Math Problem Statement

Find the arc length of the curve r(t) = ⟨10√2t, e^(10t), e^(-10t)⟩, 0 ≤ t ≤ 1.

Solution

To find the arc length of a parametric curve r(t)=102t,e10t,e10t\mathbf{r}(t) = \langle 10\sqrt{2}t, e^{10t}, e^{-10t} \rangle for 0t10 \leq t \leq 1, we use the formula for the arc length of a vector function r(t)\mathbf{r}(t) over the interval [a,b][a, b]:

L=abr(t)dtL = \int_a^b \|\mathbf{r}'(t)\| \, dt

Step 1: Compute r(t)\mathbf{r}'(t)

First, differentiate each component of r(t)\mathbf{r}(t) with respect to tt.

  • ddt(102t)=102\frac{d}{dt}(10\sqrt{2}t) = 10\sqrt{2}
  • ddt(e10t)=10e10t\frac{d}{dt}(e^{10t}) = 10e^{10t}
  • ddt(e10t)=10e10t\frac{d}{dt}(e^{-10t}) = -10e^{-10t}

Thus, r(t)=102,10e10t,10e10t\mathbf{r}'(t) = \langle 10\sqrt{2}, 10e^{10t}, -10e^{-10t} \rangle.

Step 2: Compute r(t)\|\mathbf{r}'(t)\|

Now, find the magnitude of r(t)\mathbf{r}'(t):

r(t)=(102)2+(10e10t)2+(10e10t)2\|\mathbf{r}'(t)\| = \sqrt{(10\sqrt{2})^2 + (10e^{10t})^2 + (-10e^{-10t})^2}

Simplify each term:

=(100×2)+(100e20t)+(100e20t)= \sqrt{(100 \times 2) + (100e^{20t}) + (100e^{-20t})} =200+100e20t+100e20t= \sqrt{200 + 100e^{20t} + 100e^{-20t}}

Notice that e20t+e20te^{20t} + e^{-20t} can be expressed using the identity:

e20t+e20t=2cosh(20t)e^{20t} + e^{-20t} = 2\cosh(20t)

Thus, we have:

r(t)=200+100×2cosh(20t)=200+200cosh(20t)\|\mathbf{r}'(t)\| = \sqrt{200 + 100 \times 2\cosh(20t)} = \sqrt{200 + 200\cosh(20t)}

r(t)=200(1+cosh(20t))\|\mathbf{r}'(t)\| = \sqrt{200(1 + \cosh(20t))}

Step 3: Compute the Arc Length Integral

Now, compute the arc length by integrating r(t)\|\mathbf{r}'(t)\| from t=0t = 0 to t=1t = 1:

L=01200(1+cosh(20t))dtL = \int_0^1 \sqrt{200(1 + \cosh(20t))} \, dt

Factor out 200\sqrt{200}:

L=200011+cosh(20t)dtL = \sqrt{200} \int_0^1 \sqrt{1 + \cosh(20t)} \, dt

This integral does not have a simple elementary form, so it would typically be evaluated using numerical methods. However, we can leave it in this form for now, or approximate it using tools like numerical integration.

Would you like more details on how to compute the integral numerically or further clarification?

5 Related Questions:

  1. What is the physical meaning of the arc length in a parametric curve?
  2. How is the hyperbolic cosine function cosh(x)\cosh(x) related to exponential functions?
  3. What are other methods for calculating arc length if the integral is difficult to solve analytically?
  4. How does arc length change if we scale the curve in one or more directions?
  5. What is the difference between arc length in 2D and 3D parametric curves?

Tip: When solving integrals that involve hyperbolic functions or exponentials, checking for simplification using identities like cosh(x)\cosh(x) can make computations easier.

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Math Problem Analysis

Mathematical Concepts

Parametric Curves
Arc Length
Vector Functions
Calculus

Formulas

Arc Length L = ∫_a^b ||r'(t)|| dt
Hyperbolic identity: e^x + e^(-x) = 2cosh(x)

Theorems

Arc Length Formula for Parametric Curves
Chain Rule for Differentiation

Suitable Grade Level

University-level (Calculus II or III)