Question

Determine whether the following statements are true and give an explanation or counterexample. a. The vectors \(\mathbf{r}(t)\) and \(\mathbf{r}^{\prime}(t)\) are parallel for all values of \(t\) in the domain. b. The curve described by the function \(\mathbf{r}(t)=\left\langle t, t^{2}-2 t, \cos \pi t\right\rangle\) is smooth, for \(-\infty

Short answer

Is the curve described by this vector function smooth for all t in \(-\infty<t<\infty\)? If \(f, g,\) and \(h\) are odd integrable functions and \(a\) is a real number, is it true that \(\int_{-a}^{a}(f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k})dt = \mathbf{0}\)?

Step-by-step solution

Definition of parallel vectors

To check if two vectors are parallel, we need to verify if one can be expressed as a scalar multiple of the other, i.e., if \(\mathbf{r}^{\prime}(t)=\lambda(t)\mathbf{r}(t)\). For this, let's first obtain the derivative of \(\mathbf{r}(t)\).

Obtaining the derivative of \(\mathbf{r}(t)\)

To find \(\mathbf{r}^{\prime}(t)\), we differentiate each component of \(\mathbf{r}(t)\) with respect to t. If \(\mathbf{r}(t)=\langle x(t),y(t),z(t) \rangle\), then \(\mathbf{r}^{\prime}(t)=\langle x'(t),y'(t),z'(t) \rangle\).

Checking if the derivative is a scalar multiple of the original

Compare \(\mathbf{r}^{\prime}(t)\) with \(\mathbf{r}(t)\) and check if there is any scalar function \(\lambda(t)\) that satisfies the parallel condition. If no such scalar function exists, we conclude that the statement is false and provide a counterexample of non-parallel vectors. #b. Analyzing the smoothness of a curve#

Definition of a smooth curve

A curve is considered smooth, if its tangent vector, i.e., the derivative \(\mathbf{r}^{\prime}(t)\), exists and it is continuous for all t in the interval. Since we already found the derivative in the previous part, let's analyze its continuity.

Checking the continuity of the derivative

Analyze the continuity of each component of \(\mathbf{r}^{\prime}(t)\) in the given domain. If it is continuous for all t in \((-\infty, +\infty)\), we can conclude the curve is smooth across this interval. #c. Odd integrable functions and vector integral evaluation#

Definition of odd functions and their properties

A function is considered odd if it satisfies the condition \(f(-x)=-f(x)\). For integrable odd functions \(f\), an important property states that \(\int_{-a}^{a} f(t) dt = 0\). We can use this property to analyze the given integral of the vector function.

Applying the property of odd function to the vector

We know that \(f, g,\) and \(h\) are odd functions, and their integral property holds true. So, we can now evaluate each of the scalar components in the given integral separately:$$\int_{-a}^{a} f(t) dt = 0,$$$$ \int_{-a}^{a} g(t) dt = 0,$$ and $$\int_{-a}^{a} h(t) dt = 0.$$

Combining the scalar components to form the zero vector

Now that we have evaluated each of the scalar integrals individually, we can combine them to obtain the resulting vector: $$\int_{-a}^{a}(f(t) \mathbf{i}+g(t) \mathbf{j}+h(t) \mathbf{k}) dt = \mathbf{0}.$$This confirms the given statement as true.

Key concepts

Parallel Vectors

In vector calculus, parallel vectors are defined by their ability to be expressed as scalar multiples of each other. When we have two vectors, such as \( \mathbf{r}(t) \) and \( \mathbf{r}^{\textprime}(t) \), we are looking to find a scalar function \( \lambda(t) \) where \( \mathbf{r}^{\textprime}(t) = \lambda(t)\mathbf{r}(t) \). If such a function exists, it means that at every point of their domain, the vectors are stretching or shrinking along the same line of action, never deviating from their parallel orientation.
Parallel vectors have important applications in physics and engineering, such as representing forces that have the same or opposite directions but may differ in magnitude. Understanding this concept is crucial when analyzing motion along a path or the equilibrium of forces.

Smooth Curve

A smooth curve in mathematics is a curve that is not only differentiable but also has a continuous derivative throughout its domain. In other words, a smooth curve has no sharp corners or cusps and the rate at which the curve is changing, represented by its tangent vector \( \mathbf{r}^{\textprime}(t) \), does not jump or break at any point.
The smoothness of a curve is an essential aspect of calculus because it ensures that all the powerful tools of calculus, such as integration and Taylor series expansion, can be applied. For example, analyzing the curvature or computing the arc length requires the curve to be smooth. When evaluating the smoothness of the curve \( \mathbf{r}(t) \) across the domain, continuity of its derivative is key in making that determination.

Odd Integrable Functions

In the realm of integrable functions, functions can be categorized based on symmetry as either odd or even. Odd integrable functions have a specific property that the function value at \( -x \) is the negative of the value at \( x \) (meaning \( f(-x) = -f(x) \)). A key consequence of this symmetry is integral cancellation over symmetric intervals centered at the origin.
For instance, if you were to evaluate the integral of an odd function over the interval \( [-a, a] \) where \( a \) is a positive real number, the result would always be zero. This property not only simplifies the integration process but also becomes a foundational concept in applications such as Fourier transforms, where odd functions help in breaking down complex signals into easier-to-understand parts.

Vector Derivatives

The concept of vector derivatives plays a crucial role in vector calculus, particularly in understanding how vector quantities change with respect to a parameter, like time. When we say \( \mathbf{r}^{\textprime}(t) \) is the derivative of \( \mathbf{r}(t) \), we are essentially taking the derivative of each component of the vector with respect to \( t \) separately.
This process results in a new vector, composed of the derivatives of the original vector's components. These derivatives give us the rates at which the components of the vector are changing. In physical terms, if \( \mathbf{r}(t) \) represents a position vector, then \( \mathbf{r}^{\textprime}(t) \) would represent the velocity vector, showing how the position is changing over time. Understanding vector derivatives is fundamental for exploring dynamics in physics, aspects of fluid flow, or gradient fields in multi-variable calculus.

Scalar Multiple

The concept of a scalar multiple in vector calculus is an operation that involves multiplying a vector by a scalar (a real number), which changes the magnitude of the vector but not its direction unless the scalar is negative. If \( \mathbf{v} \) is a vector and \( k \) is a scalar, the scalar multiple is given by \( k\mathbf{v} \), which results in either a stretching or shrinking of the original vector.
A scalar multiple becomes particularly relevant when identifying parallel vectors, as parallel vectors can always be expressed as scalar multiples of one another. Understanding this operation is also essential in areas such as scaling physical quantities, projecting vectors, and in linear transformations within linear algebra. It's worth noting that the concept extends to higher dimensions, maintaining its core characteristic across all vector spaces.