(a) Let x=rcos(θ), y=rsin(θ) to convert to polar coordinates.
(2rcos(θ)+1)2+r²sin²(θ)=1, 4r²cos²(θ)+4rcos(θ)+1+r²-r²cos²(θ)=1, 3r²cos²(θ)+4rcos(θ)+r²=0, r(3cos²(θ)+1)=-4cos(θ), r=-4cos(θ)/(3cos²(θ)+1).
The lower half of the ellipse corresponds to 0≤θ≤π/2. Now we have an exact mathematical description of the particle’s path in polar coordinates. This appears to correspond to 0≤t≤2, therefore t=4θ/π, and:
r(t)=-4cosπ(πt/4)/(3cos²(πt/4)+1). However, this gives us r(0)=-1 and r(2)=0, and we want r(0)=0 and r(2)=-1.
Replace t with 2-t:
r(t)=-4cos(π(2-t)/4)/(3cos²(π(2-t)/4)+1), which can also be written:
r(t)=-4cos(π(2-t)/4)/(4-3sin²(π(2-t)/4)); r(0)=0, r(2)=-1.
Arc infinitesimal, ds=rdθ and so ∫ds=∫rdθ gives us the line integral, that is, the integral along the arc path.
x=rcos(π(2-t)/4)=-4cos²(π(2-t)/4)/(3cos²(π(2-t)/4)+1),
y=-4sin(π(2-t)/4)cos(π(2-t)/4)/(3cos²(π(2-t)/4)+1)=-2sin(π(2-t)/2)/(3cos²(π(2-t)/4)+1).
These are the parametric equations for x and y.
[When t=1, x=-⅘, y=-⅘ and it can be seen that these satisfy the given equation of the ellipse:
(-0.6)²+(-0.8)²=1, so (-⅘,-⅘) lies on the ellipse (lower part).]
(b) The arc length for the entire path is given by S=∫ds=∫rdθ, as shown earlier.
r can be written r=-4cos(θ)/(4-3sin²(θ)), so S=-4∫cos(θ)dθ/(4-3sin²(θ)).
Let u=sin(θ), then du=cos(θ)dθ, and S=-4∫du/(4-3u²).
When θ=0, u=0 and when θ=π/2, u=1, so the limits for u are [0,1].
We can write this integral in terms of partial fractions:
S=-∫du(1/(2-u√3)+1/(2+u√3) for 0≤u≤1
S=(1/√3)(ln|2-u√3|-ln|2+u√3|)=(√3/3)(ln|(2-u√3)/(2+u√3)| between the limits [0,1],
S=(√3/3)ln(2-√3)² or (2√3/3)ln(2-√3)=-1.5207 approx. We can lose the negative because we are measuring an arc length. So the arc length is approx 1.52 units.
(c) The area of an infinitesimal sector is ½r²dθ, so the total area of the lower part of the given ellipse is A=½∫r²dθ for 0≤θ≤π/2. Like the arc length this will be a magnitude, that is, the absolute value of the integral. Therefore A=½∫[16cos²(θ)/(4-3sin²(θ))²]dθ=8∫[cos²(θ)/(4-3sin²(θ))²]dθ. However, there seems little point in using this form of the integral to find A, which would involve perhaps multiple substitutions to evaluate it, when there are much simpler (and shorter) methods of finding the required area. The area of an ellipse is πab where a and b are the lengths of the semi-major and semi-minor axes, which we know from the original equation: a=1 (from y-axis) and b=½ (from x-axis), so the area is π/2. The area of the lower half (or the upper half) is therefore π/4 square units.
But, since y=±√(1-(2x+1)² we can use the positive or negative versions to find the required area using A= ∫ydx for -1≤x≤0:
Let 2x+1=sin(t), then 2dx=cos(t)dt, dx=cos(t)dt/2. A=½∫cos²(t)dt=¼∫(cos(2t)+1)dt. When x=-1, sin(t)=-1, t=3π/2, and when x=0, sin(t)=1, t=π/2 so the limits of integration are [π/2,3π/2].
A=¼[sin(2t)/2+t]π/23π/2=¼(sin(3π)+3π/2-sin(π)-π/2)=π/4.
