remember that x = rcosθ and y = rsinθ. so rewrite the polar equation as 2 parametric equations x = (1+sinθ)cosθ and y = (1+sinθ)sinθ.
now if you take dy/dθ and divide by dx/dθ the dθ's cancel out and you will get dy/dx which is what we want. so use the product rule on both the x equation and the y equation and you will get dy/dθ = (1+sinθ)cosθ + sinθ(cosθ) and dx/dθ = -(1+sinθ)(sinθ) + cosθ(cosθ).
dy/dx = (1+sinθ)cosθ + sinθ(cosθ) / -(1+sinθ)(sinθ) + cosθ(cosθ) and we set this equal to 0 since we want to find horizontal tangent lines. a fraction equals 0 only when it's numerator is equal to 0 so we can ignore the bottom part and focus on (1+sinθ)cosθ + sinθ(cosθ) = 0. multiply everything out so you get cosθ + 2sinθcosθ = 0. then factor out a cosθ so you have (cosθ)(1 + 2sinθ) = 0 and set each part equal to 0. when cosθ = 0, θ = π/2, 3π/2...
when (1 + 2sinθ) = 0, sinθ = -1/2 and θ = 7π/6, 11π/6...
so horizontal tangent lines will exist where θ = π/2 + πn, 7π/6 + 2πn, 11π/6 + 2πn (where n is an integer).
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r = 1 + sin θ
r ' = cos θ
when tangent is horizontal, slope is zero
cos θ = 0
θ = π/2, 3π/2
The first thing to remember, is that the slope of the line is horizontal when dy/dx = 0, not when dr/dθ = 0.
Remember the identities.
y = rsinθ
x = rcosθ
Solve for dy/dθ and dx/dθ.
dy/dθ = rcosθ + sinθ(dr/dθ)
dx/dθ = -rsinθ + cosθ(dr/dθ)
We also have:
r = 1 + sinθ
dr/dθ = cosθ
Now solve dy/dθ and dx/dθ.
dy/dθ = (1 + sinθ)cosθ + sinθ(cosθ) = cosθ(1 + 2sinθ)
dx/dθ = -(1 + sinθ)sinθ + cosθ(cosθ) = cos²θ - sin²θ - 1
Solve for dy/dx and set equal to zero.
dy/dx = (dy/dθ) / (dx/dθ)
dy/dx = [cosθ(1 + 2sinθ)] / [cos²θ - sin²θ - 1] = 0
cosθ(1 + 2sinθ) = 0
The solutions are:
cosθ = 0
θ = π/2, 3π/2
However, an examination of the graph shows that θ = 3π/2 is an extraneous solution. So the only valid solution is:
θ = π/2
And:
1 + 2sinθ = 0
2sinθ = -1
sinθ = -1/2
θ = 7π/6, 11π/6
So all the solutions on the interval [0, 2π) for which the tangent line is horizontal are:
θ = π/2, 7π/6, 11π/6
remember that x = rcosθ and y = rsinθ. so rewrite the polar equation as 2 parametric equations x = (1+sinθ)cosθ and y = (1+sinθ)sinθ.
now if you take dy/dθ and divide by dx/dθ the dθ's cancel out and you will get dy/dx which is what we want. so use the product rule on both the x equation and the y equation and you will get dy/dθ = (1+sinθ)cosθ + sinθ(cosθ) and dx/dθ = -(1+sinθ)(sinθ) + cosθ(cosθ).
dy/dx = (1+sinθ)cosθ + sinθ(cosθ) / -(1+sinθ)(sinθ) + cosθ(cosθ) and we set this equal to 0 since we want to find horizontal tangent lines. a fraction equals 0 only when it's numerator is equal to 0 so we can ignore the bottom part and focus on (1+sinθ)cosθ + sinθ(cosθ) = 0. multiply everything out so you get cosθ + 2sinθcosθ = 0. then factor out a cosθ so you have (cosθ)(1 + 2sinθ) = 0 and set each part equal to 0. when cosθ = 0, θ = π/2, 3π/2...
when (1 + 2sinθ) = 0, sinθ = -1/2 and θ = 7π/6, 11π/6...
so horizontal tangent lines will exist where θ = π/2 + πn, 7π/6 + 2πn, 11π/6 + 2πn (where n is an integer).
Use the fact that horizontal tangent lines have a gradient of zero
ie. dr/dθ = 0
dr/dθ = cosθ = 0
so θ = 90,270, 450 etc...
or just think of the sine curve shifted up the y axis by 1. You can see that the top & bottom of the curve will be at 90, 270, 450 etc