Answer:
[tex]\displaystyle AL = 2sinh(10)[/tex]
General Formulas and Concepts:
Pre-Calculus
Calculus
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]
Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Exponential Differentiation
Integration
- Integrals
- Integration Constant C
- Definite Integrals
Parametric Integration
Vector Value Functions
Arc Length Formula [Vector]: [tex]\displaystyle AL = \int\limits^b_a {\sqrt{[i'(t)]^2 + [j'(t)]^2 + [k'(t)]^2}} \, dt[/tex]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle \vec{r}(t) = <10\sqrt{2}t , e^{10t} , e^{-10t} >[/tex]
Interval [0, 1]
Step 2: Find Arclength
- Rewrite vector value function: [tex]\displaystyle r(t) = 10\sqrt{2}t \textbf i + e^{10t} \textbf j + e^{-10t} \textbf k[/tex]
- Substitute in variables [Arc Length Formula - Vector]: [tex]\displaystyle AL = \int\limits^1_0 {\sqrt{\bigg[\frac{d}{dt}[10\sqrt{2}t \textbf i]\bigg]^2 + \bigg[\frac{d}{dt}[e^{10t} \textbf j]\bigg]^2 + \bigg[\frac{d}{dt}[e^{-10t} \textbf k ]\bigg]^2}} \, dt[/tex]
- [Integrand] Differentiate [Respective Differentiation Rules]: [tex]\displaystyle AL = \int\limits^1_0 {\sqrt{[10\sqrt{2} \textbf i]^2 + [10e^{10t} \textbf j]^2 + [-10e^{-10t} \textbf k]^2}} \, dt[/tex]
- [Integrand] Simplify: [tex]\displaystyle AL = \int\limits^1_0 {\sqrt{200 \textbf i + 100e^{20x} \textbf j + 100e^{-20x} \textbf k}} \, dt[/tex]
- [Integral] Evaluate: [tex]\displaystyle AL = 2sinh(10)[/tex]
Topic: AP Calculus BC (Calculus I + II)
Unit: Vector Value Functions
Book: College Calculus 10e
