos(π/3), π/3 ) = (4 - 4( 1/2), π/3 ) = ( 2, π/3 ) & (4 - 4cos(5 π/3), 5π/3 ) = (4 - 4( 1/2), 5 π/3 ) = ( 2, 5π/3 ) & (4 - 4cosπ, 5π/3 ) = (4 - 4(- 1), π ) = ( 8, π )

Singular points Singular points occur for all θ in the set: { θ: dy/dθ = 0 and dx/dθ = 0 } = { θ: " cosθ = 1 or cosθ = - 1/2 " & " sinθ ≠ 0 0r cosθ ≠ 1/2 " } = { 0, 2π, π – π/3, π +π/3 } ∩ { 0, π, π, 2π, π/3, 2π - π/3 } = { 0, 2π } Thus, there is only one singular point (0,0), Why? )

Homework:

I. I. Find dy/dx 1. r = 1 + cos θ 2. r = sin2θ 3. r = 1 + sin θ 4. r = csc θ 5. r = θ 6. r = 2/(1 – sinθ)

II. Find the slope of the tangent to the curve at the given point θ 0 1. r = sin3θ, θ 0 = π/6 2. r = 4sinθ, θ 0 = π/3 3. r = 1 + sinθ, θ 0 = π/4

III. Find the equation of the tangent to the curve at the given point (r 0,θ 0 ) 1. r = cos2θ, (r 0,θ 0 ) = ( -1, π/2) 2. r = 4sin 2 x, (r 0,θ 0 ) = ( 1, 5π/6)

Note If the representation (0, θ 1 ) satisfies the equation r = f(θ), & dr/dθ is nonzero at θ 1, then the line θ = θ 1 is a tangent to the curve at the pole.