Question

"Find the arc length of the following functions on the indicated

intervals:
(a) \( r=6 \cos \theta \) on on \( 0 \leq \theta \leq \frac{\pi}{2} \) (b) \( r=e^{\theta} \) on \( 0 \leq \theta \leq 2 \)"(a) r=6cosθ on on 0≤θ≤ 2
π
​
(b) r=e θ
on 0≤θ≤2

Answer 1

(a) r = 6 * cos(θ) on 0 ≤ θ ≤ π/2

To find the length of a curve in polar coordinates, we use the formula:

L = ∫ sqrt(r² + (dr/dθ)²) dθ

where r is the function in terms of θ, and dr/dθ is the derivative of r with respect to θ.

For the function r = 6 cos(θ), the derivative dr/dθ = -6 * sin(θ).

Now, let's substitute r and dr/dθ into the length formula and evaluate it over the interval from 0 to π/2.

L = ∫ sqrt[(6 cos(θ))² + (-6 sin(θ))²)] dθ from 0 to π/2
= ∫ sqrt[36 cos^2(θ) + 36 sin^2(θ)] dθ from 0 to π/2
= ∫ sqrt[36(cos^2(θ) + sin^2(θ))] dθ from 0 to π/2
= ∫ sqrt[36*1] dθ from 0 to π/2
= ∫ 6 dθ from 0 to π/2
= 6 θ | from 0 to π/2
= 6*(π/2) - 6*0
= 3π (Answer for part a)

(b) r = e^θ on 0 ≤ θ ≤ 2

Similarly, for the function r = e^θ, the derivative dr/dθ = e^θ.

Now, let's substitute r and dr/dθ into the length formula and evaluate it over the interval from 0 to 2.

L = ∫ sqrt[(e^θ)^2 + (e^θ)^2)] dθ from 0 to 2
= ∫ sqrt[2e^2θ] dθ from 0 to 2
= ∫ √2 * e^θ dθ from 0 to 2
= √2 * e^θ | from 0 to 2
= √2 * e^2 - √2 * e^0
= √2 * e^2 - √2
= √2e^2 - √2 (Answer for part b)

In conclusion, the length of arc for the function r = 6cos(θ) from 0 to π/2 is 3π and for the function r = e^θ from 0 to 2 is √2e^2 - √2.