Definition of Derivatives of Polar Functions:
Derivatives of Polar Functions: ( r , θ ) (r, \theta) ( r , θ ) r r r θ \theta θ r r r θ \theta θ
The formula for the derivative of a polar function is:
d y d x = d y d θ d x d θ = r ′ ( θ ) . sin ( θ ) + r ( θ ) . cos ( θ ) r ′ ( θ ) . cos ( θ ) − r ( θ ) . sin ( θ ) \boxed{\frac{dy}{dx} = \dfrac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{r'(\theta). \sin(\theta) + r(\theta). \cos(\theta)}{r'(\theta) .\cos(\theta) - r(\theta). \sin(\theta)}} d x d y = d θ d x d θ d y = r ′ ( θ ) . cos ( θ ) − r ( θ ) . sin ( θ ) r ′ ( θ ) . sin ( θ ) + r ( θ ) . cos ( θ )
Where:
r ( θ ) r(\theta) r ( θ ) r ′ ( θ ) r'(\theta) r ′ ( θ ) r ( θ ) r(\theta) r ( θ ) θ \theta θ d y d x \dfrac{dy}{dx} d x d y
This formula allows us to convert the polar derivative into a Cartesian slope by applying trigonometric identities.
Formula and Explanation:
The derivative of polar functions is a powerful concept for converting between polar and Cartesian coordinates.
The formula above helps in differentiating complex polar equations by providing a clear path from polar to Cartesian derivatives.
This rule is used in areas like physics (e.g., motion in polar coordinates), engineering , and astronomy where objects move in circular paths or rotational motion.
Derivative Examples with Solutions:
Example 1: Derivative of a Polar Curve
Differentiate the polar equation r ( θ ) = 2 + 3 sin ( θ ) r(\theta) = 2 + 3 \sin(\theta) r ( θ ) = 2 + 3 sin ( θ )
Solution:
Using the formula for the derivative of polar functions:
r ′ ( θ ) = d d θ ( 2 + 3 sin ( θ ) ) = 3 cos ( θ ) r'(\theta) = \dfrac{d}{d\theta} \big(2 + 3 \sin(\theta)\big) = 3 \cos(\theta) r ′ ( θ ) = d θ d ( 2 + 3 sin ( θ ) ) = 3 cos ( θ )
Now apply the formula for d y d x \dfrac{dy}{dx} d x d y
d y d x = r ′ ( θ ) . sin ( θ ) + r ( θ ) . cos ( θ ) r ′ ( θ ) . cos ( θ ) − r ( θ ) . sin ( θ ) \dfrac{dy}{dx} = \dfrac{r'(\theta). \sin(\theta) + r(\theta). \cos(\theta)}{r'(\theta). \cos(\theta) - r(\theta). \sin(\theta)} d x d y = r ′ ( θ ) . cos ( θ ) − r ( θ ) . sin ( θ ) r ′ ( θ ) . sin ( θ ) + r ( θ ) . cos ( θ )
Substitute r ( θ ) = 2 + 3 sin ( θ ) r(\theta) = 2 + 3 \sin(\theta) r ( θ ) = 2 + 3 sin ( θ ) r ′ ( θ ) = 3 cos ( θ ) r'(\theta) = 3 \cos(\theta) r ′ ( θ ) = 3 cos ( θ )
d y d x = ( 3 cos ( θ ) ) sin ( θ ) + ( 2 + 3 sin ( θ ) ) cos ( θ ) ( 3 cos ( θ ) ) cos ( θ ) − ( 2 + 3 sin ( θ ) ) sin ( θ ) \dfrac{dy}{dx} = \dfrac{(3 \cos(\theta)) \sin(\theta) + (2 + 3 \sin(\theta)) \cos(\theta)}{(3 \cos(\theta)) \cos(\theta) - (2 + 3 \sin(\theta)) \sin(\theta)} d x d y = ( 3 cos ( θ )) cos ( θ ) − ( 2 + 3 sin ( θ )) sin ( θ ) ( 3 cos ( θ )) sin ( θ ) + ( 2 + 3 sin ( θ )) cos ( θ )
Simplifying:
d y d x = 3 sin ( θ ) cos ( θ ) + 2 cos ( θ ) + 3 sin ( θ ) cos ( θ ) 3 cos 2 ( θ ) − 2 sin ( θ ) − 3 sin ( θ ) cos ( θ ) \dfrac{dy}{dx} = \dfrac{3 \sin(\theta) \cos(\theta) + 2 \cos(\theta) + 3 \sin(\theta) \cos(\theta)}{3 \cos^2(\theta) - 2 \sin(\theta) - 3 \sin(\theta) \cos(\theta)} d x d y = 3 cos 2 ( θ ) − 2 sin ( θ ) − 3 sin ( θ ) cos ( θ ) 3 sin ( θ ) cos ( θ ) + 2 cos ( θ ) + 3 sin ( θ ) cos ( θ )
Thus, the final answer is:
6 sin ( θ ) cos ( θ ) + 2 cos ( θ ) 3 cos 2 ( θ ) − 2 sin ( θ ) − 3 sin ( θ ) cos ( θ ) \boxed{\frac{6 \sin(\theta) \cos(\theta) + 2 \cos(\theta)}{3 \cos^2(\theta) - 2 \sin(\theta) - 3 \sin(\theta) \cos(\theta)}} 3 cos 2 ( θ ) − 2 sin ( θ ) − 3 sin ( θ ) cos ( θ ) 6 sin ( θ ) cos ( θ ) + 2 cos ( θ )
Example 2: Derivative of a Polar Spiral
Differentiate the polar function r ( θ ) = θ 2 r(\theta) = \theta^2 r ( θ ) = θ 2
Solution:
First, compute the derivative of r ( θ ) r(\theta) r ( θ )
r ′ ( θ ) = 2 θ r'(\theta) = 2\theta r ′ ( θ ) = 2 θ
Now apply the formula for d y d x \frac{dy}{dx} d x d y
d y d x = r ′ ( θ ) sin ( θ ) + r ( θ ) cos ( θ ) r ′ ( θ ) cos ( θ ) − r ( θ ) sin ( θ ) \dfrac{dy}{dx} = \dfrac{r'(\theta) \sin(\theta) + r(\theta) \cos(\theta)}{r'(\theta) \cos(\theta) - r(\theta) \sin(\theta)} d x d y = r ′ ( θ ) cos ( θ ) − r ( θ ) sin ( θ ) r ′ ( θ ) sin ( θ ) + r ( θ ) cos ( θ )
Substitute r ( θ ) = θ 2 r(\theta) = \theta^2 r ( θ ) = θ 2 r ′ ( θ ) = 2 θ r'(\theta) = 2\theta r ′ ( θ ) = 2 θ
d y d x = ( 2 θ ) sin ( θ ) + θ 2 cos ( θ ) ( 2 θ ) cos ( θ ) − θ 2 sin ( θ ) \dfrac{dy}{dx} = \dfrac{(2\theta) \sin(\theta) + \theta^2 \cos(\theta)}{(2\theta) \cos(\theta) - \theta^2 \sin(\theta)} d x d y = ( 2 θ ) cos ( θ ) − θ 2 sin ( θ ) ( 2 θ ) sin ( θ ) + θ 2 cos ( θ )
Thus, the final answer is:
2 θ sin ( θ ) + θ 2 cos ( θ ) 2 θ cos ( θ ) − θ 2 sin ( θ ) \boxed{\frac{2\theta \sin(\theta) + \theta^2 \cos(\theta)}{2\theta \cos(\theta) - \theta^2 \sin(\theta)}} 2 θ cos ( θ ) − θ 2 sin ( θ ) 2 θ sin ( θ ) + θ 2 cos ( θ )
Example 3: Finding the Tangent Line
Find the equation of the tangent line to the polar curve r ( θ ) = 1 + cos ( θ ) r(\theta) = 1 + \cos(\theta) r ( θ ) = 1 + cos ( θ ) θ = π 3 \theta = \dfrac{\pi}{3} θ = 3 π
Solution:
First, compute r ′ ( θ ) r'(\theta) r ′ ( θ )
r ′ ( θ ) = d d θ ( 1 + cos ( θ ) ) = − sin ( θ ) r'(\theta) = \dfrac{d}{d\theta} \big(1 + \cos(\theta)\big) = -\sin(\theta) r ′ ( θ ) = d θ d ( 1 + cos ( θ ) ) = − sin ( θ )
Now apply the formula for d y d x \dfrac{dy}{dx} d x d y
d y d x = r ′ ( θ ) sin ( θ ) + r ( θ ) cos ( θ ) r ′ ( θ ) cos ( θ ) − r ( θ ) sin ( θ ) \dfrac{dy}{dx} = \dfrac{r'(\theta) \sin(\theta) + r(\theta) \cos(\theta)}{r'(\theta) \cos(\theta) - r(\theta) \sin(\theta)} d x d y = r ′ ( θ ) cos ( θ ) − r ( θ ) sin ( θ ) r ′ ( θ ) sin ( θ ) + r ( θ ) cos ( θ )
Substitute r ( θ ) = 1 + cos ( θ ) r(\theta) = 1 + \cos(\theta) r ( θ ) = 1 + cos ( θ ) r ′ ( θ ) = − sin ( θ ) r'(\theta) = -\sin(\theta) r ′ ( θ ) = − sin ( θ )
d y d x = ( − sin ( θ ) . sin ( θ ) + ( 1 + cos ( θ ) ) . cos ( θ ) ) ( − sin ( θ ) . cos ( θ ) − ( 1 + cos ( θ ) ) . sin ( θ ) ) \dfrac{dy}{dx} = \dfrac{\big(-\sin(\theta). \sin(\theta) + (1 + \cos(\theta)). \cos(\theta)\big)}{\big(-\sin(\theta). \cos(\theta) - (1 + \cos(\theta)). \sin(\theta)\big)} d x d y = ( − sin ( θ ) . cos ( θ ) − ( 1 + cos ( θ )) . sin ( θ ) ) ( − sin ( θ ) . sin ( θ ) + ( 1 + cos ( θ )) . cos ( θ ) )
At θ = π 3 \theta = \dfrac{\pi}{3} θ = 3 π
r ( π 3 ) = 1 + cos ( π 3 ) = 1 + 1 2 = 3 2 r\left(\frac{\pi}{3}\right) = 1 + \cos\left(\dfrac{\pi}{3}\right) = 1 + \dfrac{1}{2} = \dfrac{3}{2} r ( 3 π ) = 1 + cos ( 3 π ) = 1 + 2 1 = 2 3
r ′ ( π 3 ) = − sin ( π 3 ) = − 3 2 r'\left(\dfrac{\pi}{3}\right) = -\sin\left(\dfrac{\pi}{3}\right) = -\dfrac{\sqrt{3}}{2} r ′ ( 3 π ) = − sin ( 3 π ) = − 2 3
Substitute these values into the derivative formula:
d y d x = − 3 2 ⋅ 3 2 + 3 2 ⋅ 1 2 − 3 2 ⋅ 1 2 − 3 2 ⋅ 3 2 = − 3 4 + 3 4 − 3 4 − 3 3 4 \dfrac{dy}{dx} = \dfrac{-\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{3}{2} \cdot \frac{1}{2}}{-\frac{\sqrt{3}}{2} \cdot \frac{1}{2} - \frac{3}{2} \cdot \frac{\sqrt{3}}{2}} = \dfrac{-\frac{3}{4} + \frac{3}{4}}{-\frac{\sqrt{3}}{4} - \frac{3\sqrt{3}}{4}} d x d y = − 2 3 ⋅ 2 1 − 2 3 ⋅ 2 3 − 2 3 ⋅ 2 3 + 2 3 ⋅ 2 1 = − 4 3 − 4 3 3 − 4 3 + 4 3
Simplify:
d y d x = 0 \dfrac{dy}{dx} = \boxed{0} d x d y = 0
So the final derivative and tangent equation can be used.
Example 4: Derivative of a Circle
Differentiate the equation of a circle r ( θ ) = 2 cos ( θ ) r(\theta) = 2 \cos(\theta) r ( θ ) = 2 cos ( θ )
Solution:
First, compute r ′ ( θ ) r'(\theta) r ′ ( θ )
r ′ ( θ ) = − 2 sin ( θ ) r'(\theta) = -2 \sin(\theta) r ′ ( θ ) = − 2 sin ( θ )
Now apply the formula for d y d x \dfrac{dy}{dx} d x d y
d y d x = r ′ ( θ ) . sin ( θ ) + r ( θ ) . cos ( θ ) r ′ ( θ ) . cos ( θ ) − r ( θ ) . sin ( θ ) \dfrac{dy}{dx} = \dfrac{r'(\theta). \sin(\theta) + r(\theta). \cos(\theta)}{r'(\theta). \cos(\theta) - r(\theta). \sin(\theta)} d x d y = r ′ ( θ ) . cos ( θ ) − r ( θ ) . sin ( θ ) r ′ ( θ ) . sin ( θ ) + r ( θ ) . cos ( θ )
Substitute r ( θ ) = 2 cos ( θ ) r(\theta) = 2 \cos(\theta) r ( θ ) = 2 cos ( θ ) r ′ ( θ ) = − 2 sin ( θ ) r'(\theta) = -2 \sin(\theta) r ′ ( θ ) = − 2 sin ( θ )
d y d x = ( − 2 sin ( θ ) . sin ( θ ) + 2 cos ( θ ) . cos ( θ ) ) ( − 2 sin ( θ ) . cos ( θ ) − 2 cos ( θ ) . sin ( θ ) ) \dfrac{dy}{dx} = \dfrac{\big(-2 \sin(\theta). \sin(\theta) + 2 \cos(\theta). \cos(\theta)\big)}{\big(-2 \sin(\theta). \cos(\theta) - 2 \cos(\theta). \sin(\theta)\big)} d x d y = ( − 2 sin ( θ ) . cos ( θ ) − 2 cos ( θ ) . sin ( θ ) ) ( − 2 sin ( θ ) . sin ( θ ) + 2 cos ( θ ) . cos ( θ ) )
Simplify:
d y d x = − 2 sin 2 ( θ ) + 2 cos 2 ( θ ) − 4 sin ( θ ) cos ( θ ) \dfrac{dy}{dx} = \boxed{\frac{-2 \sin^2(\theta) + 2 \cos^2(\theta)}{-4 \sin(\theta) \cos(\theta)}} d x d y = − 4 sin ( θ ) cos ( θ ) − 2 sin 2 ( θ ) + 2 cos 2 ( θ )
This is the derivative of the polar equation of the circle .
Conclusion:
The derivatives of polar functions help in calculating the slope of curves in polar coordinates.
These are widely used in physics , especially in orbital mechanics , astronomy , and engineering .
This rule enables the transformation of polar equations into Cartesian derivatives for further analysis.
