Free math worksheets

- The radius gives the distance from the origin and the angle locates the direction. - Think about the horizontal and vertical components of a length \(r\) at angle \(\theta\).

1. Use \(x=r\cos\theta\) and \(y=r\sin\theta\). 2. \(x=6\cos\frac{\pi}{3}=3\) and \(y=6\sin\frac{\pi}{3}=3\sqrt{3}\).

\((3, 3\sqrt{3})\)

- Because the radius is positive, use the stated angle without reversing the direction. - Locate the quadrant containing an angle of \(\frac{5\pi}{6}\).

1. The radius is positive, so the point lies in the direction given directly by \(\theta=\frac{5\pi}{6}\). 2. The angle \(\frac{5\pi}{6}\) is \(150^\circ\), which lies in Quadrant II. 3. On the diagram, the Quadrant II point is labeled \(B\).

Point \(B\)

- The radius is the distance from the point to the origin. - Identify the standard angle for the negative y-axis in the required interval.

1. The point is \(6\) units from the origin, so \(r=6\). 2. It lies on the negative y-axis, whose angle in \([0,2\pi)\) is \(\theta=\frac{3\pi}{2}\).

\(\left(6, \frac{3\pi}{2}\right)\)

- Find the distance from the point to the origin first. - Use the signs of the rectangular coordinates to determine the quadrant before choosing the angle. - Check that the final angle lies in the required interval.

1. The radius is \(r=\sqrt{(-4)^2+(4\sqrt{3})^2}=8\). 2. The point is in Quadrant II. Its reference angle is \(\frac{\pi}{3}\), so \(\theta=\frac{2\pi}{3}\).

\(\left(8, \frac{2\pi}{3}\right)\)

- A negative radius places the point in the direction opposite the stated angle. - Ask which angle points exactly opposite \(\frac{\pi}{3}\). - Keep the final angle in the required interval.

1. Changing a negative radius to a positive radius reverses the direction of the ray, which changes the angle by \(\pi\). 2. \(\frac{\pi}{3}+\pi=\frac{4\pi}{3}\), so an equivalent pair is \(\left(4, \frac{4\pi}{3}\right)\).

\(\left(4, \frac{4\pi}{3}\right)\)

- With the same radius sign, angles differing by a whole revolution describe the same direction. - With the opposite radius sign, the angle must point in the opposite direction. - Check each option as a complete pair rather than comparing only the angles.

1. Adding or subtracting \(2\pi\) from an angle keeps the same point when the radius is unchanged, so A is equivalent to the given pair. 2. Changing the sign of the radius requires changing the direction by \(\pi\). For B, \(-\frac{\pi}{6}+\pi=\frac{5\pi}{6}\), so B is also equivalent. 3. C reverses the radius without changing the angle by \(\pi\), and D uses a different direction with the same positive radius, so neither represents the given point.

A and B

- First locate the direction of the stated angle without using the radius sign. - A negative radius places the point on the ray opposite that direction. - Use the quadrant of the reversed direction to choose among the labeled points.

1. The angle \(\frac{\pi}{6}\) points into Quadrant I, but the radius is negative, so the point lies in the opposite direction. 2. The opposite direction is \(\frac{\pi}{6}+\pi=\frac{7\pi}{6}\), which is in Quadrant III. 3. On the diagram, the Quadrant III point is \(A\).

Point \(A\). A negative radius reverses the direction by \(\pi\), so the point lies along \(\theta=\frac{7\pi}{6}\).

- Use the horizontal and vertical components of the radius. - Determine the signs from the quadrant before evaluating the reference-angle values.

1. Use \(x=r\cos\theta\) and \(y=r\sin\theta\). 2. \(x=5\cos\frac{4\pi}{3}=-\frac{5}{2}\). 3. \(y=5\sin\frac{4\pi}{3}=-\frac{5\sqrt{3}}{2}\).

\(\left(-\frac{5}{2}, -\frac{5\sqrt{3}}{2}\right)\)

- Find the distance from the origin before choosing the angle. - Use the signs of both coordinates to identify the quadrant. - Check that the angle representative satisfies the required interval.

1. \(r=\sqrt{(-5)^2+(-5)^2}=5\sqrt{2}\). 2. The point lies in Quadrant III and has reference angle \(\frac{\pi}{4}\). 3. Therefore, \(\theta=\pi+\frac{\pi}{4}=\frac{5\pi}{4}\).

\(\left(5\sqrt{2}, \frac{5\pi}{4}\right)\)

- Check whether the standard rectangular-polar identity gives the radius itself or its square. - After using the identity, remember what sign a distance from the origin must have.

1. The correct identity is \(r^2=x^2+y^2\), not \(r=x^2+y^2\). 2. For \((-8, 6)\), \(r^2=(-8)^2+6^2=100\). 3. A polar radius used as a distance is nonnegative, so \(r=10\).

The error is omitting the square on \(r\). Since \(r^2=x^2+y^2=100\), \(r=10\).

- Read the rectangular coordinates of \(P\) from the grid before converting. - The positive radius is the distance from the origin to the point. - A negative-radius representation uses the direction opposite the positive-radius representation.

1. The graph places \(P\) at \((-2, 2)\). Its distance from the origin is \(r=\sqrt{(-2)^2+2^2}=2\sqrt{2}\). 2. With positive radius, the point is in Quadrant II on the line \(y=-x\), so \(\theta=\frac{3\pi}{4}\). 3. For a negative radius, use the opposite direction: \(\frac{3\pi}{4}+\pi=\frac{7\pi}{4}\) after reducing modulo \(2\pi\). Thus \(\left(-2\sqrt{2}, \frac{7\pi}{4}\right)\) is equivalent.

a) \(\left(2\sqrt{2}, \frac{3\pi}{4}\right)\) b) \(\left(-2\sqrt{2}, \frac{7\pi}{4}\right)\)

- The two radii and the angle between them form a triangle with the desired distance as its third side. - Find the angular separation before choosing a distance method. - Keep the calculation exact rather than converting the angles to decimals.

1. Both points are \(4\) units from the origin, and the angle between their position rays is \(\frac{5\pi}{6}-\frac{\pi}{6}=\frac{2\pi}{3}\). 2. By the Law of Cosines, \(AB^2=4^2+4^2-2\cdot4\cdot4\cos\frac{2\pi}{3}=48\). 3. Therefore, \(AB=\sqrt{48}=4\sqrt{3}\).

\(4\sqrt{3}\)

- Substitute \(r=0\) into both rectangular conversion formulas. - Ask whether changing only the angle can change either rectangular coordinate when the radius is zero. - Use the distance identity to connect the pole with \(r=0\).

1. When \(r=0\), \(x=0\cdot\cos\theta=0\) and \(y=0\cdot\sin\theta=0\) for every \(\theta\). Thus every pair \((0, \theta)\) represents the origin. 2. For example, \((0, 0)\), \(\left(0, \frac{\pi}{2}\right)\), and \((0, 7\pi)\) are three distinct polar pairs for the origin. 3. The identity \(r^2=x^2+y^2\) gives \(r=0\) at the origin, while both coordinate formulas become zero regardless of \(\theta\). Therefore, no unique angle is determined there.

a) \((0, 0)\) b) For example, \((0, 0)\), \(\left(0, \frac{\pi}{2}\right)\), and \((0, 7\pi)\) c) When \(r=0\), both \(x=r\cos\theta\) and \(y=r\sin\theta\) equal \(0\) for every \(\theta\), so the pole has no unique polar angle.

- Relate the directions used by positive and negative radii for the same point. - Use the positive y-coordinate to restrict the possible positive-radius angle. - Combine the angle relationship with the stated sum only after choosing the correct branch in \([0,2\pi)\).

1. Because \(r>0\) and the point has positive y-coordinate, \(0<\theta<\pi\). 2. Changing the sign of the radius requires reversing the direction by \(\pi\). In the required interval, \(\phi=\theta+\pi\). 3. Use the given sum: \(\theta+(\theta+\pi)=2\pi\). Thus \(2\theta=\pi\), so \(\theta=\frac{\pi}{2}\) and \(\phi=\frac{3\pi}{2}\).

\(\theta=\frac{\pi}{2}\), \(\phi=\frac{3\pi}{2}\)