Arc Length of Vector Function r̅(t) = ⟨9t², 6t, ln(t)⟩ from t = 1 to t = 4

Math Problem Statement

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Solution

To find the length of the curve given by the vector function $\vec{r}(t) = \langle 9t^2, 6t, \ln(t) \rangle$ over the interval $1 \leq t \leq 4$ , we can use the formula for the arc length of a vector function:

$L = \int_{a}^{b} \left\| \frac{d\vec{r}(t)}{dt} \right\| dt$

Step 1: Find $\frac{d\vec{r}(t)}{dt}$

We differentiate each component of $\vec{r}(t)$ with respect to $t$ :

$\frac{d}{dt} (9t^2) = 18t$
$\frac{d}{dt} (6t) = 6$
$\frac{d}{dt} (\ln(t)) = \frac{1}{t}$

Thus, the derivative of $\vec{r}(t)$ is: $\frac{d\vec{r}(t)}{dt} = \langle 18t, 6, \frac{1}{t} \rangle$

Step 2: Find the magnitude of $\frac{d\vec{r}(t)}{dt}$

Now, we find the magnitude of the derivative vector: $\left\| \frac{d\vec{r}(t)}{dt} \right\| = \sqrt{(18t)^2 + 6^2 + \left(\frac{1}{t}\right)^2}$ $\left\| \frac{d\vec{r}(t)}{dt} \right\| = \sqrt{324t^2 + 36 + \frac{1}{t^2}}$

Step 3: Set up the arc length integral

Now, we set up the integral for the length of the curve from $t = 1$ to $t = 4$ : $L = \int_{1}^{4} \sqrt{324t^2 + 36 + \frac{1}{t^2}} \, dt$

Step 4: Evaluate the integral

The integral: $L = \int_{1}^{4} \sqrt{324t^2 + 36 + \frac{1}{t^2}} \, dt$ does not have a simple antiderivative, so we would need to evaluate it numerically.

Let's compute this numerically.

The length of the curve $\vec{r}(t) = \langle 9t^2, 6t, \ln(t) \rangle$ over the interval $1 \leq t \leq 4$ is approximately $136.39$ units.

Would you like to explore any part of this process in more detail? Here are some related questions:

How do we derive the formula for arc length of a vector function?
Can you explain why the magnitude of the derivative is used in the arc length formula?
How would this problem change if the limits of integration were different?
What numerical methods are used to evaluate integrals like this one?
How does the function $\vec{r}(t)$ describe the curve geometrically?

Tip: When dealing with complex integrals, numerical methods such as Simpson's rule or Gaussian quadrature are often used to approximate the solution efficiently.

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Math Problem Analysis

Mathematical Concepts

Calculus
Arc Length of a Curve
Vector Functions
Integration

Formulas

Arc Length Formula: L = ∫[a, b] ||dr̅(t)/dt|| dt
Derivative of Vector Function: dr̅(t)/dt = ⟨18t, 6, 1/t⟩
Magnitude of Vector: ||dr̅(t)/dt|| = √(324t² + 36 + 1/t²)

Theorems

Arc Length Formula for Vector Functions

Suitable Grade Level

University (Calculus II or Multivariable Calculus)

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