Free math worksheets

- In polar coordinates, \(r\) measures distance from the pole. - Ask what set of points has one fixed distance from a single center.

1. The equation \(r=2\) says every point on the graph is exactly \(2\) units from the pole. 2. Therefore, the graph is a circle centered at the origin with radius \(2\).

A circle centered at \((0,0)\) with radius \(2\)

- Multiplying the polar equation by \(r\) can create expressions that have standard Cartesian replacements. - Look for \(r^2\) and \(r\cos\theta\) before converting. - After conversion, rewrite the quadratic equation in a form that displays its geometric features.

1. Multiply by \(r\): \(r^2=4r\cos\theta\). 2. Use \(r^2=x^2+y^2\) and \(r\cos\theta=x\) to get \(x^2+y^2=4x\). 3. Complete the square: \((x-2)^2+y^2=4\). 4. The graph is a circle centered at \((2,0)\) with radius \(2\).

\((x-2)^2+y^2=4\); a circle centered at \((2,0)\) with radius \(2\)

- For an even multiplier of \(\theta\), positive and negative radius values produce distinct petals. - Petal centers occur in directions where the magnitude of \(r\) reaches its maximum. - Account for the fact that a negative radius points in the opposite direction.

1. For a rose \(r=a\sin(n\theta)\) with even \(n\), the graph has \(2n\) petals. Here \(n=2\), so there are \(4\) petals. 2. Maximum distance occurs when \(|r|=3\). The resulting petal directions are the diagonal directions \(\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\).

a) \(4\) petals b) \(\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\)

- Compare the equation at \(\theta\) and \(-\theta\) to test symmetry. - A point is at the pole exactly when its radius is zero. - The farthest point occurs where the radius is as large as possible.

1. Because replacing \(\theta\) by \(-\theta\) leaves \(\cos\theta\) unchanged, the graph is symmetric about the polar axis. 2. The graph reaches the pole when \(r=0\): \(2+2\cos\theta=0\), so \(\theta=\pi\). 3. The largest radius occurs when \(\cos\theta=1\), at \(\theta=0\). Then \(r=4\), giving the point \((4,0)\).

a) The polar axis b) \(\theta=\pi\) c) \((4,0)\)

- Each symmetry has a corresponding angle substitution that can leave a polar equation unchanged. - Simplify the cosine expressions using evenness and periodicity rather than trying to sketch first. - Treat the three symmetry tests independently.

1. For the polar axis, replace \(\theta\) by \(-\theta\): \(3\cos(-2\theta)=3\cos(2\theta)\), so the equation is unchanged. 2. For the line \(\theta=\frac{\pi}{2}\), replace \(\theta\) by \(\pi-\theta\): \(3\cos(2\pi-2\theta)=3\cos(2\theta)\), so the equation is unchanged. 3. For the pole, replace \(\theta\) by \(\theta+\pi\): \(3\cos(2\theta+2\pi)=3\cos(2\theta)\), so the equation is unchanged.

a) Yes b) Yes c) Yes

- Passing through the pole means the radius is zero. - Check the sign of \(r\) for angles between the two zero-radius angles. - For \(x\)-axis intersections, consider the polar-axis directions and remember what a negative radius does.

1. Set \(r=0\): \(1+2\cos\theta=0\), so \(\cos\theta=-\frac{1}{2}\). Thus \(\theta=\frac{2\pi}{3}\) and \(\theta=\frac{4\pi}{3}\). 2. For angles near \(\pi\), \(\cos\theta<-\frac{1}{2}\), so \(r<0\). Those negative-radius values reverse direction and trace the inner loop. 3. At \(\theta=0\), \(r=3\), giving \((3,0)\). At \(\theta=\pi\), \(r=-1\), which gives \((1,0)\). The two zero-radius angles give the pole \((0,0)\).

a) \(\theta=\frac{2\pi}{3}\) and \(\theta=\frac{4\pi}{3}\) b) Between those angles, some radius values are negative, which creates the inner loop. c) \((0,0)\), \((1,0)\), and \((3,0)\)

- Compare the total angular change with the angle in one full revolution. - Substitute the angle reached after one revolution into the radius equation. - The final angle is coterminal with the positive \(x\)-axis.

1. The angle increases from \(0\) to \(4\pi\). Since one revolution is \(2\pi\), the curve makes \(2\) complete revolutions. 2. After the first revolution, \(\theta=2\pi\), so \(r=\frac{2\pi}{\pi}=2\). 3. At the final value \(\theta=4\pi\), \(r=4\). Thus \(x=4\cos4\pi=4\) and \(y=4\sin4\pi=0\).

a) \(2\) b) \(2\) c) \((4,0)\)

- First use the number of petals to narrow the equation family. - Then compare where a maximum radius occurs for sine versus cosine. - A single known petal direction is enough to distinguish the two rose orientations.

1. A four-petal rose is produced by a sine or cosine equation with \(2\theta\), so C and D do not match the graph type. 2. For \(r=2\cos(2\theta)\), \(r=2\) at \(\theta=0\), so one petal is centered on the positive \(x\)-axis; symmetry produces the other petals on the coordinate axes. 3. For \(r=2\sin(2\theta)\), the petals are centered on the diagonal directions instead. Therefore, A matches the graph.

A. \(r=2\cos(2\theta)\)