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 5 answersLast reply by: satwinder kaurMon Jan 4, 2016 5:19 PMPost by adam matthews on December 10, 2012i got dy/dx is -4/sqroot 2 or -2.414, also that is what my graphing calc says{??

### Polar Differentiation

• Solving for slope:
• Convert polar equation to parametric equations
• Differentiate the 2 parametric parts separately
• Divide

### Polar Differentiation

Given r = 2sinθ, what is it's parametric form?
• Apply definitions
• x = rcosθ
• = 2sinθcosθ
• y = rsinθ
• = 2sin2θ
(2sinθcosθ,2sin2θ)
Given (3cosθ+ cos2θ, − sinθcosθ), what is the slope at θ = [(7π)/8]
• Find [dy/dx] by finding [dx/(dθ)] and [dy/(dθ)]
• x = 3cosθ+ cos2θ
• [dx/(dθ)] = − 3sinθ− 2cosθsinθ
• y = − sinθcosθ
• [dy/(dθ)] = sin2θ− cos2θ
• Divide [dy/(dθ)] by [dx/(dθ)]
• [dy/dx] = [(sin2θ− cos2θ)/( − 3sinθ− 2cosθsinθ)]
• Solve for θ = [(7π)/8]
• [(sin2θ− cos2θ)/( − 3sinθ− 2cosθsinθ)] = [(sin2[(7π)/8] − cos2[(7π)/8])/( − 3sin[(7π)/8] − 2cos[(7π)/8]sin[(7π)/8])]
= 0.81
If r = sinθ, what is the y parametric form, and [dy/(dθ)]?
• Apply definition
• y = rsinθ
• = sin2θ
Find the derivative with Power - Reducing identity
sin2θ = [(1 − cos(2θ))/2] = [1/2] − [(cos(2θ))/2][dy/(dθ)] = sin2θ
If r = sinθ, what is the x parametric form, and [dx/(dθ)]?
• Apply definition
• x = rcosθ
• = sinθcosθ
Find the derivative with Product Rule
[dx/(dθ)] = sinθ( − sinθ) + cosθ(cosθ)
If r = sinθ, what is its slope at θ = [( − 5π)/3]
• Divide [dy/(dθ)] by [dx/(dθ)] to find [dy/dx]
• [dy/dx] = [(sin2θ)/( − sin2θ+ cos2θ)]
• Solve for θ = [( − 5π)/3]
• [(sin2θ)/( − sin2θ+ cos2θ)] = [(sin2([( − 5π)/3]))/( − sin2([( − 5π)/3]) + cos2([( − 5π)/3]))]
= − √3
If r = √3 (1 − cos θ), what is the y parametric form, and [dx/(dθ)]?
• Apply definition
• x = rcosθ
• = √3 (1 − cosθ)(cosθ)
• = √3 cosθ− cos2θ
Find the derivative with Product Rule
√3 cosθ− cos2θ = √3 cosθ− [(1 + cos(2θ))/2] = √3 cosθ− [1/2] − [(cos(2θ))/2][dx/(dθ)] = − √3 sinθ+ sin2θ
If r = √3 (1 − cosθ), what is the y parametric form, and [dy/(dθ)]?
• Apply definition
• y = rsinθ
• = √3 (1 − cosθ)(sinθ)
• = √3 sinθ− sinθcosθ
Find the derivative
[dy/(dθ)] = √3 cosθ− cos(2θ)
If r = sinθ, what is its slope at θ = [7p/4]
• Divide [dy/(dθ)] by [dx/(dθ)] to find [dy/dx]
• [dy/dx] = [(√3 cosθ− cos(2θ))/( − √3 sinθ+ sin2θ)]
• Solve for θ = [7p/4]
• [(√3 cosθ− cos(2θ))/( − √3 sinθ+ sin2θ)] = [(√3 cos( [7p/4] ) − cos((2)( [7p/4] )))/( − √3 sin( [7p/4] ) + sin((2)( [7p/4] )))]
• = [(√6 )/(√6 − 2)]
= 5.45
If r = eθ cos(θ), what is [dy/dx]?
• Find f′(θ)
• f(θ) = eθ cos(θ)
• f′(θ) = − eθ sinθ+ eθ cosθ
• = eθ (cosθ− sinθ)
• Apply definitions
• x = rcosθ
• = ecosθcosθ
• = ecos2θ
• y = rsinθ
• = ecosθsinθ
• Find [dy/(dθ)]
• [dy/(dθ)] = f(θ)( − sinθ) + f′(θ)(cosθ)
• = − eθ cosθsinθ+ ( − eθ sinθ+ eθ cosθ)(cosθ)
• = − eθ cosθsinθ− eθ sinθcosθ+ eθ cos2θ
• Find [dx/(dθ)]
• [dx/(dθ)] = f(θ)(cosθ) + f′(θ)( − sinθ)
• = eθ cosθsinθ(cosθ) − ( − eθ sinθ+ eθ cosθ)(sinθ)
• = eθ cos2θsinθ+ eθ sin2θ+ eθ cosθsinθ
• Divide [dy/(dθ)] by [dx/(dθ)] to find [dy/dx]
• [dy/dx] = [( − eθ cosθsinθ− eθ sinθcosθ+ eθ cos2θ)/(eθ cos2θsinθ+ eθ sin2θ+ eθ cosθsinθ)]
= [( − cosθsinθ− sinθcosθ+ cos2θ)/(cos2θsinθ+ sin2θ+ cosθsinθ)]
If r = eθ cos(θ), what is its slope at θ = [(π)/2]?
• Divide [dy/(dθ)] by [dx/(dθ)] to find [dy/dx]
• [( − cosθsinθ− sinθcosθ+ cos2θ)/(cos2θsinθ+ sin2θ+ cosθsinθ)] = [( − cos[(π)/2]sin[(π)/2] − sin[(π)/2]cos[(π)/2] + cos2[(π)/2])/(cos2[(π)/2]sin[(π)/2] + sin2[(π)/2] + cos[(π)/2]sin[(π)/2])]
• = [( − 0 − 0 + 0)/(0 + 1 + 0)]
= 0

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.