Question

Sketch the space curve and find its length over the given interval. $$ \mathbf{r}(t)=a \cos t \mathbf{i}+a \sin t \mathbf{j}+b t \mathbf{k} $$ $$ [0,2 \pi] $$

Short answer

The curve is a helix spiraling upwards. The length of the curve over the interval [0, 2π] is \(2\pi s\)

Step-by-step solution

Visualization of the Curve

Visualization of the space curve involves plotting the parametric equations. From the vector equation \(\mathbf{r}(t)=a \cos t \mathbf{i}+a \sin t \mathbf{j}+b t \mathbf{k}\), we can deduce the x, y, and z components. These are \(x=a \cos t\), \(y=a \sin t\), and \(z=b t\). Intuitively, we can realize that \(x^2+y^2=a^2\), which describes a circle in the x-y plane, and \(z\) increases linearly with time \(t\). So, the curve is a helix spiraling upwards.

Calculation of the Arc Length of the Curve

The arc length of a curve from \(t=a\) to \(t=b\) is given by the integral \(\int_{a}^{b} |\mathbf{r}'(t)| dt\), where \(\mathbf{r}'(t)\) is the derivative of the vector function \(\mathbf{r}(t)\). In our case, we compute the derivative of the function \(\mathbf{r}(t)\) to get \(\mathbf{r}'(t)=-a \sin t \mathbf{i}+a \cos t \mathbf{j}+b \mathbf{k}\). By squaring and summing all components of \(\mathbf{r}'(t)\), we get \(|-a \sin t|^2+ |a \cos t|^2 + |b|^2 = a^2+b^2\). Note that \(a^2 + b^2\) is constant and its square root (which we term as \(s\)) is the magnitude of \(\mathbf{r}'(t)\).

Integration to Find the Arc Length

Following the formula for the arc length and noting that the magnitude of \(\mathbf{r}'(t)\) is constant, we have that the arc length \(\int_{0}^{2\pi} |\mathbf{r}'(t)| dt\) equals \(s *(2\pi - 0) = 2\pi s\).

Key concepts

Parametric Equations

Parametric equations are a powerful mathematical tool used to describe the path of a point in space. In a typical Cartesian system, you would express the path as a function of y in terms of x, or vice versa. However, parametric equations introduce an independent variable, often denoted as 't', which is known as a parameter. This parameter traces the path of the point as it varies over time.

For instance, the set of parametric equations for the curve provided in the exercise are:\[ x = a \times \text{cos}(t), \quad y = a \times \text{sin}(t), \quad z = b \times t \]. Here, 'a' and 'b' are constants, and 't' varies from 0 to \(2\text{pi}\). This set intimately describes the x, y, and z coordinates of a point on the space curve. Through parametric equations, complex curves, like helices and parabolas, which are challenging to represent in a function form, are more easily visualized and understood.

Vector Functions

Vector functions extend parametric equations into a multidimensional space. They are an elegant way to describe the motion of an object along a path in space, with each component function representing a dimension in space. A vector function encompasses multiple pieces of information: direction, magnitude, and the path an object takes.

In our problem, the vector function is \[ \mathbf{r}(t) = a \times \text{cos}(t) \mathbf{i} + a \times \text{sin}(t) \mathbf{j} + b \times t \mathbf{k} \]. Each component of this vector represents a parametric equation for the corresponding axis. The \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are unit vectors along the x, y, and z axes, respectively. When you differentiate this vector function with respect to 't', you obtain a new vector that depicts the rate of change of \( \mathbf{r}(t) \), indicating the instantaneous velocity of the point as it moves along the curve.

Calculus Integration

Calculus integration is an essential process used to calculate the area under a curve, the volume of a solid, and as we see in this exercise, the length of a curve. The integration of differentiable functions allows us to accumulate the tiny changes in a function to compute these large-scale measurements.

For the arc length, we use the integral of the magnitude of the vector function's derivative, \(|\mathbf{r}'(t)|\), over the interval of 't'. In mathematical terms, the arc length 's' from 't = a' to 't = b' could be expressed as:\[ s = \int_{a}^{b} |\mathbf{r}'(t)| \, dt \]. The resulting integral produces the scalar arc length of the curve over the given interval. In this exercise, since the magnitude of the derivative is constant, the integration simplifies to the product of this constant magnitude and the length of the interval over which it is integrated.