Problem 22 - Thomas Calculus

Question

Use Simpson's Rule with \(n=10\) to approximate the length of arc of \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t^{3} \mathbf{k}\) from the origin to the point \((2,4,8) .\)

Step-by-Step Solution

Answer

Using Simpson's Rule, the approximate arc length is calculated to be 10.3611 units.

Step 1: Determine the parameter interval

To find the length of the arc from the origin to the point (2,4,8), we must first determine the values of the parameter \( t \) such that \( \mathbf{r}(t) = (2, 4, 8) \). From the given parametric equations: \( x = t, \ y = t^2, \ z = t^3 \), we let \( t = 2 \), since \( x = t = 2 \). This corresponds to the point (2, 4, 8). Therefore, the interval for \( t \) is \([0, 2]\).

Step 2: Find the derivative of the vector function

The next step is to find the derivative of the vector function \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \). The derivative is \( \mathbf{r}'(t) = \dfrac{d}{dt} t \mathbf{i} + \dfrac{d}{dt} t^2 \mathbf{j} + \dfrac{d}{dt} t^3 \mathbf{k} = \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \).

Step 3: Calculate the magnitude of the derivative

Now, find the magnitude of \( \mathbf{r}'(t) \):\[ \| \mathbf{r}'(t) \| = \sqrt{(1)^2 + (2t)^2 + (3t^2)^2} = \sqrt{1 + 4t^2 + 9t^4} \].

Step 4: Set up Simpson's Rule for approximation

Simpson's Rule for arc length states:\[ \frac{\Delta t}{3} \left[ f(t_0) + 4f(t_1) + 2f(t_2) + \, \ldots \, + 4f(t_{n-1}) + f(t_n) \right] \], where \( f(t) = \| \mathbf{r}'(t) \| \). Here, \( n = 10 \), thus \( \Delta t = \frac{2-0}{10} = 0.2 \).

Step 5: Approximate the arc length using Simpson's Rule

Calculate the values of \( t_i = 0, 0.2, 0.4, \ldots, 2 \) and compute \( \sqrt{1 + 4t_i^2 + 9t_i^4} \) for each \( t_i \). Plug these values into Simpson's Rule formula:\[ L \approx \frac{0.2}{3} \left[ \sqrt{1 + 4(0)^2 + 9(0)^4} + 4\sqrt{1 + 4(0.2)^2 + 9(0.2)^4} + 2\sqrt{1 + 4(0.4)^2 + 9(0.4)^4} + ... + 4\sqrt{1 + 4(1.8)^2 + 9(1.8)^4} + \sqrt{1 + 4(2)^2 + 9(2)^4} \right] \].

Key Concepts

Arc Length

The arc length is simply the "distance" along a curve or path. Unlike straight lines, the arc length involves curvature, making their calculations more challenging. For straight lines, the distance between two points is found with the simple distance formula. However, along a curve defined by parametric equations or vector functions, the arc length involves integrating the function's derivative over a specified interval. In this context, numerical methods like Simpson's Rule can approximate the arc length when computing an exact value by integration becomes cumbersome. Simpson's Rule simplifies this by using piecewise parabolic segments to estimate the total length of the curve, providing an approached solution that balances accuracy and simplicity.

Vector Function

A vector function expresses a vector quantity for different values of a parameter, usually denoted by letters like t. For a three-dimensional curve, a vector function would look like \[ \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k} \]where \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) are the unit vectors along the x, y, and z axes respectively. This compact form allows us to represent not only a position in space but also changes in direction and magnitude along the curve. Such representations are useful when analyzing curves in physics and engineering. In our problem, the function \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \) shows how the curve's position evolves as the parameter \( t \) changes from 0 to 2.

Parametric Equations

Parametric equations offer an incredibly flexible way of describing curves in a plane or space. Instead of expressing y as a function of x (or vice versa), parametric equations express both x and y (and z, in three dimensions) as functions of a third variable, often t, which acts as the parameter. This allows modeling of complex curves where a single variable function would be insufficient. In the given vector function \[ \mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j} + t^3\mathbf{k} \] we use parametric equations \( x = t \), \( y = t^2 \), and \( z = t^3 \) to represent the point on the curve corresponding to each value of t. It is these parametric equations that let us handle complex motion and curves easily by splitting the multidimensional problem into manageable parts.

Derivative of Vector Function

The derivative of a vector function provides crucial insight into the curve's geometry and how it changes over t. Obtaining the derivative \[ \mathbf{r}'(t) = \frac{d}{dt}(t\mathbf{i}) + \frac{d}{dt}(t^2\mathbf{j}) + \frac{d}{dt}(t^3\mathbf{k}) = \mathbf{i} + 2t\mathbf{j} + 3t^2\mathbf{k} \]involves differentiating each component with respect to the parameter, t. This derivative, \( \mathbf{r}'(t) \), signifies the velocity of an object traveling along the path described by \( \mathbf{r}(t) \), showing the rate and direction of change at any point t. Calculating its magnitude \[ \|\mathbf{r}'(t)\| = \sqrt{(1)^2 + (2t)^2 + (3t^2)^2} \]is vital for determining the infinitesimal arc length, as it signifies how fast the curve 'moves' through space. This magnitude integrates over the parameter interval to derive the total arc length.