Find the length in the first quadrant of the circle described by the polar equation
r=(2 sin theta)+(4 cos theta)
A. (sqrt2)(pi)
B. (sqrt5)(pi)
C. (2)(pi)
D. (5)(pi)
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To find the length of the circle described by the polar equation r = (2 sin θ) + (4 cos θ) in the first quadrant, we need to find the arc length.
The arc length of a polar curve is given by the formula:
L = ∫[a,b] √[r(θ)² + (dr/dθ)²] dθ
In this case, the polar equation is r = (2 sin θ) + (4 cos θ), and we need to find the arc length in the first quadrant, which means θ ranges from 0 to π/2.
To find the length, we need to evaluate the integral:
L = ∫[0,π/2] √[(2 sin θ + 4 cos θ)² + (d(2 sin θ + 4 cos θ)/dθ)²] dθ
First, we need to find the derivative of (2 sin θ + 4 cos θ) with respect to θ. Using the chain rule, we get:
d(2 sin θ + 4 cos θ)/dθ = 2 cos θ - 4 sin θ
Now, let's substitute the values into the arc length integral:
L = ∫[0,π/2] √[(2 sin θ + 4 cos θ)² + (2 cos θ - 4 sin θ)²] dθ
Simplifying the expression inside the square root:
L = ∫[0,π/2] √[ 4 sin²θ + 8 sin θ cos θ + 16 cos²θ + 4 cos²θ - 2*4sinθcosθ + 16 sin²θ ] dθ
L = ∫[0,π/2] √[ 20 sin²θ + 12 sin θ cos θ + 12 cos²θ ] dθ
L = ∫[0,π/2] √[ 4(5 sin²θ + 3 sin θ cos θ + 3 cos²θ) ] dθ
L = ∫[0,π/2] √[ 4( 5(sin²θ + cos²θ) + 3 sin θ cos θ ) ] dθ
L = ∫[0,π/2] √[ 4(5 + 3 sin θ cos θ ) ] dθ
L = ∫[0,π/2] 2√(5 + 3 sin θ cos θ) dθ
Now, let's solve this integral.