Question

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Answer 1

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Answer 2

15. The same polar coordinate as point P will be arrived if: (1) a full rotation of 2nπ is performed or (2) a rotation of (2n+1)π is performed and r is negated. Among the choices, we can see exactly one option with coordinates that follow these two rules, and it is choice A.

ANSWER: A. (1, pi divided by 3 + 2nπ) or (-1, pi divided by 3 + (2n+1)π)

16. To convert the coordinate (3, -3) to polar, we just let x be rcosθ and y be rsinθ. We can get r by solving $r^{2}=x^{2}+y^{2}$ while we can get θ by solving for $\theta=arctan(\frac{y}{x})$.

$r^{2}=(3)^{2}+(-3)^{2}$
$r=3sqrt(2)$

$\theta=arctan(\frac{-3}{3})=315degrees$

Therefore, one pair of polar coordinates would be (3 square root of 2, 315 degrees) and another one would be (-3 square root of 2, 135 degrees) [note the rule we stated in number 15].

ANSWER: A. (3 square root of 2, 315 degrees), (-3 square root of 2, 135 degrees)

17. Based on your definition that the graph is "circular" with some indentation on the left toward the origin, we can only deduce that the limacon with the form r = a + bcosθ has a value of a that is greater than b. Looking at the choices, our options will be narrowed down to 2 (B and D). Among these, a graph that will look like a circle would be a convex limacon and a limacon of this kind would require that the value of a should be greater than or equal to the value of 2b. Therefore, this will narrow down the answer to option D.

ANSWER: D. r = 4 + cos θ

18. To test for symmetry about the x-axis, we replace the variables r and θ with r and -θ respectively or -r and π-θ. The equation will be symmetric if it will be unchanged (i.e. the same points will still satisfy the new equation).

$r=4-4cos\theta$
$r=4-4cos(-\theta)$
$-r=4-4cos(\pi-\theta)$

If you examine closely, the same set of points will satisfy the equations above. Therefore, the equation is symmetric about the x axis.

For symmetry about the y-axis, we replace r and θ with -r and -θ.

$r=4-4cos\theta$
$-r=4-4cos(-\theta)$

Unfortunately, the equation was changed upon substitution therefore we know that it is not symmetric about the y-axis.

For symmetry about the origin, we just replace r and θ with -r and θ respectively.

$r=4-4cos\theta$
$-r=4-4cos\theta$

These two equations are not also similar so no symmetry about the origin is exhibited. 

ANSWER: B. x-axis only

19. Let's consider the general form of the ellipse: $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$.

From the problem we know that a is equal to 58 since we are given the fact that the height of the tunnel is 58ft. To find b, we use the fact that the point (21,29) is a point on the ellipse as stated in the problem.

$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{58^{2}}=1$
$\frac{21^{2}}{b^{2}}+\frac{29^{2}}{58^{2}}=1$
$\frac{21^{2}}{b^{2}}+0.25=1$
$441=0.75b^{2}$
$b^{2}=588$

ANSWER: $\frac{x^{2}}{588}+\frac{y^{2}}{3364}=1$

20. For this item, we follow similar rules stated in item #18. To test for symmetry, we need to examine if the equation will remain unchanged after performing substitutions. 

$r=4cos5\theta$

x-axis:
$r=4cos(-5\theta)$
$-r=4-4cos[5(\pi-\theta)]$

y-axis:
$-r=4cos(-5\theta)$

origin:
$-r=4cos5\theta$

If you examine the equations, you'll see that it is only symmetric at the x-axis.

ANSWER: The equation is symmetric about the x-axis.