To find the length of the curve $\overline{r}(t)$ over the interval $1 \leq t \leq 5$, we need to apply the formula for the arc length of a parametric curve:

$L = \int_{t_1}^{t_2} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt$

In this case, the parametric curve is given by:

$\overline{r}(t) = \langle 4t^2, 4t, \ln(t) \rangle$

Step 1: Compute the derivatives

First, we compute the derivatives of each component of $\overline{r}(t)$:

• $x(t) = 4t^2$, so $\frac{dx}{dt} = 8t$
• $y(t) = 4t$, so $\frac{dy}{dt} = 4$
• $z(t) = \ln(t)$, so $\frac{dz}{dt} = \frac{1}{t}$

Step 2: Substitute into the arc length formula

Now we substitute these derivatives into the formula for the arc length:

$L = \int_{1}^{5} \sqrt{(8t)^2 + (4)^2 + \left(\frac{1}{t}\right)^2} \, dt$

Simplifying inside the square root:

$L = \int_{1}^{5} \sqrt{64t^2 + 16 + \frac{1}{t^2}} \, dt$

Step 3: Compute the integral

To find the exact value of the length of the curve, we would now compute the integral:

$L = \int_{1}^{5} \sqrt{64t^2 + 16 + \frac{1}{t^2}} \, dt$

This integral doesn't have a simple antiderivative, so it would be evaluated numerically. Let me compute this for you.It seems like I can’t do more advanced data analysis right now. Please try again later.

Would you like to proceed with a more detailed step-by-step explanation of how to set up the integral for numerical evaluation, or do you have any other questions?

Here are 5 related questions for further exploration:

1. How would you compute the arc length if one of the coordinates had a more complex function, like $e^t$?
2. What are some methods to approximate integrals like this one numerically?
3. Can the arc length formula be extended to curves in higher-dimensional spaces?
4. How does the arc length change when the curve is reparametrized?
5. What is the geometric interpretation of the arc length of a curve?

Tip: Numerical integration is often the go-to technique for arc length problems when the integral does not have a simple analytical solution.