the spiraling polar curve
r=5e^20
from 0 to 2pi
how do you find the length?
If you call a length element on the curve ds, then we have:
(ds)^2 = (dr)^2 + r^2 (dtheta)^2 (1)
dr is the component of the length element in the radial direction and r d theta is the component in the direction orthogonal to that. So, by Pythagoras' therem you have to ad up the squares of both to find the squared length element.
From (1) it follws that:
ds = sqrt[(dr)^2 + r^2 (dtheta)^2] =
sqrt[r^2 + (dr/dtheta)^2] dtheta
If you integrate this from theta = 0 to 2 pi, you get the curve length:
s = Integral from zero to 2 pi of
sqrt[r^2 + (dr/dtheta)^2] dtheta
To find the length of a curve given in polar coordinates, you can use the arc length formula for polar curves. The formula is:
L = ∫[a, b] √(r² + (dr/dθ)²) dθ
In this case, the given polar curve is r = 5e^20, and we need to find the length from θ = 0 to θ = 2π.
First, let's find dr/dθ by taking the derivative of r with respect to θ:
dr/dθ = d/dθ (5e^20) = 0 (since e^20 is a constant)
Now, we can plug the values into the arc length formula:
L = ∫[0, 2π] √(r² + (dr/dθ)²) dθ
= ∫[0, 2π] √((5e^20)² + 0²) dθ
= ∫[0, 2π] √(25e^40) dθ
= ∫[0, 2π] 5e^20 dθ (taking the square root of (25e^40) gives us 5e^20)
Integrating 5e^20 with respect to θ gives:
L = 5e^20θ |[0, 2π]
= 5e^20(2π - 0)
= 10πe^20
Therefore, the length of the spiraling polar curve r = 5e^20 from θ = 0 to θ = 2π is 10πe^20.