Question

A rotating water sprinkler makes one revolution every 15 s. The water reaches a distance of \(5 \mathrm{m}\) from the sprinkler. a) What is the arc length of the sector watered when the sprinkler rotates through \(\frac{5 \pi}{3} ?\) Give your answer as both an exact value and an approximate measure, to the nearest hundredth. b) Show how you could find the area of the sector watered in part a). c) What angle does the sprinkler rotate through in 2 min? Express your answer in radians and degrees.

Short answer

a) \(\frac{25\pi}{3} \approx 26.18\text{ m}\. b) \(\frac{125\pi}{6} \approx 65.45 \text{m}^2.\ c) \(16\pi\text{ radians} = 2880^\circ.\

Step-by-step solution

- Calculate the Angular Speed

Find the angular speed of the sprinkler. Given it makes one full revolution in 15 s, an entire revolution corresponds to an angle of \[2\pi \text{ radians.}\]Thus, the angular speed \(\omega\) is \(\frac{2\pi}{15}\) radians per second.

- Determine the Arc Length for Angle \(\frac{5\pi}{3} \)- Part (a)

Using the formula for arc length, \( L = r \cdot \theta \), where \( r = 5 \mathrm{m} \) and \( \theta = \frac{5\pi}{3} \). Therefore, \[ L = 5 \mathrm{m} \cdot \frac{5\pi}{3} = \frac{25\pi}{3} \mathrm{m}.\] The approximate measure is \[ L \approx 26.18 \mathrm{m}.\]

- Calculate the Area of the Sector - Part (b)

The area of the sector can be found using \( A = \frac{1}{2} r^2 \theta \), where \( r = 5 \mathrm{m} \) and \( \theta = \frac{5\pi}{3} \). Therefore, \[ A = \frac{1}{2} (5 \mathrm{m})^2 \cdot \frac{5\pi}{3} = \frac{25}{2} \cdot \frac{5\pi}{3} = \frac{125\pi}{6} \mathrm{m}^2.\] The approximate measure of the area is \[ A \approx 65.45 \mathrm{m}^2.\]

- Calculate the Angle Rotation in 2 min - Part (c)

Convert 2 minutes to seconds: \[2 \text{ min} = 120 \text{ s}.\] Using the angular speed calculated in Step 1 (\(\omega = \frac{2\pi}{15}\) radians per second), the angle rotated through in 120 s is \(\theta = \omega \cdot t = \frac{2\pi}{15} \cdot 120 = 16\pi \text{ radians}.\) Convert this to degrees: \[16\pi \times \frac{180}{\pi} = 2880^\circ.\]

Key concepts

Angular Speed

Angular speed is the rate at which an object rotates or revolves around a point, often measured in radians per second. For our water sprinkler, which completes one full revolution every 15 seconds, we find the angular speed as follows: A full revolution is equivalent to an angle of \(2\pi\) radians. Therefore, the angular speed \(\omega\) is calculated by dividing the total radians by the time for one revolution. This gives us \(\omega = \frac{2\pi}{15}\) radians per second.

Arc Length

Arc length is the distance along the curved path of a circle between two points. To determine it, we use the formula \(L = r \cdot \theta\). In this exercise, our water sprinkler sprays water up to a distance of 5 meters, making \(r = 5\) meters. When the sprinkler rotates through an angle of \(\frac{5\pi}{3}\), the arc length is calculated as follows: \(L = 5 \cdot \frac{5\pi}{3} = \frac{25\pi}{3} \) meters, which is approximately 26.18 meters. This tells us how far the water sprays along the path.

Area of Sector

The area of a sector is the region enclosed by two radii and the arc they form. To find it, we use the formula \(A = \frac{1}{2} r^2 \theta\). Here, the radius \(r = 5\) meters and the angle \(\theta = \frac{5\pi}{3}\). The calculation is: \(A = \frac{1}{2} \cdot 5^2 \cdot \frac{5\pi}{3} = \frac{125\pi}{6}\) square meters, which is approximately 65.45 square meters. This tells us the area watered by the sprinkler.

Radian Measure

Radians are a way of measuring angles based on the radius of a circle. One complete revolution (360 degrees) is equivalent to \(2\pi\) radians. This measure simplifies many calculations in trigonometry and calculus. For instance, when determining the angle rotated in 2 minutes by the sprinkler, we first convert 2 minutes to seconds (120 seconds). Using the angular speed from earlier (\(\omega = \frac{2\pi}{15}\) radians per second), we find the angle \(\theta = \frac{2\pi}{15} \times 120 = 16\pi\) radians, which is equivalent to 2880 degrees. This shows the sprinkler's rotation over the given time span.