rubiabori1473 rubiabori1473
  • 27-06-2018
  • Mathematics
contestada

Find the slope of the tangent line to the polar curve r=2−sin(θ)r=2−sin⁡(θ) at the point specified by θ=π/3

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Fluxions
Fluxions Fluxions
  • 07-07-2018
First recall that

x = rcosθ  and  y = rsinθ  

Thus

x = (2 - sinθ)(cosθ)  and  y = (2 - sinθ)(sinθ)

Now

dx/d
θ = 2(sinθ)^2 - 2sinθ - 1  and  dy/dθ = 2cosθ - 2sinθcosθ

Now evaluating each derivative above @  θ = π/3,  we obtain

dx/dθ = (1 - 2√3)/2  and  dy/dθ = (2 - √3)/2

Now

dy/dx = (dy/d
θ)(dθ/dx)

Thus @  θ = π/3,

dy/dx = [(2 - 
√3)/2][2/(1 - 2√3)]

dy/dx = (2 - 
√3)/(1 - 2√3)  is the slope of the required tangent

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