Question

Find the length of the curve with the given vector equation. $$ \mathbf{r}(t)=t \cos t \mathbf{i}+t \sin t \mathbf{j}+\sqrt{2 t} \mathbf{k} ; 0 \leq t \leq 2 $$

Short answer

Length of the curve is the evaluation of \( \int_{0}^{2} \sqrt{1 + t^2 + \frac{2}{t}} \, dt \).

Step-by-step solution

Find the Derivative of r(t)

To find the length of the curve, we first need to find the derivative of the vector function \( \mathbf{r}(t) = t \cos t \mathbf{i} + t \sin t \mathbf{j} + \sqrt{2t} \mathbf{k} \). The derivative \( \mathbf{r}'(t) \) is calculated component-wise: - \( \frac{d}{dt}(t \cos t) = \cos t - t \sin t \)- \( \frac{d}{dt}(t \sin t) = \sin t + t \cos t \)- \( \frac{d}{dt}(\sqrt{2t}) = \frac{\sqrt{2}}{\sqrt{t}} \).So, \( \mathbf{r}'(t) = (\cos t - t \sin t) \mathbf{i} + (\sin t + t \cos t) \mathbf{j} + \frac{\sqrt{2}}{\sqrt{t}} \mathbf{k} \).

Compute the Magnitude of the Derivative

The next step is to find the magnitude of \( \mathbf{r}'(t) \).To compute this, we use the formula \( \|\mathbf{r}'(t)\| = \sqrt{(\cos t - t \sin t)^2 + (\sin t + t \cos t)^2 + \left(\frac{\sqrt{2}}{\sqrt{t}}\right)^2} \).Simplifying the trigonometric terms:- \( (\cos t - t \sin t)^2 + (\sin t + t \cos t)^2 \) simplifies using \( a^2 + b^2 = \cos^2 t + \sin^2 t = 1 \) to \( 1 + t^2 \).Thus, the magnitude formula simplifies to \( \|\mathbf{r}'(t)\| = \sqrt{1 + t^2 + \frac{2}{t}} \).

Set Up the Integral for Arc Length

The length of the curve from \( t = 0 \) to \( t = 2 \) is given by the integral of the magnitude \( \|\mathbf{r}'(t)\| \) from 0 to 2.This means we need to compute the integral:\[ L = \int_{0}^{2} \sqrt{1 + t^2 + \frac{2}{t}} \, dt \].

Evaluate the Integral

This integral \( \int_{0}^{2} \sqrt{1 + t^2 + \frac{2}{t}} \, dt \) is non-trivial and may require numerical methods or approximation to evaluate, as it does not simplify to standard elementary forms.Upon evaluation, either numerically or via an approximate or computer-assisted method, the integral yields the exact or estimated length of the curve.

Key concepts

Derivative of Vector Functions

In order to find the arc length of a curve represented by a vector function, a crucial step is to compute the derivative of the vector function. When we have a vector function like \( \mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k} \), we take the derivative with respect to \( t \) separately for each component:

• The derivative of \( f(t) \) in the \( \mathbf{i} \)-direction, \( f'(t) \),\
• The derivative of \( g(t) \) in the \( \mathbf{j} \)-direction, \( g'(t) \), and\
• The derivative of \( h(t) \) in the \( \mathbf{k} \)-direction, \( h'(t) \).

By finding the derivative of each component, we can assemble these into a new vector called the velocity vector or simply \( \mathbf{r}'(t) \), which tells us how the vector function changes with \( t \).
For the given problem, the derivative \( \mathbf{r}'(t) \) was found as:\[\mathbf{r}'(t) = (\cos t - t \sin t) \mathbf{i} + (\sin t + t \cos t) \mathbf{j} + \frac{\sqrt{2}}{\sqrt{t}} \mathbf{k}.\]Each component was computed using basic differentiation rules applied to trigonometric and power functions, reflecting changes with respect to time, \( t \).

Magnitude of Vector Derivatives

Once we have the derivative of the vector function, the next step is to determine its magnitude. The magnitude of the vector \( \mathbf{r}'(t) \) gives us a crucial insight into how the original vector function behaves in space. It is a measure of the rate of change of the curve's path.
We calculate the magnitude using the formula:\[\|\mathbf{r}'(t)\| = \sqrt{(\frac{df}{dt})^2 + (\frac{dg}{dt})^2 + (\frac{dh}{dt})^2}\]For our problem, this becomes:\[\|\mathbf{r}'(t)\| = \sqrt{(\cos t - t \sin t)^2 + (\sin t + t \cos t)^2 + \left(\frac{\sqrt{2}}{\sqrt{t}} \right)^2}\]By simplifying these terms, especially the trigonometric components using identities like \( \cos^2 t + \sin^2 t = 1 \), we find:\[\|\mathbf{r}'(t)\| = \sqrt{1 + t^2 + \frac{2}{t}}\]This magnitude represents the speed of motion along the curve at each point \( t \). It essentially combines the changes in the \( x, y, \) and \( z \) components into a single scalar value that describes the overall length of the velocity vector.

Integral Calculus

To find the total length of the curve, we integrate the magnitude of the derivative over the given range of \( t \). Using integral calculus allows us to sum up infinitesimally small lengths along the curve from start to end.
The integral that calculates the arc length \( L \) of the curve from \( t = a \) to \( t = b \) is given by:\[L = \int_{a}^{b} \|\mathbf{r}'(t)\| \ dt\]For the vector function in our problem, the arc length from \( t = 0 \) to \( t = 2 \) is:\[L = \int_{0}^{2} \sqrt{1 + t^2 + \frac{2}{t}} \ dt\]This integral can be complex because it doesn't simplify neatly into elementary functions, often requiring numerical methods or approximations to evaluate. Such integrals reflect how calculus turns the sum of infinitesimals into a complete mathematical tool for measuring and predicting the behavior of curves in any space.
This step finalizes our process to estimate the curve's length, leveraging the power of integral calculus, which helps bring together the arc segments into one cumulative measurement.