Math Problem Statement

Write the line segment given by
r ( t )

e 3 t

i − 8 e 3 t

j − 4 e 3 t

k as a function
r ( s ) parameterized by the arc length
s from the point where
t

0 .

Solution

We are given the line segment r(t)=e3ti−8e3tj−4e3tk\mathbf{r}(t) = e^{3t} \mathbf{i} - 8 e^{3t} \mathbf{j} - 4 e^{3t} \mathbf{k}, and we need to re-parameterize it in terms of the arc length ss starting from the point where t=0t = 0.

Step 1: Find the magnitude of the velocity vector

First, differentiate r(t)\mathbf{r}(t) with respect to tt to find the velocity vector r′(t)\mathbf{r}'(t).

= 3e^{3t} \mathbf{i} - 24 e^{3t} \mathbf{j} - 12 e^{3t} \mathbf{k}$$ The magnitude of $$\mathbf{r}'(t)$$ is: $$|\mathbf{r}'(t)| = \sqrt{(3e^{3t})^2 + (-24e^{3t})^2 + (-12e^{3t})^2} = e^{3t} \sqrt{3^2 + (-24)^2 + (-12)^2} = e^{3t} \sqrt{9 + 576 + 144} = e^{3t} \sqrt{729} = 27e^{3t}$$ ### Step 2: Find the arc length $$s$$ as a function of $$t$$ The arc length $$s$$ is the integral of the magnitude of the velocity vector from $$t = 0$$ to $$t$$: $$s = \int_0^t |\mathbf{r}'(u)| \, du = \int_0^t 27e^{3u} \, du$$ To solve this integral: $$s = 27 \int_0^t e^{3u} \, du = 27 \left[ \frac{1}{3} e^{3u} \right]_0^t = 9 \left( e^{3t} - e^0 \right) = 9 (e^{3t} - 1)$$ Thus, the arc length is: $$s = 9 (e^{3t} - 1)$$ ### Step 3: Solve for $$t$$ in terms of $$s$$ We need to express $$t$$ as a function of $$s$$. Solving for $$t$$: $$s = 9 (e^{3t} - 1)$$ $$\frac{s}{9} = e^{3t} - 1$$ $$e^{3t} = \frac{s}{9} + 1$$ Taking the natural logarithm of both sides: $$3t = \ln\left( \frac{s}{9} + 1 \right)$$ $$t = \frac{1}{3} \ln\left( \frac{s}{9} + 1 \right)$$ ### Step 4: Re-parameterize $$\mathbf{r}(t)$$ in terms of $$s$$ Now that we have $$t$$ in terms of $$s$$, we substitute this expression into $$\mathbf{r}(t)$$. Recall: $$\mathbf{r}(t) = e^{3t} \mathbf{i} - 8 e^{3t} \mathbf{j} - 4 e^{3t} \mathbf{k}$$ Since $$e^{3t} = \frac{s}{9} + 1$$, we can write $$\mathbf{r}(s)$$ as: $$\mathbf{r}(s) = \left( \frac{s}{9} + 1 \right) \mathbf{i} - 8 \left( \frac{s}{9} + 1 \right) \mathbf{j} - 4 \left( \frac{s}{9} + 1 \right) \mathbf{k}$$ Thus, the re-parameterized line segment in terms of arc length $$s$$ is: $$\mathbf{r}(s) = \left( \frac{s}{9} + 1 \right) \mathbf{i} - 8 \left( \frac{s}{9} + 1 \right) \mathbf{j} - 4 \left( \frac{s}{9} + 1 \right) \mathbf{k}$$ --- Would you like more details or have any questions? ### Related Questions: 1. What is the significance of re-parameterizing by arc length? 2. How does the parameterization simplify if the vector field has a constant magnitude? 3. Can the same procedure apply to curves in different coordinate systems (e.g., polar or cylindrical)? 4. What are the conditions under which arc length parameterization is valid? 5. How does this arc length parameterization relate to the Frenet-Serret formulas? **Tip:** Re-parameterizing curves by arc length helps to understand the geometry of the curve, as arc length gives a natural way to describe distances along the curve.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Arc Length Parameterization
Exponential Functions

Formulas

r(t) = e^{3t} i − 8 e^{3t} j − 4 e^{3t} k
r'(t) = 3e^{3t} i − 24 e^{3t} j − 12 e^{3t} k
Arc length s = ∫ |r'(u)| du
t = (1/3) ln(s/9 + 1)

Theorems

Arc Length Parameterization
Chain Rule in Differentiation

Suitable Grade Level

Undergraduate Calculus

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