What is the slope of the tangent line of r=2θ2−3θcos(2θ−π3) at θ=−5π3?

1 Answer
AltairSafir
Apr 9, 2018

Derivative with Polar Coordinates is ∂x∂y=∂r∂θsin(θ)+rcos(θ)∂r∂θcos(θ)−rsin(θ)

Explanation:

∂r∂θ=4θ−3⋅cos(2θ−π3)−3θ⋅(−sin(2θ−π3))⋅2=
=4θ−3⋅cos(2θ−π3)+6θ⋅(sin(2θ−π3))
The numerator:
∂r∂θ⋅sin(θ)=
=((4θ−3⋅cos(2θ−π3)+6θ⋅(sin(2θ−π3)))⋅sin(θ)
r⋅cos(θ)=(2θ2−3θcos(2θ−π3))⋅cos(θ)

The denominator:
∂r∂θ⋅cos(θ)=
4θ−3⋅cos(2θ−π3)+6θ⋅(sin(2θ−π3))⋅cos(θ)
r⋅sin(θ)=(2θ2−3θcos(2θ−π3))⋅sin(θ)
You are calculating the slope at θ=−5π3, insert θ=−5π3 in the formulae above:

The numerator:
∂r∂θ⋅sin(θ)=−42.999
r⋅cos(θ)=31.343

The denominator:
∂r∂θ⋅cos(θ)=−24.825
r⋅sin(θ)=54.287

∂x∂y(θ=−5π3)=
=−42.999+31.343−24.825−54.287=0.15

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