1. Polar Differentiation

2. & 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
&

3. 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

4. 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

5. x= (1 + cosθ) cosθ = cosθ+ cos2θ1.
Solution:
x= (1 + cosθ) cosθ = cosθ+ cos2θ
y= (1 + cosθ) sinθ = sinθ+ sinθ cosθ= sinθ + (1/2) sin2θ
dx/dθ = - sinθ cosθ sinθ
dy/dθ = cosθ + cos2θ

7. 2.
The slope of the tangent at the point
( 1 , π/2) is equal to:

8. 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

9. Homework:

10. 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)

11. 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

12. 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)