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)

I am confused also. That does not look to me like the polar equation for a circle.

It is the equation of a circle.

r = 2sinθ + 4cosθ
r*r = 2rsinθ + 4rcosθ

x^2 + y^2 = 2y + 4x

(x^2 - 4x + 4) + (y^2 - 2y + 1) = 4+1
(x-2)^2 + (y-1)^2 = 5

However, we need to find the area using polar coordinates:

As I noted in my earlier posting, the first quadrant means 0 <= θ <= pi/2

So, we integrate

A = 1/2 Int(r^2 dθ)[0,pi/2]
= 1/2 Int(4sin^2θ + 16sinθ cosθ + 16 cos^2θ dθ)[0,pi/2]

Recalling some trig identities:

sin^2θ + cos^2θ = 1
sin 2θ = 2sinθ cosθ
cos^2θ = (1 + cos2θ)/2

A = 1/2 Int(4 + 8sin2θ + 6(1 + cos2θ) dθ)[0,pi/2]
= 1/2 Int(10 + 8sin2θ + 6cos2θ)[0,pi/2]

= 1/2(10θ - 4cos2θ + 3sin2θ)[0,pi/2]
= (5θ - 2cos2θ + 3/2 sin2θ)[0,pi/2]
= (5*pi/2 - 2(-1) + 0) - (0 - 2(1) + 0)
= 5pi/2 + 4
= 11.854

I don't get any of the given choices. And, my answer agrees with wolfram dot com:

integrate .5*((2 sin theta)+(4 cos theta) )^2 dtheta, theta=0..pi/2

Take off the limits of integration to see that their indefinite integral also agrees with mine.

To find the length in the first quadrant of the circle described by the polar equation r = (2 sin θ) + (4 cos θ), we need to convert the equation into Cartesian coordinates and then find the arc length.

Let's start by converting the polar equation into Cartesian coordinates. We know that r = √(x^2 + y^2) and x = r cos θ, y = r sin θ. By substituting these values into the given polar equation, we can solve for x and y.

r = (2 sin θ) + (4 cos θ)
√(x^2 + y^2) = (2y) + (4x)
(x^2 + y^2) = 4y + 4x^2

Simplifying this equation, we have:
x^2 + y^2 - 4x^2 - 4y = 0
-3x^2 - 4y + y^2 = 0

Now, we can rewrite this equation in standard form:
-3x^2 - 4y + y^2 = 0
(y^2 - 4y) - 3x^2 = 0
(y^2 - 4y + 4) - 3x^2 = 4
(y - 2)^2 - 3x^2 = 4

It's clear that this equation represents an ellipse. To find the length in the first quadrant, we need to find the arc length of the ellipse within the range of 0 ≤ θ ≤ π/2.

To find the arc length, we can use the formula:

L = ∫ sqrt(1 + (dy/dx)^2) dx,

where dy/dx represents the derivative of the curve.

Taking the derivative of the ellipse equation, we get:
dy/dx = 3x / (y - 2)

Now, plugging in this derivative to the arc length formula, we have:
L = ∫ sqrt(1 + ((3x / (y - 2))^2)) dx.

To evaluate this integral, we need to express it in terms of x only. We can rewrite y in terms of x using the equation of the ellipse:

(y - 2)^2 - 3x^2 = 4
(y - 2)^2 = 4 + 3x^2
y - 2 = √(4 + 3x^2)
y = √(4 + 3x^2) + 2.

Substituting this expression into the arc length formula, we get:

L = ∫ sqrt(1 + ((3x / (√(4 + 3x^2) + 2))^2)) dx.

To find the exact value of the arc length, we can evaluate this integral. However, it is quite complicated to integrate analytically. We can use numerical methods such as numerical integration or approximation techniques like Riemann sums to obtain an approximate value for the arc length.

Given the answer choices, it seems that the best approach is to rely on a graphing calculator or software to plot the polar equation and calculate the length of the curve within the first quadrant. This will provide a more accurate and efficient solution, as it avoids the complexity of the integral calculation.

Therefore, the best way to find the length in the first quadrant of the circle described by the polar equation r = (2 sin θ) + (4 cos θ) would be to use a graphing calculator or software to plot the curve and calculate the arc length.