The equation can be simplified to
r = 1 + sin(2theta)r=1+sin(2θ)
using trigonometric identities.
The polar plot is shown here
![enter image source here]()
http://www.wolframalpha.com/input/?i=r%3D1%2Bsin%282theta%29
To find the tangent line, you need to know a point it passes though and its gradient, frac{dy}{dx}dydx
We know that at theta=-{3pi}/8θ=−3π8,
x=rcosthetax=rcosθ
= (1+sin(2(-{3pi}/8)))cos(-{3pi}/8)=(1+sin(2(−3π8)))cos(−3π8)
= (2-sqrt2)^{3/2}/4=(2−√2)324
~~ 0.112≈0.112
y=rsinthetay=rsinθ
= (1+sin(2(-{3pi}/8)))sin(-{3pi}/8)=(1+sin(2(−3π8)))sin(−3π8)
= -1/2*sqrt{{2-sqrt2}/2}=−12⋅√2−√22
~~ -0.271≈−0.271
The tangent line passes through (0.112,-0.271)(0.112,−0.271).
We also know that at theta=-{3pi}/8θ=−3π8,
frac{dx}{d theta} = frac{d}{d theta}(r sintheta)dxdθ=ddθ(rsinθ)
= frac{d}{d theta}((1+sin(2theta))*cos(theta))=ddθ((1+sin(2θ))⋅cos(θ))
= frac{-2sin(theta)+cos(theta)+3cos(3theta)}{2}=−2sin(θ)+cos(θ)+3cos(3θ)2
Substituting theta=-{3pi}/8θ=−3π8, we have
frac{dx}{d theta}_{theta=-{3pi}/8} = -1/2*sqrt{{2-sqrt2}/2}dxdθθ=−3π8=−12⋅√2−√22
~~ -0.271≈−0.271
Similarly,
frac{dy}{d theta} = frac{d}{d theta}(r sintheta)dydθ=ddθ(rsinθ)
= frac{d}{d theta}((1+sin(2theta))*sin(theta))=ddθ((1+sin(2θ))⋅sin(θ))
= 2sin(theta)cos(2theta) + (sin(2theta)+1)*cos(theta)=2sin(θ)cos(2θ)+(sin(2θ)+1)⋅cos(θ)
Substituting theta=-{3pi}/8θ=−3π8, we have
frac{dy}{d theta}_{theta=-{3pi}/8} = 1/2*sqrt{frac{26-7sqrt{2}}{2}}dydθθ=−3π8=12⋅√26−7√22
~~ 1.42≈1.42
From the chain rule, we can calculate the gradient as
frac{dy}{dx} = frac{ frac{dy}{d theta} }{ frac{dx}{d theta} }dydx=dydθdxdθ
= -sqrt{19+6sqrt{2}}=−√19+6√2
~~ 5.24≈5.24
Equation of tangent line:
frac{y-(-1/2*sqrt{{2-sqrt2}/2})}{x-(2-sqrt2)^{3/2}/4}=-sqrt{19+6sqrt{2}}y−(−12⋅√2−√22)x−(2−√2)324=−√19+6√2
