Question

Calculate the root-mean-square value of each of the following functions across the interval stated: (a) \(f(t)=t,[0,10]\) (b) \(f(t)=\cos 2 t,[0, \pi]\) (c) \(f(t)=\sin 2 t,[0, \pi]\) (d) \(f(t)=10 \sin 50 \pi t,[0,0.01]\)

Short answer

Question: Determine the root-mean-square (RMS) value of each function on the given interval: (a) \(f(t) = t\) on [0, 10] (b) \(f(t) = \cos(2t)\) on [0, π] (c) \(f(t) = \sin(2t)\) on [0, π] (d) \(f(t) = 10\sin(50πt)\) on [0, 0.01] Answer: (a) RMS = \(\dfrac{10\sqrt{3}}{3}\) (b) RMS = \(\dfrac{\sqrt{2}}{2}\) (c) RMS = \(\dfrac{\sqrt{2}}{2}\) (d) RMS = \(5\sqrt{2}\)

Step-by-step solution

(a) Calculate RMS value for \(f(t) = t\) on \([0, 10]\)#

1. Square the function: \(f^2(t) = t^2\) 2. Find the average value of \(t^2\) over the interval \([0, 10]\): \( = \dfrac{1}{10 - 0} \int_{0}^{10} t^2 dt\) 3. Take the square root of the average value: RMS = \(\sqrt{}\)

Solution for (a)#

1. \(f^2(t) = t^2\) 2. \( = \dfrac{1}{10} \int_{0}^{10} t^2 dt = \dfrac{1}{10}\left[\frac{1}{3}t^3\right]_{0}^{10} = \dfrac{1}{10}\left( \frac{1000}{3} \right) = \dfrac{100}{3}\) 3. RMS = \(\sqrt{} = \sqrt{\dfrac{100}{3}} = \dfrac{10\sqrt{3}}{3}\)

(b) Calculate RMS value for \(f(t) = \cos(2t)\) on \([0, \pi]\)#

1. Square the function: \(f^2(t) = \cos^2(2t)\) 2. Find the average value of \(\cos^2(2t)\) over the interval \([0, \pi]\): \( = \dfrac{1}{\pi - 0} \int_{0}^{\pi} \cos^2(2t) dt\) 3. Take the square root of the average value: RMS = \(\sqrt{}\)

Solution for (b)#

1. \(f^2(t) = \cos^2(2t)\) 2. Replace \(\cos^2(2t)\) with \(\frac{1+\cos(4t)}{2}\) (double angle formula) to simplify the integration: \( = \dfrac{1}{\pi} \int_{0}^{\pi} \frac{1+\cos(4t)}{2} dt = \dfrac{1}{2\pi}\left[t+\frac{\sin(4t)}{4}\right]_0^{\pi} = \dfrac{1}{2}\) 3. RMS = \(\sqrt{} = \sqrt{\dfrac{1}{2}} = \dfrac{\sqrt{2}}{2}\)

(c) Calculate RMS value for \(f(t) = \sin(2t)\) on \([0, \pi]\)#

1. Square the function: \(f^2(t) = \sin^2(2t)\) 2. Find the average value of \(\sin^2(2t)\) over the interval \([0, \pi]\): \( = \dfrac{1}{\pi - 0} \int_{0}^{\pi} \sin^2(2t) dt\) 3. Take the square root of the average value: RMS = \(\sqrt{}\)

Solution for (c)#

1. \(f^2(t) = \sin^2(2t)\) 2. Replace \(\sin^2(2t)\) with \(\frac{1-\cos(4t)}{2}\) (double angle formula) to simplify the integration: \( = \dfrac{1}{\pi} \int_{0}^{\pi} \frac{1-\cos(4t)}{2} dt = \dfrac{1}{2\pi}\left[t-\frac{\sin(4t)}{4}\right]_0^{\pi} = \dfrac{1}{2}\) 3. RMS = \(\sqrt{} = \sqrt{\dfrac{1}{2}} = \dfrac{\sqrt{2}}{2}\)

(d) Calculate RMS value for \(f(t) = 10\sin(50\pi t)\) on \([0, 0.01]\)#

1. Square the function: \(f^2(t) = 100\sin^2(50\pi t)\) 2. Find the average value of \(100\sin^2(50\pi t)\) over the interval \([0, 0.01]\): \( = \dfrac{1}{0.01 - 0} \int_{0}^{0.01} 100\sin^2(50\pi t) dt\) 3. Take the square root of the average value: RMS = \(\sqrt{}\)

Solution for (d)#

1. \(f^2(t) = 100\sin^2(50\pi t)\) 2. Replace \(100\sin^2(50\pi t)\) with \(50(1-\cos(100\pi t))\) (double angle formula) to simplify the integration: \( = \dfrac{1}{0.01}\int_{0}^{0.01} 50(1-\cos(100\pi t)) dt = 50\left[t-\frac{\sin(100\pi t)}{100\pi}\right]_0^{0.01} = 50\) 3. RMS = \(\sqrt{} = \sqrt{50} = 5\sqrt{2}\)

Key concepts

Calculus and the RMS Value

Calculus is the mathematical study of continuous change. It provides us with the tools to calculate the root-mean-square (RMS) value, which is a measure of the magnitude of a varying quantity. It is particularly used in physics and engineering fields to define the effective voltage or current of an alternating current (AC) circuit. Calculus steps in with the operations of differentiation and integration, which help in understanding the behavior of functions over a specific interval.

When we calculate the RMS value of a function, we first square the function, integrate the resulting function over the interval, and then divide by the interval's length to find the average squared value. Finally, we take the square root to find the RMS value. These steps involve the use of derivatives (for squaring the function) and integrals (for finding the average value), showcasing how calculus is essential in computing the RMS value. In our exercise, calculus enabled us to determine the RMS values of various functions over their respective intervals.

Trigonometric Functions and Their Characteristics

Trigonometric functions, such as sine and cosine, are fundamental in mathematics, with applications that span from geometry to periodic phenomena in sciences and engineering. These functions are periodic, meaning they repeat their values in regular intervals, which makes them especially important when dealing with alternating currents and waves in physics.

In the calculation of the RMS value, we encounter trigonometric functions like \(\cos(2t)\) and \(\sin(2t)\). To integrate these functions over a period, we can use trigonometric identities, such as the double-angle formulas: \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\) and \(\sin^2(x) = \frac{1 - \cos(2x)}{2}\). These identities help simplify the squared trigonometric functions, turning what could be a complicated integration process into a more manageable one. Our exercise showcases the use of these identities in the computation of the RMS values for \(\cos(2t)\) and \(\sin(2t)\) across the interval \( [0, \pi] \).

The Definite Integral in Computing Average Values

The definite integral is a concept in calculus that computes the accumulation of quantities, such as areas under curves, and is essential in finding average values of functions. In the context of RMS, once we have squared our function, we need to determine the average value of this squared function over a specified interval. This is done using a definite integral, which integrates the squared function over the interval and then divides by the interval's length.

For example, to calculate the average value of \(f^2(t)\) from our exercises, we used definite integrals such as \( \int_{0}^{10} t^2 dt \) or \( \int_{0}^{\pi} \cos^2(2t) dt \), followed by division by the length of the interval to normalize the value. This process unfolds the mean of the function's squared value, which is then square-rooted to obtain the RMS. The definitive integral functionality is crucial in determining RMS values of different functions, which is exemplified in the step-by-step solutions for each function given in the original exercise.