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Presentation transcript:

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) cosθ and y = f(θ) sinθ &

Example 1 Let r = sin3θ Find dy/dx Solution: x= sin3θ cosθ & y= sin3θ sinθ dx/dθ = - sin3θ sinθ + 3cos3θ cosθ dy/dθ = sin3θ cosθ + 3cos3θ 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

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

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

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

Homework:

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

II. Find the slope of the tangent to the curve at the given point θ0 1. r = sin3θ , θ0 = π/6 2. r = 4sinθ , θ0 = π/3 3. r = 1 + sinθ , θ0 = π/4

III. Find the equation of the tangent to the curve at the given point (r0,θ0) 1. r = cos2θ , (r0,θ0) = ( -1 , π/2) 2. r = 4sin2x , (r0,θ0) = ( 1 , 5π/6)