Math Problem Statement

find the arc-length parameterization for r(t) = <e^2t, e^-2t, 2sqrt(2)t> for t>=0

Solution

To find the arc-length parameterization for the given vector function r(t)=e2t,e2t,22t\mathbf{r}(t) = \langle e^{2t}, e^{-2t}, 2\sqrt{2}t \rangle, we need to follow these steps:

Step 1: Compute the speed v(t)v(t)

The speed is the magnitude of the velocity vector r(t)\mathbf{r}'(t), which is the derivative of r(t)\mathbf{r}(t).

  1. First, find the derivative of each component of r(t)\mathbf{r}(t):

r(t)=ddte2t,ddte2t,ddt(22t)=2e2t,2e2t,22\mathbf{r}'(t) = \langle \frac{d}{dt} e^{2t}, \frac{d}{dt} e^{-2t}, \frac{d}{dt} (2\sqrt{2}t) \rangle = \langle 2e^{2t}, -2e^{-2t}, 2\sqrt{2} \rangle

  1. Now, compute the magnitude of r(t)\mathbf{r}'(t), which gives us the speed:

r(t)=(2e2t)2+(2e2t)2+(22)2|\mathbf{r}'(t)| = \sqrt{(2e^{2t})^2 + (-2e^{-2t})^2 + (2\sqrt{2})^2} r(t)=4e4t+4e4t+8|\mathbf{r}'(t)| = \sqrt{4e^{4t} + 4e^{-4t} + 8}

Step 2: Set up the arc-length formula

The arc-length function s(t)s(t) is given by the integral of the speed:

s(t)=0tr(u)dus(t) = \int_0^t |\mathbf{r}'(u)| \, du

Substitute the expression for r(t)|\mathbf{r}'(t)|:

s(t)=0t4e4u+4e4u+8dus(t) = \int_0^t \sqrt{4e^{4u} + 4e^{-4u} + 8} \, du

This integral is challenging to evaluate directly, but this expression represents the formula for the arc-length function. There may be methods to simplify this integral, but as it stands, this is the general approach to find the arc-length.

Step 3: Solve for tt as a function of ss

Once s(t)s(t) is computed, the final step is to invert the function to express tt in terms of ss, thereby finding the arc-length parameterization t(s)t(s).

Would you like a more detailed breakdown of simplifying the integral or other details?

Here are 5 questions you may find useful based on this process:

  1. What is the interpretation of the arc-length parameterization geometrically?
  2. How does the speed function relate to the curvature of the curve?
  3. Could we use numerical methods to approximate the arc-length function?
  4. What would change if we were to restrict the domain of tt further?
  5. How would we handle the same problem for a curve in higher dimensions?

Tip: In cases where an integral is hard to solve analytically, numerical integration can provide approximate solutions for arc-length.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Arc-Length Parameterization
Calculus of Parametric Curves

Formulas

Speed function |r'(t)| = sqrt((dr_x/dt)^2 + (dr_y/dt)^2 + (dr_z/dt)^2)
Arc-Length function s(t) = ∫ |r'(u)| du

Theorems

Fundamental Theorem of Calculus
Arc-Length Parameterization Theorem

Suitable Grade Level

College Level - Calculus III