Question

Explain why or why not 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

Answer: No, this statement is generally false. A counterexample can be given for any non-straight curve, such as the vector function \(\mathbf{r}(t)=\left\langle t, t^{2}, 0 \right\rangle\), which has a derivative \(\mathbf{r}^{\prime}(t)=\left\langle 1, 2t, 0 \right\rangle\). At \(t=0\), \(\mathbf{r}(0)=\left\langle 0, 0, 0 \right\rangle\) and \(\mathbf{r}^{\prime}(0)=\left\langle 1, 0, 0 \right\rangle\). These two vectors are not parallel as they are not in the same direction.

Step-by-step solution

a. Determine if \(\mathbf{r}(t)\) and \(\mathbf{r}^{\prime}(t)\) are parallel for all values of \(t\) in the domain.

This statement is generally false. A counterexample can be given for any non-straight curve. Let's consider the vector function \(\mathbf{r}(t)=\left\langle t, t^{2}, 0 \right\rangle\), which has derivative \(\mathbf{r}^{\prime}(t)=\left\langle 1, 2t, 0 \right\rangle\). At \(t=0\), \(\mathbf{r}(0)=\left\langle 0, 0, 0 \right\rangle\) and \(\mathbf{r}^{\prime}(0)=\left\langle 1, 0, 0 \right\rangle\). These two vectors are not parallel as they are not in the same direction.

b. Determine if the curve described by \(\mathbf{r}(t)=\left\langle t, t^{2}-2t, \cos \pi t\right\rangle\) is smooth for \(-\infty

A curve is smooth if its derivative exists and has no discontinuities. Let's find the derivative \(\mathbf{r}^{\prime}(t)\):$$\mathbf{r}^{\prime}(t)=\left\langle 1, 2t-2, -\pi\sin(\pi t)\right\rangle.$$There are no discontinuities in this derivative for any value of \(t\) in the domain \((-\infty,\infty)\). Therefore, the curve described by \(\mathbf{r}(t)\) is smooth for \(-\infty<t<\infty\).

c. Determine if \(\int_{-a}^{a}(f(t) \mathbf{i}+g(t) \mathbf{j}+h(t) \mathbf{k}) dt=\mathbf{0}\) if \(f, g,\) and \(h\) are odd integrable functions and \(a\) is a real number.

Consider the integral $$\int_{-a}^{a}(f(t) \mathbf{i}+g(t) \mathbf{j}+h(t) \mathbf{k}) dt.$$We can break this integral into three separate integrals:$$(\int_{-a}^{a}f(t)dt)\mathbf{i}+(\int_{-a}^{a}g(t)dt)\mathbf{j}+(\int_{-a}^{a}h(t)dt)\mathbf{k}.$$Since \(f, g,\) and \(h\) are odd functions, by the property of odd functions, their integrals from \(-a\) to \(a\) are zero:$$\int_{-a}^{a}f(t)dt=0, \quad \int_{-a}^{a}g(t)dt=0, \quad \int_{-a}^{a}h(t)dt=0.$$Thus,$$(\int_{-a}^{a}(f(t) \mathbf{i}+g(t) \mathbf{j}+h(t) \mathbf{k}) dt)=0\mathbf{i}+0\mathbf{j}+0\mathbf{k}=\mathbf{0}.$$Therefore, the statement is true.

Key concepts

Vectors and Vector-Valued Functions

Vector-valued functions are crucial in describing physical phenomena such as force, velocity, and acceleration, where both magnitude and direction are important. Unlike ordinary functions that map to real numbers, vector-valued functions output vectors.

Consider the vector-valued function \( \mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle \), where \( f(t) \), \( g(t) \), and \( h(t) \) are scalar functions of \( t \). These functions give the components of the vector in each dimension. For instance, \( f(t) \) could represent the x-coordinate, \( g(t) \) the y-coordinate, and \( h(t) \) the z-coordinate of a point in space as a function of time.

In calculus, we often explore how these vectors change as the parameter \( t \) varies, which leads to the concept of the derivative of the vector-valued function, denoted \( \mathbf{r}^{\textprime}(t) \). These derivatives themselves are vector functions that can reveal instantaneous rates of change like velocity or acceleration in physics.

Smooth Curves in Calculus

A curve is said to be smooth if it is continuously differentiable. This means the curve has a tangent at every point without any abrupt changes in direction, which is represented mathematically by a derivative that is continuous everywhere on the curve's domain.

From the calculus standpoint, the smoothness of a curve described by a vector-valued function like \( \mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle \) depends on the derivatives of the component functions. If each of \( f^{\textprime}(t) \), \( g^{\textprime}(t) \), and \( h^{\textprime}(t) \) exists and is continuous over the domain of interest, the curve traced out by \( \mathbf{r}(t) \) is smooth. This concept is important not only in theoretical mathematics but also in practical applications like computer graphics and motion planning.

Properties of Odd Functions

An odd function has a unique symmetry: it is symmetric about the origin, which means that its graph creates a matching pattern in all four quadrants of the Cartesian plane. Mathematically, a function \( f(t) \) is odd if \( f(-t) = -f(t) \) for all \( t \) in the domain.

One crucial property of odd functions is that their integrals over an interval symmetric about the origin, from \( -a \) to \( a \), are zero. To understand why, consider that the area under the curve from \( -a \) to \( 0 \) is the negative reflection of the area from \( 0 \) to \( a \), hence they cancel out when added together. This property greatly simplifies solving integrals of odd functions, especially in vector calculus when dealing with components of vectors.

Integrals of Vector Functions

When integrating a vector-valued function, we integrate each of the component functions independently. For \( \mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle \), with respect to \( t \), the result is another vector function representing the accumulation of the components:

\( \int \mathbf{r}(t) \, dt = \langle \int f(t) \, dt, \int g(t) \, dt, \int h(t) \, dt \rangle \).

This operation can be used to find quantities such as displacement or area under a curve in three-dimensional space. When the component functions are odd and the limits of integration are symmetric about the origin, as in the property of odd functions, the integrals of the components are all zero. Consequently, the vector integral over such an interval yields the zero vector \( \mathbf{0} \), as the integrated vector function has no magnitude in any direction.