Question

Investigation Consider the graph of the vector-valued function \(\mathbf{r}(t)=t \mathbf{i}+\left(4-t^{2}\right) \mathbf{j}+t^{3} \mathbf{k}\) on the interval \([0,2] .\) (a) Approximate the length of the curve by finding the length of the line segment connecting its endpoints. (b) Approximate the length of the curve by summing the lengths of the line segments connecting the terminal points of the vectors \(\mathbf{r}(0), \mathbf{r}(0.5), \mathbf{r}(1), \mathbf{r}(1.5),\) and \(\mathbf{r}(2)\) (c) Describe how you could obtain a more accurate approximation by continuing the processes in parts (a) and (b). (d) Use the integration capabilities of a graphing utility to approximate the length of the curve. Compare this result with the answers in parts (a) and (b).

Short answer

The lengths of the approximated curve for parts (a) and (b) can be obtained by using the distance formula. A more accurate approximation could be obtained by dividing the interval into smaller sub-intervals and summing the lengths of line segments connecting the terminal points of these sub-intervals. The result from integration may vary depending on the number of segments used, and it may be more closer to the actual length than the earlier segment-wise approximations.

Step-by-step solution

Connect the Endpoints (Part a)

Draw a straight line segment connecting the endpoints of the curve. The coordinates of the endpoints are \( \mathbf{r}(0) \) and \( \mathbf{r}(2) \). Find the length of the line segment using the distance formula \(d = \sqrt { (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2} \)

Connect Key Points (Part b)

Divide the curve into smaller segments, taking the points \( \mathbf{r}(0) \), \( \mathbf{r}(0.5) \), \( \mathbf{r}(1) \), \( \mathbf{r}(1.5) \) and \( \mathbf{r}(2) \). Find the lengths of the line segments connecting these points, again using the distance formula, and sum them up.

Improving Accuracy (Part c)

To obtain a more accurate approximation, the interval \([0,2]\) can be divided into smaller sub-intervals. The length of the line segments connecting the terminal points of these sub-intervals can then be summed, leveraging the fact that a closer approximation of the curve is achieved when more segments are used.

Approximation Using Integration (Part d)

To approximate the length of the curve using integration, find the derivative of the vector function \( \mathbf{r}(t) \), hence the magnitude of \( \mathbf{r'}(t) \), and then integrate the magnitude over the interval \([0,2]\). Compare this result with the answers in parts (a) and (b).

Key concepts

Curve Length Approximation

When dealing with the geometry of curves, particularly those described by vector-valued functions, it's often necessary to estimate their length. The process of curve length approximation involves a few methods, ranging from simple straight-line measures to sophisticated integral calculus techniques.

As seen in the problem's strategy for part (a), you start with the most rudimentary approach by connecting the endpoints of the curve with a straight line. The length of this line segment gives you a very crude approximation of the curve's length. This is like stretching a rubber band between the first and last point on the curve; it doesn't follow the bends of the path, but it's quick and easy to compute. However, if you want a more precise estimate, as in part (b), you can add more points along the curve and connect them with straight line segments. This is akin to putting bends into your rubber band to help it follow the curve's shape more closely. In real-world applications, like in navigation or computer graphics, more accurate representations of curved paths are crucial, and that’s why methods to improve the approximation have been developed.

Imagine each added point as a step towards the true length of the curve. The more points you include, the closer your 'piecewise straight line' will follow the wiggle of the original curve. Especially when faced with functions that curve significantly, like the function given by \( \mathbf{r}(t) \) in the exercise, this refinement is important. The refinement process in part (c) is fairly straightforward: continue to decrease the size of the intervals between points, increasing the number of segments, and thus, getting a closer and closer approximation to the actual length of the curve. It’s like increasing the resolution of an image; more pixels give better clarity and detail.

Distance Formula

The distance formula is a crucial mathematical tool that allows us to calculate the straight-line distance between two points in Euclidean space. When dealing with vector-valued functions, these points are of course represented in terms of their vector coordinates. For the case of three dimensions as presented in the exercise, this formula is an extension of the Pythagorean theorem and takes the form \(d = \sqrt { (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2} \).

Using this formula for part (a) and (b) effectively transforms the complex problem of traversing a curved path into a series of manageable straight lines. The simplicity behind the distance formula is what makes it so powerful; one can easily take the coordinates of two points, plug in the values, and quickly determine the distance between them without needing intricate mathematical operations. However, the use of the distance formula in part (a) and (b) of the exercise is like taking measurements with a ruler instead of a flexible tape measure. It doesn't capture the entire story when the path isn't straight but does provide a stepping stone towards more accurate methods.

Integration of Vector Functions

The integration of vector functions, such as the one shown in step 4, takes our understanding of curve length to its mathematical pinnacle. Integration is a fundamental concept in calculus that allows for the accumulation of quantities, and when applied to vector functions, it enables us to encapsulate continuous and complex motion along curves. By finding the derivative of a vector function to get \( \mathbf{r'}(t) \) and then taking the magnitude of this derivative, we grasp the instantaneous rate of change at each point along the curve.

The integral of this magnitude over a certain interval, like \( [0,2] \) in the exercise, provides not just an approximation, but the true length of the curve within that interval. This procedure is likened to using a flexible tape that bends seamlessly along every twist and turn of the function’s path, capturing the exact length. In the grand scheme of calculus, this application of integration is a crowning achievement as it surmounts the limitations of the aforementioned approximation methods, delivering the precision required for advanced physics, engineering, and computer graphics calculations. Especially with the advent of graphing utilities, employing numerical methods to evaluate these integrals brings high accuracy to our fingertips, further displaying the power of integration for vector functions.