Math: Precalculus: Polar Coordinates and Graphs
Convert, interpret, and graph equations in polar form
Math: Precalculus: Polar Coordinates and Graphs
Convert, interpret, and graph equations in polar form
Math - Grade 9-12
- 1
Convert the polar point (6, pi/6) to rectangular coordinates.
Use x = r cos(theta) and y = r sin(theta).
The rectangular coordinates are (3sqrt(3), 3) because x = 6cos(pi/6) = 3sqrt(3) and y = 6sin(pi/6) = 3. - 2
Convert the rectangular point (-3, 3sqrt(3)) to polar coordinates with r > 0 and 0 <= theta < 2pi.
The polar coordinates are (6, 2pi/3) because r = sqrt((-3)^2 + (3sqrt(3))^2) = 6 and the point is in Quadrant II with reference angle pi/3. - 3
Plot the polar point (4, 5pi/6). State the quadrant where the point lies.
Locate the angle first, then move outward from the pole by the radius.
The point lies in Quadrant II because 5pi/6 is between pi/2 and pi. It is 4 units from the pole along that direction. - 4
Give two different polar coordinate pairs that represent the same point as (5, pi/4).
Two equivalent coordinate pairs are (5, 9pi/4) and (-5, 5pi/4). Adding 2pi to the angle keeps the same point, and changing the sign of r moves the point pi radians in the opposite direction. - 5
Convert the polar equation r = 8cos(theta) to rectangular form and describe its graph.
Multiply both sides by r and use r^2 = x^2 + y^2 and x = r cos(theta).
The rectangular form is x^2 + y^2 = 8x, or (x - 4)^2 + y^2 = 16. The graph is a circle with center (4, 0) and radius 4. - 6
Convert the rectangular equation x^2 + y^2 = 6y to a polar equation.
The polar equation is r = 6sin(theta). This comes from r^2 = 6r sin(theta), then dividing by r for points other than the pole. - 7
Identify the graph of r = 3 + 3cos(theta). Name the type of limaçon and state its main direction.
For r = a + bcos(theta), compare a and b and use cosine to determine left or right orientation.
The graph is a cardioid, which is a special limaçon with no inner loop. It opens to the right because the equation uses cos(theta) with a positive coefficient. - 8
For the polar equation r = 2 - 5sin(theta), identify whether the limaçon has an inner loop, a dimple, or no dimple. State its main direction.
The graph has an inner loop because |a| < |b| for r = a + bsin(theta). It is mainly directed downward because the sine term is negative. - 9
Find the maximum value of r for r = 4 + 2cos(theta), and state the angle where it occurs.
The largest value occurs when the cosine term is as large as possible.
The maximum value of r is 6, and it occurs at theta = 0 because cos(theta) has its maximum value 1 at theta = 0. - 10
Determine the symmetry of r = 7sin(2theta). Test for symmetry about the polar axis, the line theta = pi/2, and the pole.
The graph is symmetric about the pole because replacing theta with theta + pi gives the same equation. It is also symmetric about both the polar axis and the line theta = pi/2 for this rose curve. - 11
For the rose curve r = 5cos(3theta), state the number of petals and the length of each petal.
For rose curves, odd n gives n petals and even n gives 2n petals.
The graph has 3 petals because n = 3 is odd in r = a cos(ntheta). Each petal has length 5 because the maximum value of r is 5. - 12
For the rose curve r = 4sin(6theta), state the number of petals and the length of each petal.
The graph has 12 petals because n = 6 is even in r = a sin(ntheta). Each petal has length 4 because the maximum value of r is 4. - 13
Find the slope of the tangent line to r = 2cos(theta) at theta = pi/4. Use dy/dx = (r' sin(theta) + r cos(theta)) divided by (r' cos(theta) - r sin(theta)).
Substitute r, r', sin(theta), and cos(theta) carefully before simplifying.
The slope is -1. For r = 2cos(theta), r' = -2sin(theta). At theta = pi/4, r = sqrt(2) and r' = -sqrt(2), so dy/dx = 0 divided by -2 = 0. Correction: the numerator is r' sin(theta) + r cos(theta) = -1 + 1 = 0, so the slope is 0. - 14
Find the area enclosed by one petal of r = 6sin(3theta).
The area of one petal is 3pi. One petal is traced from theta = 0 to theta = pi/3, so the area is (1/2) integral from 0 to pi/3 of 36sin^2(3theta) dtheta = 18(pi/6) = 3pi. - 15
A polar graph has equation r = 2theta for theta >= 0. Describe the graph and explain how r changes as theta increases.
In an Archimedean spiral, the radius is proportional to the angle.
The graph is an Archimedean spiral. As theta increases, r increases at a constant rate of 2 units per radian, so the spiral moves steadily farther from the pole.