Question

Show that if the vectors \(\mathbf{u}\) and \(\mathbf{v}\) are not both \(\mathbf{0}\) and \(\beta \neq 0\) then the vector functions $$ \mathbf{y}_{1}=e^{\alpha t}(\mathbf{u} \cos \beta t-\mathbf{v} \sin \beta t) \quad \text { and } \quad \mathbf{y}_{2}=e^{\alpha t}(\mathbf{u} \sin \beta t+\mathbf{v} \cos \beta t) $$ are linearly independent on every interval. HINT: There are two cases to consider: (i) \(\\{\mathbf{u}, \mathbf{v}\\}\) linearly independent, and (ii) \(\\{\mathbf{u}, \mathbf{v}\\}\) linearly dependent. In either case, exploit the the linear independence of \(\\{\cos \beta t, \sin \beta t\\}\) on every interval.

Short answer

Show that the given vector functions are linearly independent on every interval when the conditions \(\mathbf{u}\) and \(\mathbf{v}\) are not both \(\mathbf{0}\) and \(\beta \neq 0\) are satisfied. Solution: We consider two cases - when \(\{\mathbf{u}, \mathbf{v}\}\) are linearly independent and when they are linearly dependent. In each case, we use the linear independence of \(\{\cos \beta t, \sin \beta t\}\) to show that \(c_1\mathbf{y}_{1} + c_2\mathbf{y}_{2} = \mathbf{0}\) implies \(c_1=c_2=0\). Thus, the given vector functions \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\) are linearly independent on every interval.

Step-by-step solution

Case 1: \(\{\mathbf{u},\mathbf{v}\}\) are linearly independent

In this case, let's assume that $$c_1\mathbf{y}_{1} + c_2\mathbf{y}_{2} = c_1e^{\alpha t}(\mathbf{u} \cos \beta t-\mathbf{v} \sin \beta t) + c_2e^{\alpha t}(\mathbf{u} \sin \beta t+\mathbf{v} \cos \beta t) = \mathbf{0}$$ Now, let's factor out \(e^{\alpha t}\): $$e^{\alpha t} (c_1(\mathbf{u} \cos \beta t-\mathbf{v} \sin \beta t) + c_2(\mathbf{u} \sin \beta t+\mathbf{v} \cos \beta t)) = \mathbf{0}$$ Since \(e^{\alpha t} \neq 0\) for all \(t\), we have $$c_1(\mathbf{u} \cos \beta t-\mathbf{v} \sin \beta t) + c_2(\mathbf{u} \sin \beta t+\mathbf{v} \cos \beta t) = \mathbf{0}$$ This implies $$\mathbf{u}(c_1\cos \beta t + c_2\sin \beta t) + \mathbf{v}(-c_1\sin \beta t + c_2\cos \beta t) = \mathbf{0}$$ Taking into account that \(\mathbf{u}\) and \(\mathbf{v}\) are linearly independent, we obtain: $$ c_1\cos \beta t + c_2\sin \beta t = 0 \\ -c_1\sin \beta t + c_2\cos \beta t = 0$$ As \(\{\cos \beta t, \sin \beta t\}\) are linearly independent, this implies \(c_1 = c_2 = 0\), which means \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\) are linearly independent in this case.

Case 2: \(\{\mathbf{u},\mathbf{v}\}\) are linearly dependent

In this case, we have \(\mathbf{v} = k\mathbf{u}\) for some non-zero constant \(k\). Let's substitute this into the expressions for \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\): $$ \mathbf{y}_{1}=e^{\alpha t}(\mathbf{u} \cos \beta t-k\mathbf{u} \sin \beta t) \\ \mathbf{y}_{2}=e^{\alpha t}(\mathbf{u} \sin \beta t+k\mathbf{u} \cos \beta t)$$ By factoring out \(e^{\alpha t}\mathbf{u}\), we obtain: $$ \mathbf{y}_{1}=e^{\alpha t}\mathbf{u}(\cos \beta t-k\sin \beta t) \\ \mathbf{y}_{2}=e^{\alpha t}\mathbf{u}(\sin \beta t+k\cos \beta t)$$ Now let's suppose that \(c_1\mathbf{y}_{1} + c_2\mathbf{y}_{2} = \mathbf{0}\): $$ c_1e^{\alpha t}\mathbf{u}(\cos \beta t-k\sin \beta t) + c_2e^{\alpha t}\mathbf{u}(\sin \beta t+k\cos \beta t) = \mathbf{0}$$ Since \(e^{\alpha t}\mathbf{u}\neq0\), we obtain: $$ c_1(\cos \beta t-k\sin \beta t) + c_2(\sin \beta t+k\cos \beta t) = 0$$ As \(\{\cos \beta t, \sin \beta t\}\) are linearly independent, we have \(c_1 = c_2 = 0\), which implies that \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\) are linearly independent in this case as well. In conclusion, both cases show that the given vector functions \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\) are linearly independent on every interval.

Key concepts

Vector Functions

Vector functions are mathematical expressions where each vector component is a function of one or more variables, often time. These functions are useful for describing paths and movements, among other applications.

In this exercise, we explore two vector functions that depend on an exponential factor combined with trigonometric expressions:

• extbf{\(\mathbf{y}_{1}\)} is defined as \(e^{\alpha t}(\mathbf{u} \cos \beta t - \mathbf{v} \sin \beta t)\).
• extbf{\(\mathbf{y}_{2}\)} is defined as \(e^{\alpha t}(\mathbf{u} \sin \beta t + \mathbf{v} \cos \beta t)\).

These functions can represent changes over time, with \(t\) being the variable. The inclusion of vector quantities \(\mathbf{u}\) and \(\mathbf{v}\) allows these functions to encapsulate direction and magnitude.

Linear Dependence

Linear dependence refers to a situation where one vector in a set can be expressed as a linear combination of others. This is significant because it implies that the vectors do not span the entire space. If all vectors in a set are linearly dependent, then they do not form a basis.

In the exercise at hand, we consider the cases where:

• The vectors \(\{\mathbf{u}, \mathbf{v}\}\) are linearly dependent, meaning \(\mathbf{v} = k\mathbf{u}\) with a non-zero constant \(k\).
• The vector functions \(\{\mathbf{y}_{1}, \mathbf{y}_{2}\}\) still remain linearly independent due to the independence of the trigonometric components \(\{\cos \beta t, \sin \beta t\}\).

This highlights a fascinating aspect: linear independence in some components can be strong enough to preserve the overall independence of complex vector functions.

Basis Functions

Basis functions are a set of linearly independent functions that can represent any function in a given space through linear combinations. In trigonometry, \(\cos \beta t\) and \(\sin \beta t\) serve as fundamental basis functions on any interval, due to their ability to describe all periodic functions that share the interval.

In this scenario, these trigonometric functions help maintain the linear independence of \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\), even if \(\mathbf{u}\) and \(\mathbf{v}\) themselves are dependent. They achieve this by ensuring that no non-trivial linear combination results in the zero function, thus preserving independence.

This property is crucial when considering solutions to differential equations and analyzing systems in physics and engineering.

Exponential Functions

Exponential functions are those where a constant base is raised to a variable exponent, appearing frequently in mathematical modeling of natural phenomena. In our exercise, \(e^{\alpha t}\) is the exponential function applied to both vector functions \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\).

The presence of the exponential component impacts the growth or decay of the vectors over time. Despite its influence, the fundamental linear independence still chiefly stems from the trigonometric components.

In many real-world scenarios, exponential functions describe processes such as radioactive decay or population growth, and similarly, they can be applied in complex vector functions to analyze dynamic systems.