To find the exact length of the polar curve defined by the equation r = e^(3θ), we can use the formula for the arc length of a polar curve. The arc length L from θ = a to θ = b is given by:
L = \int_a^b \sqrt{\left(\frac{dr}{dθ}\right)^2 + r^2} \, dθ.
In our case, we are tasked with finding the length for the interval 0 ≤ θ ≤ 2π. First, we need to calculate \frac{dr}{dθ} for the given r = e^(3θ):
\frac{dr}{dθ} = 3 e^{3θ}.
Now substituting r and \frac{dr}{dθ} into the arc length formula results in:
L = \int_0^{2π} \sqrt{(3 e^{3θ})^2 + (e^{3θ})^2} \, dθ.
We can simplify this expression:
L = \int_0^{2π} \sqrt{9 e^{6θ} + e^{6θ}} \, dθ = \int_0^{2π} \sqrt{10 e^{6θ}} \, dθ.
This further simplifies to:
L = \int_0^{2π} \sqrt{10} e^{3θ} \, dθ = \sqrt{10} \int_0^{2π} e^{3θ} \, dθ..
Now we can compute the integral:
\int e^{3θ} \, dθ = \frac{1}{3} e^{3θ} + C.
Substituting the limits from 0 to 2π gives us:
L = \sqrt{10} \left[ \frac{1}{3} e^{3(2π)} - \frac{1}{3} e^{3(0)} \right] = \sqrt{10} \left[ \frac{1}{3} e^{6π} - \frac{1}{3} \right].
Finally, collecting everything together, we find:
L = \frac{\sqrt{10}}{3} (e^{6π} - 1).
So, the exact length of the polar curve given by r = e^{3θ} from 0 to 2π is:
L = \frac{\sqrt{10}}{3} (e^{6π} – 1).