To find the length of the curve defined by the vector function

$\mathbf{r}(t) = 2t\,\mathbf{i} + e^t\,\mathbf{j} + e^{-t}\,\mathbf{k}, \quad 0 \leq t \leq 4,$

we use the formula for the arc length of a vector-valued function:

$L = \int_a^b \left\|\mathbf{r}'(t)\right\| \, dt,$

where $\mathbf{r}'(t)$ is the derivative of $\mathbf{r}(t)$, and $\left\|\mathbf{r}'(t)\right\|$ is its magnitude.

Step 1: Find the derivative $\mathbf{r}'(t)$

$\mathbf{r}(t) = 2t\,\mathbf{i} + e^t\,\mathbf{j} + e^{-t}\,\mathbf{k}$

Taking the derivative with respect to $t$:

$\mathbf{r}'(t) = \frac{d}{dt}(2t)\,\mathbf{i} + \frac{d}{dt}(e^t)\,\mathbf{j} + \frac{d}{dt}(e^{-t})\,\mathbf{k} = 2\,\mathbf{i} + e^t\,\mathbf{j} - e^{-t}\,\mathbf{k}.$

Step 2: Find the magnitude $\left\|\mathbf{r}'(t)\right\|$

$\left\|\mathbf{r}'(t)\right\| = \sqrt{(2)^2 + (e^t)^2 + (-e^{-t})^2} = \sqrt{4 + e^{2t} + e^{-2t}}.$

Step 3: Evaluate the integral for arc length

$L = \int_0^4 \sqrt{4 + e^{2t} + e^{-2t}} \, dt.$

This integral is not straightforward to solve analytically due to the complexity of the expression inside the square root. We can solve it numerically.

Let's compute the numerical value of this integral.The length of the curve $\mathbf{r}(t) = 2t\,\mathbf{i} + e^t\,\mathbf{j} + e^{-t}\,\mathbf{k}$ from $t = 0$ to $t = 4$ is approximately:

$L \approx 55.31.$

Would you like further details or have any questions?

Related Questions:

1. How do you find the arc length of a curve given by parametric equations?
2. Can you explain the process of numerical integration?
3. What are other methods to solve arc length integrals when exact integration is difficult?
4. How would this problem change if the limits of integration were different?
5. How do you interpret the magnitude of the derivative of a vector function?

Tip:

For complex integrals that are difficult to solve analytically, numerical methods like Simpson’s Rule, Trapezoidal Rule, or software tools are often used to approximate the result efficiently.