1. ## Curve Help (various)

Hi, I was wondering if someone can help me complete this multi-step problem. The parts I've done are in blue. I'll post the whole of it, first (sorry, I don't know how to post an image of the curve):

The curve above is drawn in the xy-plane and is described by the equation in polar coordinates r = θ + sin(2 θ) for 0 ≤ θ ≤ π, where r is measured in meters and θ is measured in radians. The derivative of r with respect to θ is given by dr/dθ = 1 + 2cos(2θ).

a) find the area bounded by the curve and the x-axis.

So, for this, the area formula would be 1/2
∫ r^2 dθ, no? So it would end up being:
1/2 ∫( θ + sin(2 θ) )^2, from 0 to pi, which yields?

b) find the angle θ that corresponds to the point on the curve with x-coordinate -2.

?

I thik it means that since x = r cos θ = f(θ) cos θ , you get this:

-2 =
(θ + sin(2 θ)) cos θ

Is this right? Am I just supposed to solve for
θ?

c) For π/3 < θ < 2π/3, dr/dθ is negative. What does this fact say about r? What does this fact say about the curve?

I am guessing that since dr/d
θ is negative, 0>dr/dθ, which makes r decreasing. And because it is decreasing, the graph gets closer to the origin. Is this right?

d) Find the value of θ in the interval 0 ≤ θ ≤ π/2 that corresponds to the point on the curve in the first quadrant with greatest distance from the origin. Justify your answer.

? I still don't know how to do this.

2. ## Re: Curve Help

a. The integral appears correct. To solve it, we take
$A= \frac{1}{2} \int\limits_{0}^{\pi}(\theta + \sin (2 \theta))^2\, dx$
$= \frac{1}{2} \int\limits_{0}^{\pi}(\theta^2 + 2* \theta \sin (2 \theta) + \sin^2 (2 \theta))\, dx$
Then integrate seperately the three terms of the integral:
i) The antiderivative of $\theta^2$ is $\frac{1}{3} \theta^3$

ii) The antiderivative of $2* \theta \sin (2 \theta)$ can be determined using integration by parts or by consulting a table of integrals; the result is $\frac{1}{2} \sin (2 \theta) - \theta \cos (2 \theta)$

iii) The antiderivative of $\sin^2 (2 \theta)$ can be found using the trig identity $\sin^2 x = \dfrac{1 - \cos (2x)}{2}$; the result here is $\dfrac{\theta}{2} + \dfrac{\sin (4 \theta)}{8}$

Combine these to get the total antiderivative, evaluate at the limits, and don't forget the 1/2 out front of the integral

b. Yes, that is the equation you need to solve for θ. It's not an easy one, though, and looks like it will need to be solved numerically

c. Right on.

d. Remember that the distance from the origin is r, so we're looking for the point in 0 ≤ θ ≤ π/2 where r is maximum. First, we look for a local maximum for r in this range, which will be where $\frac{dr}{d\theta} = 0$.
As $\frac{dr}{d\theta} = 1 + 2 \cos (2 \theta)$, the zero occurs where $\cos (2 \theta) = -\frac{1}{2}$, giving 2θ=2π/3, or θ=π/3. At this value, $r = \dfrac{\pi}{3} + \dfrac{\sqrt{3}}{2}$. We just need to check that r at the endpoints of our interval 0 ≤ θ ≤ π/2 is not larger than this to check if this is the absolute maximum for the interval.

--Kevin C.