1. h
2. b
3. e
4. g
5. h
6. c
7. a
8. g
9.b
10. d
1) There are some families of Polar Equations that we can identify by the trig function, the number of terms.
2) So examining that table we can state that
1. r=8-8cos(θ)
Note that this can be written from this pattern:
[tex]r=a\pm b\cos (\theta),a>0,b>0[/tex]So this is a Limaçon, but since a= 8 and b=8 we can see a special case, for Limaçons: a Cardiod
2. r=5sin(4θ)
In this equation, we have a Rose Curve for the value of the coefficient of that sinusoidal function is a constant than the value of the angle.
3. r=9+4cos(θ)
We have here a similar case to the first one, note that in this case, the quantity of "a" is greater than the coefficient "b".
So this is a Dimpled Limaçon
4. r=3
If we convert this to a rectangular form, we'll clearly see the equation of the circle.
[tex]\begin{gathered} x^2+y^2=3^2 \\ x^2+y^2=9 \end{gathered}[/tex]So this is a Circle
5. r=5-3sin(θ)
This is clearly a Limaçon, but this time a> b, so that will make a change in the Limaçon, again we can see a Cardiod
6. θ=2π
This is another case, which we can see better when we convert it to the Rectangular coordinates:
Plotting that we can see a line.
[tex]\begin{gathered} \tan (\theta)=2\pi \\ (\theta)=\tan ^{-1}(2\pi)_{} \end{gathered}[/tex]7. r²=25sin(2θ)
Note that generally, we can write this one as r²=a²sin(2θ) and this is the formula for a Lemniscate
8. r=6sin(θ)
This equation when plotted yields a circle, for whenever we plot r=asin(θ) or r=bcos(θ) we end up tracing Circles
9. r=-3cos(5θ)
Similarly, to the 2nd item, this produces a Rose Curve note that this follows the same pattern.
10. r=2+6sin(θ)
This is also a Limaçon, as in some previous cases but since the coefficient a=2 and b=6 and a Limaçon with a Loop
2) Thus the answer is:
1. h
2. b
3. e
4. g
5. h
6. c
7. a
8. g
9.b
10. d
