15. Ans: (A)
The general forms of finding all the polar coordinates are:
1) When r >= 0(meaning positive):(r, θ + 2nπ) where, n = integer
2) When r < 0(meaning negative): (-r, θ + (2n+1)π) where, n = integer
Since r = +1, -1(ordered pair)
θ(given) = [tex] \frac{ \pi }{3} [/tex]
When r = +1(r>0):(1, [tex] \frac{ \pi }{3} [/tex] + 2nπ)
When r = -1(r<0):(-1, [tex] \frac{ \pi }{3} [/tex] + (2n+1)π)
Therefore, the correct option is (A) (1, pi divided by 3 + 2nπ) or (-1, pi divided by 3 + (2n + 1)π)
16. Ans: (A)
In polar coordinates,[tex]r = \sqrt{x^2 + y^2} [/tex]
Since x = 3, y=-3; therefore,
[tex]r = \sqrt{(3)^2 + (-3)^2} = 3 \sqrt{2} [/tex]
To find the angle,
tanθ = y/x = -3/3 = -1
=> θ = -45°
=> θ = -45°+360° = 315° (when [tex]r = 3 \sqrt{2} [/tex])
If r = -r = [tex]3 \sqrt{2} [/tex], then,
θ = -45° + 180° = 135°
Therefore, the correct option is (A) (3 square root of 2 , 315°), (-3 square root of 2 , 135°)
17. Ans: (A)
(Question-17 missing Image is attached below) The general form of the limacon curve is:
r = b + a cosθ
If b < a, the curve would have inner loop.
As you can see in the image attached(labeled Question-17), the limacon curve graph has the inner loop. Therefore, the correct option is (A) r = 2 + 3 cosθ, since b = 2, and a = 3; and the condition b < a (2 < 3) is met.
18. Ans: (B)
Let's find out!
1. If we replace θ with -θ, we would get:
r = 4 - 4*cos(-θ )
Since, cos(-θ) = +cosθ, therefore,
r = 4 - 4*cos(θ)
Same as the original, therefore, graph is symmetric to x-axis.
2. If we replace r with -r, we would get:
-r = 4 - 4*cos(θ )
r = -4 + 4*cos(θ)
NOT same as original, therefore, graph is NOT symmetric to its origin.
3. If we replace θ with -θ and r with -r, we would get:
-r = 4 - 4*cos(-θ )
Since, cos(-θ) = +cosθ, therefore,
r = -4 + 4*cos(θ)
NOT same as original, therefore, graph is NOT symmetric to y-axis.
Ans: The graph is symmetric to: x-axis only!
19. Ans: [tex] \frac{x^2}{588} + \frac{y^2}{3364} = 1[/tex]
Explanation:
As the question suggests that it is a horizontal ellipse, therefore, the equation for the horizontal ellipse is:
[tex] \frac{x^2}{a^2} + \frac{y^2}{b^2} =1[/tex] -- (A)
Since,
x = 21ft,
y = 29ft,
b = 58ft,
[tex]a^2[/tex] = ?
Plug-in the values in equation (A),
(A)=> [tex] \frac{441}{a^2} + \frac{841}{3364} = 1[/tex]
=> [tex]a^2[/tex] = 588
Therefore, the equation becomes,Ans: [tex] \frac{x^2}{588} + \frac{y^2}{3364} = 1[/tex]
20. Ans: x-axis only
Let's find out!
1. If we replace θ with -θ, we would get:
r = 4*cos(-5θ )
Since, cos(-θ) = +cosθ, therefore,
r = +4*cos(5θ) = Same as original
Therefore, graph is symmetric to x-axis.
2. If we replace r with -r, we would get:
-r = 4*cos(5θ )
r = -4*cos(5θ) = Not same
NOT same as original, therefore, graph is NOT symmetric to its origin.
3. If we replace θ with -θ and r with -r, we would get:
-r = 4*cos(-5θ )
Since, cos(-θ) = +cosθ, therefore,
r = -4*cos(5θ) = Not Same
NOT same as original, therefore, graph is NOT symmetric to y-axis.
Ans: The graph is symmetric to: x-axis only!
