Find the length of the spiraling polar curve r = 3e^(4theta)
From 0 to 2pi.
From 0 to 2pi.
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I will give you a preliminary argument first, which is that 3 e^(4 θ) becomes so large for θ = 2π, and does so much of its growth near the end of its path, so that I can roughly approximate it just looking at the start and end points and imagining a straight line between them: 3 e^(8π) − 3. In fact, 3 e^(8π) ≈ 2.5 · 10¹¹ is a huge number in its own right, so we might even ignore this starting point as essentially zero.
Now to give you the mathematics. Anything of this kind is usually an integral of the form:
s = ∫ ds
Arclengths are usually given by the Pythagorean theorem as ds = √(dy² + dx²), but in polar coordinates our orthogonal lengths are not dy and dx, but dr and r dθ. Thus the arclength is instead in polar coordinates:
ds = √(dr² + r² dθ²)
We can pull out the r dθ term from the square root and instead write this as:
ds = r dθ √[1 + u²].
Where u = (1/r) · (dr/dθ).
In your case, dr/dθ = 4 r(θ), by the chain rule, and so u = (1/r) · (dr/dθ) = 4. This means that for this curve, the stuff in the square root simplifies nicely to √(1 + 4²) = √(17). The integral therefore becomes:
∫ ds = ∫ r dθ √(17).
The integral s = ∫ 3 e^(4 θ) √(17) dθ = √(17) ¼ [3 e^(4 θ)]_{θ: 0 → 2π}
s = √(17/16) · [3 e^(8π) − 3]
Thus you see that I was right, we're only off by about 1/32 ≈ 3% using the first approximation. The growth is mostly concentrated in this last turning.
Now to give you the mathematics. Anything of this kind is usually an integral of the form:
s = ∫ ds
Arclengths are usually given by the Pythagorean theorem as ds = √(dy² + dx²), but in polar coordinates our orthogonal lengths are not dy and dx, but dr and r dθ. Thus the arclength is instead in polar coordinates:
ds = √(dr² + r² dθ²)
We can pull out the r dθ term from the square root and instead write this as:
ds = r dθ √[1 + u²].
Where u = (1/r) · (dr/dθ).
In your case, dr/dθ = 4 r(θ), by the chain rule, and so u = (1/r) · (dr/dθ) = 4. This means that for this curve, the stuff in the square root simplifies nicely to √(1 + 4²) = √(17). The integral therefore becomes:
∫ ds = ∫ r dθ √(17).
The integral s = ∫ 3 e^(4 θ) √(17) dθ = √(17) ¼ [3 e^(4 θ)]_{θ: 0 → 2π}
s = √(17/16) · [3 e^(8π) − 3]
Thus you see that I was right, we're only off by about 1/32 ≈ 3% using the first approximation. The growth is mostly concentrated in this last turning.