Question

Rewrite the function \(y(t)\) in the form \(y(t)=R e^{a t} \cos (\beta t-\delta)\), where \(0 \leq \delta<2 \pi\). Use this representation to sketch a graph of the given function, on a domain sufficiently large to display its main features. $$ y(t)=-e^{-t} \cos t+\sqrt{3} e^{-t} \sin t $$

Short answer

$$ R = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2 $$ and $$ \delta = \arctan \frac{\sqrt{3}}{-1} = \arctan (-\sqrt{3}) = \frac{5\pi}{3}. $$ Therefore, we can rewrite the function \(y(t)\) as: $$ y(t) = 2e^{-t}\cos\left(t - \frac{5\pi}{3}\right). $$ #tag_title#Step 2: Sketch the graph of the function#tag_content# To sketch the graph, we first observe the main features of the rewritten function. 1. Amplitude: \(R = 2\) 2. Decay factor: \(a = -1\) 3. Frequency: \(\beta = 1\) 4. Phase shift: \(\delta = \frac{5\pi}{3}\) Now, let's sketch the graph using these features: 1. The amplitude is 2, so the maximum and minimum values of the function will be 2 and -2, respectively, at any point in time. 2. The decay factor is -1, meaning the function will decay exponentially over time as the envelope of the cosine function with a decay factor of \(e^{-t}\). 3. The frequency of oscillation is 1, which means there are no additional oscillations in the function, and it oscillates in its usual sinusoidal form. 4. The phase shift is \(\frac{5\pi}{3}\), which means the function is shifted to the right by \(\frac{5\pi}{3}\) units. Taking all these features into account, we can now sketch the graph of the given function. The graph will look like a damped sinusoidal wave with a decay envelope of an exponential function and a phase shift to the right by \(\frac{5\pi}{3}\) units.

Step-by-step solution

Rewrite the given function in the desired form

To rewrite \(y(t)\) in the form \(y(t)=Re^{at}\cos(\beta t-\delta)\), we will use the identity $$ A\cos(\alpha) + B\sin(\alpha) = R\cos(\alpha - \delta), $$ where \(R = \sqrt{A^2 + B^2}\) and \(\delta = \arctan \frac{B}{A}\). The function we have is: $$ y(t) = -e^{-t}\cos t + \sqrt{3}e^{-t}\sin t $$ Notice that we can write: $$ y(t) = e^{-t}\left(-\cos t + \sqrt{3}\sin t\right) $$ Now, we can apply the trigonometric identity stated above:

Key concepts

Trigonometric Identities

Trigonometric identities are essential tools in mathematics that help simplify and solve trigonometric equations. They allow us to express trigonometric functions in different ways, making them easier to work with in calculus, algebra, and many applied fields. In our problem, we're using the identity:

• \( A\cos(\alpha) + B\sin(\alpha) = R\cos(\alpha - \delta) \),
• where \( R = \sqrt{A^2 + B^2} \) and \( \delta = \arctan \frac{B}{A} \).

These identities allow us to combine sine and cosine terms into a single cosine term with a phase shift \( \delta \) and amplitude \( R \).

In the given function, \( -e^{-t}\cos t + \sqrt{3}e^{-t}\sin t \), we treat \( A = -1 \) and \( B = \sqrt{3} \). Substituting these into the formula gives us:

• \( R = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2 \),
• \( \delta = \arctan \frac{\sqrt{3}}{-1} = \arctan(-\sqrt{3}) \).

This identity simplifies the original equation into a single expression involving \( \cos(\beta t - \delta) \), making it easier to analyze and graph.

Exponential Functions

Exponential functions are a type of mathematical function that grows or decays at rates proportional to their current value. They are commonly written as \( y = ae^{bx} \), where \( e \) is the base of the natural logarithm. In our problem, the functions include terms like \( e^{-t} \), indicating exponential decay.

The entire function \( y(t) = e^{-t}(-\cos t + \sqrt{3}\sin t) \) uses an exponential term \( e^{-t} \) to multiply the trigonometric expression. This means that as time \( t \) increases, the amplitude of the entire function decreases, which is a hallmark of exponential decay.

Understanding the behavior of exponential functions helps us know how quickly the amplitude of our trigonometric function will reduce towards zero over time. In practical terms, this means that the function's oscillations will diminish as \( t \) becomes large, making it an important feature to consider when graphing.

Graphing Functions

Graphing functions involves plotting points that represent the values of a function on a coordinate grid. It provides a visual representation of how a function behaves over a certain domain. For the function \( y(t) = e^{-t}(-\cos t + \sqrt{3}\sin t) \), we want to sketch its graph over a sufficient domain to illustrate its major characteristics.

Since the function includes an exponential decay term \( e^{-t} \), we expect the function's peaks and troughs to diminish as \( t \) increases. The trigonometric components, configured by the identity, indicate that our function oscillates with a period related to \( 2\pi/\beta \), where \( \beta \) is understood in the context of the frequency of oscillation.

• Start by identifying key points: where \( \cos t \) and \( \sin t \) reach their maxima and minima.
• Observe the effect of the exponential term on these points.

This approach ensures a graph that clearly exhibits damped oscillations, an exponential decay pattern, and the influence of the phase shift over the given domain.

Phase Shift

Phase shift refers to the horizontal shifting of a periodic function on a graph. In trigonometric functions such as sine or cosine, it changes the point at which the function starts its cycle. In our expression, using the trigonometric identity, we find \( \delta = \arctan(-\sqrt{3}) \), influencing the phase shift of our function.

This phase shift \( \delta \) adjusts our cosine term \( \cos(\beta t - \delta) \) by shifting it to the right if \( \delta \) is positive, or to the left if \( \delta \) is negative. The impact of the phase shift is that the peaks and troughs of the oscillation change position relative to their usual start and stop points on the time axis.

Understanding the phase shift is important for accurately graphing and predicting values of periodic functions. The shift is vital for precise modeling because it can change how the function appears over any given interval. Consequently, recognizing and computing the phase shift helps in understanding the timing and sequence of function events and behaviors.