To find the arc length of a parametric curve defined by $x = \ln(8t)$ and $y = 5t^3$ over the interval from $t = 11$ to $t = 17$, we use the formula for the arc length of a parametric curve:

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

Step 1: Compute $\frac{dx}{dt}$

Given $x = \ln(8t)$, we differentiate with respect to $t$:

$\frac{dx}{dt} = \frac{d}{dt} \ln(8t) = \frac{1}{8t} \cdot 8 = \frac{1}{t}$

Step 2: Compute $\frac{dy}{dt}$

Given $y = 5t^3$, we differentiate with respect to $t$:

$\frac{dy}{dt} = \frac{d}{dt} 5t^3 = 15t^2$

Step 3: Plug these into the arc length formula

The arc length $L$ is:

$L = \int_{11}^{17} \sqrt{\left(\frac{1}{t}\right)^2 + \left(15t^2\right)^2} \, dt$

Simplify the expression under the square root:

$L = \int_{11}^{17} \sqrt{\frac{1}{t^2} + 225t^4} \, dt$

So, the integral that gives the arc length of the curve over the interval from $t = 11$ to $t = 17$ is:

$L = \int_{11}^{17} \sqrt{\frac{1}{t^2} + 225t^4} \, dt$

Would you like more details or have any questions?

Here are 5 related questions to further your understanding:

1. How would you compute the arc length if $x(t) = t^2$ and $y(t) = t^3$?
2. What are the steps to compute the arc length for a curve given in polar coordinates?
3. Can you explain why the arc length formula involves squaring the derivatives?
4. How would the integral change if the curve was defined by $x = \sin(t)$ and $y = \cos(t)$?
5. What is the significance of the limits of integration in the arc length formula?

Tip: When working with arc length problems, it’s crucial to carefully differentiate the parametric equations and simplify under the square root before integrating.